Re: [math-fun] sum of 2 triangular numbers is a triangular number
Hi Neil, your parameterization is more or less the same as what I came up with a week ago on Thurs. Feb 20. The purpose of the last email was to say that a similar parameterization (ranges over divisors) can be written for Pythagorean triples. The (t,u,v) form you gave does not make an entirely strong analogy. My head is hurting from recent wave of attacks and local problems, (despite having fun at arthropod museum) so I can not write too much more, but probably will anyways. To reiterate: Pythagorean is to Trithagorean as {n, d/2 - n^2/(2 d), d/2 + n^2/(2 d)} is to {n, n*(n + 1)/d - (d + 2)/4, n*(n + 1)/d + (d - 2)/4}. — Brad
On Feb 27, 2020, at 11:27 PM, Neil Sloane <njasloane@gmail.com> wrote:
PS Brad, I was not trying to "diss"you, I simply did not understand what you were saying. I proposed a formula that seemed to give all the trythagorean triples. I'm quite willing to believe either of the following: You had already given that formula two days ago My formula is wrong, and you had given the correct formula two days ago! Best regards Neil
Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com
On Thu, Feb 27, 2020 at 11:47 PM Neil Sloane <njasloane@gmail.com> wrote: Dear Grad, Trouble is, I don't speak Mathematica. Could you possibly translate what you said in that last email into human? Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com On Thu, Feb 27, 2020 at 10:46 PM Brad Klee <bradklee@gmail.com> wrote: Hi Neil, I think you may have missed this little piece of code: PythagoreanTriples[sqNn_] := Select[{sqNn, k, m} /. Solve[ MapThread[#1 == #2 &, {{(sqNn^2)/#, #}, {(m - k), (m + k)}}]][[1]] & /@ Divisors[sqNn^2][[1 ;; -1]], And[And @@ (IntegerQ /@ #), And @@ (# > 0 & /@ #)] &] (*https://oeis.org/A046079*) Length[PythagoreanTriples[#]] & /@ Range[2, 20] Out[] = {0, 1, 1, 1, 1, 1, 2, 2, 1, 1, 4, 1, 1, 4, 3, 1, 2, 1, 4} If you hack out the part that says Solve[...] and maybe put a "d" in for the "#" sign, then you can get that [n,d/2-n^2/(2*d),d/2+n^2/(2*d)], with "d" for divisor. Just to double-check for possible typo, let's try: Expand[{n, d/2 - n^2/(2 d), d/2 + n^2/(2 d)} /. {x_, y_, z_} :>x^2 + y^2 - z^2] Out[]=0 Then, we can refine the idea of an "analog", Pythagorean : Trithagorean :: {n, d/2 - n^2/(2 d), d/2 + n^2/(2 d)} : {n, n*(n + 1)/d - (d + 2)/4, n*(n + 1)/d + (d - 2)/4}. The other relevant sequence is https://oeis.org/A309507, see my message from Sat. Feb 22. --Brad On Thu, Feb 27, 2020 at 3:54 PM Neil Sloane <njasloane@gmail.com> wrote:
I mentioned this on Feb 19 2020 and several people sent very helpful replies. For squares, we have the theorem that all solutions to x^2+y^2=z^2 in integers are given by Pythagorean triples, t*[u^2-v^2, 2uv, u^2+v^2]. Expanding on the replies from Michael Collins and Rich, it looks like the following is the analog for triangular numbers T_n = n(n+1)/2. ALL solutions to T_n+T_k=T_m are given by the list of what one might call Trithagorean triples: these are the triples [n,k,m] = [n, T/Q-(Q+1)/2,T/Q+(Q-1)/2] where n >= 2, T=n(n+1)/2, and Q is any odd divisor of T less than n, plus these triples with the first two coordinates swapped. I haven't found this in the literature, but it can hardly be new. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Concerning the question about the general solution to the Trythagorean Triples problem, I happened to be speaking to Richard Guy this morning and I thought the question might cheer him up (he is in the hospital, and not doing well, I'm sorry to say. He is 104.) A colleague kindly put me on the speaker phone in his room. But he was not well enough to answer. I have all ten volumes of his Reviews in Number Theory, but I was hoping not to have to search through them. On Fri, Feb 28, 2020 at 1:19 AM Brad Klee <bradklee@gmail.com> wrote:
Hi Neil, your parameterization is more or less the same as what I came up with a week ago on Thurs. Feb 20. The purpose of the last email was to say that a similar parameterization (ranges over divisors) can be written for Pythagorean triples. The (t,u,v) form you gave does not make an entirely strong analogy. My head is hurting from recent wave of attacks and local problems, (despite having fun at arthropod museum) so I can not write too much more, but probably will anyways. To reiterate:
Pythagorean is to Trithagorean as
{n, d/2 - n^2/(2 d), d/2 + n^2/(2 d)}
is to
{n, n*(n + 1)/d - (d + 2)/4, n*(n + 1)/d + (d - 2)/4}.
— Brad
On Feb 27, 2020, at 11:27 PM, Neil Sloane <njasloane@gmail.com> wrote:
PS Brad, I was not trying to "diss"you, I simply did not understand what you were saying. I proposed a formula that seemed to give all the trythagorean triples. I'm quite willing to believe either of the following: You had already given that formula two days ago My formula is wrong, and you had given the correct formula two days ago! Best regards Neil
Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com
On Thu, Feb 27, 2020 at 11:47 PM Neil Sloane <njasloane@gmail.com> wrote: Dear Grad, Trouble is, I don't speak Mathematica. Could you possibly translate what you said in that last email into human? Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com On Thu, Feb 27, 2020 at 10:46 PM Brad Klee <bradklee@gmail.com> wrote: Hi Neil, I think you may have missed this little piece of code: PythagoreanTriples[sqNn_] := Select[{sqNn, k, m} /. Solve[ MapThread[#1 == #2 &, {{(sqNn^2)/#, #}, {(m - k), (m + k)}}]][[1]] & /@ Divisors[sqNn^2][[1 ;; -1]], And[And @@ (IntegerQ /@ #), And @@ (# > 0 & /@ #)] &] (*https://oeis.org/A046079*) Length[PythagoreanTriples[#]] & /@ Range[2, 20] Out[] = {0, 1, 1, 1, 1, 1, 2, 2, 1, 1, 4, 1, 1, 4, 3, 1, 2, 1, 4} If you hack out the part that says Solve[...] and maybe put a "d" in for the "#" sign, then you can get that [n,d/2-n^2/(2*d),d/2+n^2/(2*d)], with "d" for divisor. Just to double-check for possible typo, let's try: Expand[{n, d/2 - n^2/(2 d), d/2 + n^2/(2 d)} /. {x_, y_, z_} :>x^2 + y^2 - z^2] Out[]=0 Then, we can refine the idea of an "analog", Pythagorean : Trithagorean :: {n, d/2 - n^2/(2 d), d/2 + n^2/(2 d)} : {n, n*(n + 1)/d - (d + 2)/4, n*(n + 1)/d + (d - 2)/4}. The other relevant sequence is https://oeis.org/A309507, see my message from Sat. Feb 22. --Brad On Thu, Feb 27, 2020 at 3:54 PM Neil Sloane <njasloane@gmail.com> wrote:
I mentioned this on Feb 19 2020 and several people sent very helpful replies. For squares, we have the theorem that all solutions to x^2+y^2=z^2 in integers are given by Pythagorean triples, t*[u^2-v^2, 2uv, u^2+v^2]. Expanding on the replies from Michael Collins and Rich, it looks like the following is the analog for triangular numbers T_n = n(n+1)/2. ALL solutions to T_n+T_k=T_m are given by the list of what one might call Trithagorean triples: these are the triples [n,k,m] = [n, T/Q-(Q+1)/2,T/Q+(Q-1)/2] where n >= 2, T=n(n+1)/2, and Q is any odd divisor of T less than n, plus these triples with the first two coordinates swapped. I haven't found this in the literature, but it can hardly be new. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Neil, I'm sorry to hear that Richard is not doing well. I hope that he pulls through. In the meantime, it appears that Sierpinski has looked at the problem: MR0182602 (32 #85) Reviewed Sierpiński, W. On triangular numbers which are sums of two smaller triangular numbers. (Polish) Wiadom. Mat. (2) 7 (1963), 27–28. 10.05 The author proves the following theorem: The necessary and sufficient condition that a triangular number tn=12n(n+1)t_n=\frac 1{2}n(n+1) be the sum of two smaller triangular numbers is that n2+(n+1)2n^2+(n+1)^2 be a composite number. From the hypothesis H of A. Schinzel [Schinzel and the author, Acta Arith. 4 (1958), 185–208; erratum, ibid. 5 (1959), 259; MR0106202], it follows that there exist infinitely many triangular numbers which are not sums of two smaller triangular numbers. Reviewed by J. W. Andrushkiw and Vaidya: MR0379350 (52 #255) Reviewed Vaidya, A. M. On representing an integer as a sum of two triangular numbers. Vidya B 15 (1972), no. 2, 104–105. 10A40 The author proves that a positive integer kk is the sum of two triangular numbers if and only if 4k+1=u2+v24k+1=u^2+v^2, where u−v≥3u-v\geq 3. He also observes that for a positive integer nn we have n2+(n+1)2=u2+v2n^2+(n+1)^2=u^2+v^2 with u−v≥3u-v\geq 3 if and only if n2+(n+1)2n^2+(n+1)^2 is not a prime number. Combining these two results the author obtains the following theorem of W. Sierpiński [Elem. Math. 17 (1962), 63–65; MR0150088]: The triangular number n(n+1)/2n(n+1)/2 is the sum of two nonzero triangular numbers if and only if n2+(n+1)2n^2+(n+1)^2 is not a prime number. Reviewed by K. Szymiczek and Haggard proves something close to what we want: MR1467589 (99f:11022) Reviewed Haggard, Paul W.(1-ENC) Pythagorean triples and sums of triangular numbers. (English summary) Internat. J. Math. Ed. Sci. Tech. 28 (1997), no. 1, 109–116. 11B65 (11D09) An integer is said to be a triangular number if it is of the form tn=n(n+1)/2t_n=n(n+1)/2. It is known [see, e.g., D. M. Burton, Elementary number theory, fourth edition, McGraw-Hill, New York, 1998 (p.16)] that tx=ty+tzt_x=t_y+t_z, where x=n(n+3)/2+1x=n(n+3)/2+1, y=n+1y=n+1, z=n(n+3)/2z=n(n+3)/2 and n≥1n\geq 1. In this paper the author shows that for each primitive Pythagorean triple (A,B,C)(A,B,C), of positive integers with AA odd such that A2+B2=C2A^2+B^2=C^2 and C−B=1C-B=1, one can find a triple (a,b,c)(a,b,c) of integers such that ta=tb+tct_a=t_b+t_c. Reviewed by Jau-Shyong Shiue On Fri, Feb 28, 2020 at 3:21 PM Neil Sloane <njasloane@gmail.com> wrote:
Concerning the question about the general solution to the Trythagorean Triples problem, I happened to be speaking to Richard Guy this morning and I thought the question might cheer him up (he is in the hospital, and not doing well, I'm sorry to say. He is 104.) A colleague kindly put me on the speaker phone in his room. But he was not well enough to answer.
I have all ten volumes of his Reviews in Number Theory, but I was hoping not to have to search through them.
On Fri, Feb 28, 2020 at 1:19 AM Brad Klee <bradklee@gmail.com> wrote:
Hi Neil, your parameterization is more or less the same as what I came up with a week ago on Thurs. Feb 20. The purpose of the last email was to say that a similar parameterization (ranges over divisors) can be written for Pythagorean triples. The (t,u,v) form you gave does not make an entirely strong analogy. My head is hurting from recent wave of attacks and local problems, (despite having fun at arthropod museum) so I can not write too much more, but probably will anyways. To reiterate:
Pythagorean is to Trithagorean as
{n, d/2 - n^2/(2 d), d/2 + n^2/(2 d)}
is to
{n, n*(n + 1)/d - (d + 2)/4, n*(n + 1)/d + (d - 2)/4}.
— Brad
On Feb 27, 2020, at 11:27 PM, Neil Sloane <njasloane@gmail.com> wrote:
PS Brad, I was not trying to "diss"you, I simply did not understand what you were saying. I proposed a formula that seemed to give all the trythagorean triples. I'm quite willing to believe either of the following: You had already given that formula two days ago My formula is wrong, and you had given the correct formula two days ago! Best regards Neil
Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com
On Thu, Feb 27, 2020 at 11:47 PM Neil Sloane <njasloane@gmail.com> wrote: Dear Grad, Trouble is, I don't speak Mathematica. Could you possibly translate what you said in that last email into human? Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com On Thu, Feb 27, 2020 at 10:46 PM Brad Klee <bradklee@gmail.com> wrote: Hi Neil, I think you may have missed this little piece of code: PythagoreanTriples[sqNn_] := Select[{sqNn, k, m} /. Solve[ MapThread[#1 == #2 &, {{(sqNn^2)/#, #}, {(m - k), (m + k)}}]][[1]] & /@ Divisors[sqNn^2][[1 ;; -1]], And[And @@ (IntegerQ /@ #), And @@ (# > 0 & /@ #)] &] (*https://oeis.org/A046079*) Length[PythagoreanTriples[#]] & /@ Range[2, 20] Out[] = {0, 1, 1, 1, 1, 1, 2, 2, 1, 1, 4, 1, 1, 4, 3, 1, 2, 1, 4} If you hack out the part that says Solve[...] and maybe put a "d" in for the "#" sign, then you can get that [n,d/2-n^2/(2*d),d/2+n^2/(2*d)], with "d" for divisor. Just to double-check for possible typo, let's try: Expand[{n, d/2 - n^2/(2 d), d/2 + n^2/(2 d)} /. {x_, y_, z_} :>x^2 + y^2 - z^2] Out[]=0 Then, we can refine the idea of an "analog", Pythagorean : Trithagorean :: {n, d/2 - n^2/(2 d), d/2 + n^2/(2 d)} : {n, n*(n + 1)/d - (d + 2)/4, n*(n + 1)/d + (d - 2)/4}. The other relevant sequence is https://oeis.org/A309507, see my message from Sat. Feb 22. --Brad On Thu, Feb 27, 2020 at 3:54 PM Neil Sloane <njasloane@gmail.com> wrote:
I mentioned this on Feb 19 2020 and several people sent very helpful replies. For squares, we have the theorem that all solutions to x^2+y^2=z^2 in integers are given by Pythagorean triples, t*[u^2-v^2, 2uv, u^2+v^2]. Expanding on the replies from Michael Collins and Rich, it looks like the following is the analog for triangular numbers T_n = n(n+1)/2. ALL solutions to T_n+T_k=T_m are given by the list of what one might call Trithagorean triples: these are the triples [n,k,m] = [n, T/Q-(Q+1)/2,T/Q+(Q-1)/2] where n >= 2, T=n(n+1)/2, and Q is any odd divisor of T less than n, plus these triples with the first two coordinates swapped. I haven't found this in the literature, but it can hardly be new. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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This one's also relevant: MR2550124 (2010k:11056) Reviewed Ulas, Maciej(PL-JAGL) A note on Sierpiński's problem related to triangular numbers. (English summary) Colloq. Math. 117 (2009), no. 2, 165–173. 11D72 (11D25 11D41) The author considers the system of equations tx+ty=tp,ty+tz=tq,tx+tz=tr,t_x+t_y=t_p,\quad t_y+t_z=t_q,\quad t_x+t_z=t_r, where tn:=n(n+1)/2t_n\coloneq n(n+1)/2 denotes a triangular number. In fact the problem was suggested by W. Sierpiński, who asked the following question: Is it possible to find three different triangular numbers such that the sum of any pair of them is a triangular number? The author answers the above question by giving all integer solutions of the above system which satisfy the condition x<y<z<1000x<y<z<1000. Further, he also shows that there are infinitely many integer solutions to the system in question by constructing two one-parameter polynomial solutions of that system. Considering the rational case, the author proves that this system has a rational three-parameter solution. Finally, the above result on the rational case is used to construct a rational two-parameter solution of a related system. Reviewed by Attila Bérczes On Fri, Feb 28, 2020 at 4:26 PM Victor Miller <victorsmiller@gmail.com> wrote:
Neil, I'm sorry to hear that Richard is not doing well. I hope that he pulls through. In the meantime, it appears that Sierpinski has looked at the problem:
MR0182602 (32 #85) Reviewed Sierpiński, W. On triangular numbers which are sums of two smaller triangular numbers. (Polish) Wiadom. Mat. (2) 7 (1963), 27–28. 10.05
The author proves the following theorem: The necessary and sufficient condition that a triangular number tn=12n(n+1)t_n=\frac 1{2}n(n+1) be the sum of two smaller triangular numbers is that n2+(n+1)2n^2+(n+1)^2 be a composite number. From the hypothesis H of A. Schinzel [Schinzel and the author, Acta Arith. 4 (1958), 185–208; erratum, ibid. 5 (1959), 259; MR0106202], it follows that there exist infinitely many triangular numbers which are not sums of two smaller triangular numbers. Reviewed by J. W. Andrushkiw
and Vaidya:
MR0379350 (52 #255) Reviewed Vaidya, A. M. On representing an integer as a sum of two triangular numbers. Vidya B 15 (1972), no. 2, 104–105. 10A40
The author proves that a positive integer kk is the sum of two triangular numbers if and only if 4k+1=u2+v24k+1=u^2+v^2, where u−v≥3u-v\geq 3. He also observes that for a positive integer nn we have n2+(n+1)2=u2+v2n^2+(n+1)^2=u^2+v^2 with u−v≥3u-v\geq 3 if and only if n2+(n+1)2n^2+(n+1)^2 is not a prime number. Combining these two results the author obtains the following theorem of W. Sierpiński [Elem. Math. 17 (1962), 63–65; MR0150088]: The triangular number n(n+1)/2n(n+1)/2 is the sum of two nonzero triangular numbers if and only if n2+(n+1)2n^2+(n+1)^2 is not a prime number. Reviewed by K. Szymiczek
and Haggard proves something close to what we want:
MR1467589 (99f:11022) Reviewed Haggard, Paul W.(1-ENC) Pythagorean triples and sums of triangular numbers. (English summary) Internat. J. Math. Ed. Sci. Tech. 28 (1997), no. 1, 109–116. 11B65 (11D09)
An integer is said to be a triangular number if it is of the form tn=n(n+1)/2t_n=n(n+1)/2. It is known [see, e.g., D. M. Burton, Elementary number theory, fourth edition, McGraw-Hill, New York, 1998 (p.16)] that tx=ty+tzt_x=t_y+t_z, where x=n(n+3)/2+1x=n(n+3)/2+1, y=n+1y=n+1, z=n(n+3)/2z=n(n+3)/2 and n≥1n\geq 1. In this paper the author shows that for each primitive Pythagorean triple (A,B,C)(A,B,C), of positive integers with AA odd such that A2+B2=C2A^2+B^2=C^2 and C−B=1C-B=1, one can find a triple (a,b,c)(a,b,c) of integers such that ta=tb+tct_a=t_b+t_c. Reviewed by Jau-Shyong Shiue
On Fri, Feb 28, 2020 at 3:21 PM Neil Sloane <njasloane@gmail.com> wrote:
Concerning the question about the general solution to the Trythagorean Triples problem, I happened to be speaking to Richard Guy this morning and I thought the question might cheer him up (he is in the hospital, and not doing well, I'm sorry to say. He is 104.) A colleague kindly put me on the speaker phone in his room. But he was not well enough to answer.
I have all ten volumes of his Reviews in Number Theory, but I was hoping not to have to search through them.
On Fri, Feb 28, 2020 at 1:19 AM Brad Klee <bradklee@gmail.com> wrote:
Hi Neil, your parameterization is more or less the same as what I came up with a week ago on Thurs. Feb 20. The purpose of the last email was to say that a similar parameterization (ranges over divisors) can be written for Pythagorean triples. The (t,u,v) form you gave does not make an entirely strong analogy. My head is hurting from recent wave of attacks and local problems, (despite having fun at arthropod museum) so I can not write too much more, but probably will anyways. To reiterate:
Pythagorean is to Trithagorean as
{n, d/2 - n^2/(2 d), d/2 + n^2/(2 d)}
is to
{n, n*(n + 1)/d - (d + 2)/4, n*(n + 1)/d + (d - 2)/4}.
— Brad
On Feb 27, 2020, at 11:27 PM, Neil Sloane <njasloane@gmail.com> wrote:
PS Brad, I was not trying to "diss"you, I simply did not understand what you were saying. I proposed a formula that seemed to give all the trythagorean triples. I'm quite willing to believe either of the following: You had already given that formula two days ago My formula is wrong, and you had given the correct formula two days ago! Best regards Neil
Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com
On Thu, Feb 27, 2020 at 11:47 PM Neil Sloane <njasloane@gmail.com> wrote: Dear Grad, Trouble is, I don't speak Mathematica. Could you possibly translate what you said in that last email into human? Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com On Thu, Feb 27, 2020 at 10:46 PM Brad Klee <bradklee@gmail.com> wrote: Hi Neil, I think you may have missed this little piece of code: PythagoreanTriples[sqNn_] := Select[{sqNn, k, m} /. Solve[ MapThread[#1 == #2 &, {{(sqNn^2)/#, #}, {(m - k), (m + k)}}]][[1]] & /@ Divisors[sqNn^2][[1 ;; -1]], And[And @@ (IntegerQ /@ #), And @@ (# > 0 & /@ #)] &] (*https://oeis.org/A046079*) Length[PythagoreanTriples[#]] & /@ Range[2, 20] Out[] = {0, 1, 1, 1, 1, 1, 2, 2, 1, 1, 4, 1, 1, 4, 3, 1, 2, 1, 4} If you hack out the part that says Solve[...] and maybe put a "d" in for the "#" sign, then you can get that [n,d/2-n^2/(2*d),d/2+n^2/(2*d)], with "d" for divisor. Just to double-check for possible typo, let's try: Expand[{n, d/2 - n^2/(2 d), d/2 + n^2/(2 d)} /. {x_, y_, z_} :>x^2 + y^2 - z^2] Out[]=0 Then, we can refine the idea of an "analog", Pythagorean : Trithagorean :: {n, d/2 - n^2/(2 d), d/2 + n^2/(2 d)} : {n, n*(n + 1)/d - (d + 2)/4, n*(n + 1)/d + (d - 2)/4}. The other relevant sequence is https://oeis.org/A309507, see my message from Sat. Feb 22. --Brad On Thu, Feb 27, 2020 at 3:54 PM Neil Sloane <njasloane@gmail.com> wrote: > I mentioned this on Feb 19 2020 and several people sent very helpful > replies. > For squares, we have the theorem that all solutions to x^2+y^2=z^2 in > integers are given by Pythagorean triples, t*[u^2-v^2, 2uv, u^2+v^2]. > Expanding on the replies from Michael Collins and Rich, it looks like the > following is the analog for triangular numbers T_n = n(n+1)/2. > ALL solutions to T_n+T_k=T_m are given by the list of what one might call > Trithagorean triples: these are the triples > [n,k,m] = [n, T/Q-(Q+1)/2,T/Q+(Q-1)/2] where n >= 2, T=n(n+1)/2, and Q is > any odd divisor of T less than n, > plus these triples with the first two coordinates swapped. > I haven't found this in the literature, but it can hardly be new. > _______________________________________________ > math-fun mailing list > math-fun@mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
M. A. Nyblom, "On the representation of the integers as a difference of nonconsecutive triangular numbers", Fibonacci Quarterly 39:3 (2001), pp. 256-263. https://www.fq.math.ca/Scanned/39-3/nyblom.pdf Cf. A001227, A057716, A098033. The Tripathi paper is a later work from the same journal containing similar theorems, but no back reference? The proof of Theorem 2.1 is too fast for me in that it does not explain why the square is completed, or what happens otherwise. It turns out that this step is not necessary. I found a mistake in my implementation of Rich's solution, see instead: RichTriples[TNn_] := Sort[Select[ {TNn, k, m} /. Solve[MapThread[ #1 == #2 &, {{TNn (TNn + 1)/#, #}, {(m + k + 1), (m - k)}}]][[1]] & /@ Complement[Divisors[TNn (TNn + 1)], {TNn}], And[And @@ (IntegerQ /@ #), And @@ (# > 0 & /@ #)] &]] Length@RichTriples[#] & /@ Range[2, 10] Out[] = {1, 1, 1, 3, 3, 1, 2, 5, 3} This is the correct count when including the so-called "trivial" solutions (at first, I misunderstood that these should be excluded for no reason). I think it also needs to be said and proven that (x-y)*(x+y) = x^2 - y^2 is a unique factorization, but perhaps Fibonacci authors think that this is an obvious fact? Wrong again! --Brad On Fri, Feb 28, 2020 at 3:51 PM Victor Miller <victorsmiller@gmail.com> wrote:
This one's also relevant:
MR2550124 (2010k:11056) Reviewed Ulas, Maciej(PL-JAGL) A note on Sierpiński's problem related to triangular numbers. (English summary) Colloq. Math. 117 (2009), no. 2, 165–173. 11D72 (11D25 11D41)
The author considers the system of equations tx+ty=tp,ty+tz=tq,tx+tz=tr,t_x+t_y=t_p,\quad t_y+t_z=t_q,\quad t_x+t_z=t_r, where tn:=n(n+1)/2t_n\coloneq n(n+1)/2 denotes a triangular number. In fact the problem was suggested by W. Sierpiński, who asked the following question: Is it possible to find three different triangular numbers such that the sum of any pair of them is a triangular number? The author answers the above question by giving all integer solutions of the above system which satisfy the condition x<y<z<1000x<y<z<1000. Further, he also shows that there are infinitely many integer solutions to the system in question by constructing two one-parameter polynomial solutions of that system. Considering the rational case, the author proves that this system has a rational three-parameter solution. Finally, the above result on the rational case is used to construct a rational two-parameter solution of a related system. Reviewed by Attila Bérczes
On Fri, Feb 28, 2020 at 4:26 PM Victor Miller <victorsmiller@gmail.com> wrote:
Neil, I'm sorry to hear that Richard is not doing well. I hope that he pulls through. In the meantime, it appears that Sierpinski has looked at the problem:
MR0182602 (32 #85) Reviewed Sierpiński, W. On triangular numbers which are sums of two smaller triangular numbers. (Polish) Wiadom. Mat. (2) 7 (1963), 27–28. 10.05
The author proves the following theorem: The necessary and sufficient condition that a triangular number tn=12n(n+1)t_n=\frac 1{2}n(n+1) be the sum of two smaller triangular numbers is that n2+(n+1)2n^2+(n+1)^2 be a composite number. From the hypothesis H of A. Schinzel [Schinzel and the author, Acta Arith. 4 (1958), 185–208; erratum, ibid. 5 (1959), 259; MR0106202], it follows that there exist infinitely many triangular numbers which are not sums of two smaller triangular numbers. Reviewed by J. W. Andrushkiw
and Vaidya:
MR0379350 (52 #255) Reviewed Vaidya, A. M. On representing an integer as a sum of two triangular numbers. Vidya B 15 (1972), no. 2, 104–105. 10A40
The author proves that a positive integer kk is the sum of two triangular numbers if and only if 4k+1=u2+v24k+1=u^2+v^2, where u−v≥3u-v\geq 3. He also observes that for a positive integer nn we have n2+(n+1)2=u2+v2n^2+(n+1)^2=u^2+v^2 with u−v≥3u-v\geq 3 if and only if n2+(n+1)2n^2+(n+1)^2 is not a prime number. Combining these two results the author obtains the following theorem of W. Sierpiński [Elem. Math. 17 (1962), 63–65; MR0150088]: The triangular number n(n+1)/2n(n+1)/2 is the sum of two nonzero triangular numbers if and only if n2+(n+1)2n^2+(n+1)^2 is not a prime number. Reviewed by K. Szymiczek
and Haggard proves something close to what we want:
MR1467589 (99f:11022) Reviewed Haggard, Paul W.(1-ENC) Pythagorean triples and sums of triangular numbers. (English summary) Internat. J. Math. Ed. Sci. Tech. 28 (1997), no. 1, 109–116. 11B65 (11D09)
An integer is said to be a triangular number if it is of the form tn=n(n+1)/2t_n=n(n+1)/2. It is known [see, e.g., D. M. Burton, Elementary number theory, fourth edition, McGraw-Hill, New York, 1998 (p.16)] that tx=ty+tzt_x=t_y+t_z, where x=n(n+3)/2+1x=n(n+3)/2+1, y=n+1y=n+1, z=n(n+3)/2z=n(n+3)/2 and n≥1n\geq 1. In this paper the author shows that for each primitive Pythagorean triple (A,B,C)(A,B,C), of positive integers with AA odd such that A2+B2=C2A^2+B^2=C^2 and C−B=1C-B=1, one can find a triple (a,b,c)(a,b,c) of integers such that ta=tb+tct_a=t_b+t_c. Reviewed by Jau-Shyong Shiue
On Fri, Feb 28, 2020 at 3:21 PM Neil Sloane <njasloane@gmail.com> wrote:
Concerning the question about the general solution to the Trythagorean Triples problem, I happened to be speaking to Richard Guy this morning and I thought the question might cheer him up (he is in the hospital, and not doing well, I'm sorry to say. He is 104.) A colleague kindly put me on the speaker phone in his room. But he was not well enough to answer.
I have all ten volumes of his Reviews in Number Theory, but I was hoping not to have to search through them.
On Fri, Feb 28, 2020 at 1:19 AM Brad Klee <bradklee@gmail.com> wrote:
Hi Neil, your parameterization is more or less the same as what I came up with a week ago on Thurs. Feb 20. The purpose of the last email was to say that a similar parameterization (ranges over divisors) can be written for Pythagorean triples. The (t,u,v) form you gave does not make an entirely strong analogy. My head is hurting from recent wave of attacks and local problems, (despite having fun at arthropod museum) so I can not write too much more, but probably will anyways. To reiterate:
Pythagorean is to Trithagorean as
{n, d/2 - n^2/(2 d), d/2 + n^2/(2 d)}
is to
{n, n*(n + 1)/d - (d + 2)/4, n*(n + 1)/d + (d - 2)/4}.
— Brad
On Feb 27, 2020, at 11:27 PM, Neil Sloane <njasloane@gmail.com> wrote:
PS Brad, I was not trying to "diss"you, I simply did not understand what you were saying. I proposed a formula that seemed to give all the trythagorean triples. I'm quite willing to believe either of the following: You had already given that formula two days ago My formula is wrong, and you had given the correct formula two days ago! Best regards Neil
Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com
> On Thu, Feb 27, 2020 at 11:47 PM Neil Sloane <njasloane@gmail.com> wrote: Dear Grad, Trouble is, I don't speak Mathematica. Could you possibly translate what you said in that last email into human? Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com > On Thu, Feb 27, 2020 at 10:46 PM Brad Klee <bradklee@gmail.com> wrote: > Hi Neil, > I think you may have missed this little piece of code: > PythagoreanTriples[sqNn_] := Select[{sqNn, k, m} /. Solve[ > MapThread[#1 == #2 &, {{(sqNn^2)/#, #}, {(m - k), (m + > k)}}]][[1]] & /@ Divisors[sqNn^2][[1 ;; -1]], > And[And @@ (IntegerQ /@ #), And @@ (# > 0 & /@ #)] &] > (*https://oeis.org/A046079*) > Length[PythagoreanTriples[#]] & /@ Range[2, 20] > Out[] = {0, 1, 1, 1, 1, 1, 2, 2, 1, 1, 4, 1, 1, 4, 3, 1, 2, 1, 4} > If you hack out the part that says Solve[...] and maybe put a "d" in > for the "#" sign, then you can get that [n,d/2-n^2/(2*d),d/2+n^2/(2*d)], > with "d" for divisor. Just to double-check for possible typo, let's try: > Expand[{n, d/2 - n^2/(2 d), d/2 + n^2/(2 d)} /. {x_, y_, z_} :>x^2 + y^2 > - z^2] > Out[]=0 > Then, we can refine the idea of an "analog", > Pythagorean : Trithagorean :: > {n, d/2 - n^2/(2 d), d/2 + n^2/(2 d)} : {n, n*(n + 1)/d - (d + 2)/4, > n*(n + 1)/d + (d - 2)/4}. > The other relevant sequence is https://oeis.org/A309507, see > my message from Sat. Feb 22. --Brad > On Thu, Feb 27, 2020 at 3:54 PM Neil Sloane <njasloane@gmail.com> wrote: >> I mentioned this on Feb 19 2020 and several people sent very helpful >> replies. >> For squares, we have the theorem that all solutions to x^2+y^2=z^2 in >> integers are given by Pythagorean triples, t*[u^2-v^2, 2uv, u^2+v^2]. >> Expanding on the replies from Michael Collins and Rich, it looks like > the >> following is the analog for triangular numbers T_n = n(n+1)/2. >> ALL solutions to T_n+T_k=T_m are given by the list of what one might > call >> Trithagorean triples: these are the triples >> [n,k,m] = [n, T/Q-(Q+1)/2,T/Q+(Q-1)/2] where n >= 2, T=n(n+1)/2, and Q > is >> any odd divisor of T less than n, >> plus these triples with the first two coordinates swapped. >> I haven't found this in the literature, but it can hardly be new. >> _______________________________________________ >> math-fun mailing list >> math-fun@mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > _______________________________________________ > math-fun mailing list > math-fun@mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Slightly related, there is this nice demonstration of T(6n-1) = 12 P(n) + 3 n where P(n) is the n-th pentagonal number in the new issue of the Intelligencer https://bit.ly/2wkGQyU On Sat, Feb 29, 2020 at 8:14 PM Brad Klee <bradklee@gmail.com> wrote:
M. A. Nyblom, "On the representation of the integers as a difference of nonconsecutive triangular numbers", Fibonacci Quarterly 39:3 (2001), pp. 256-263.
https://www.fq.math.ca/Scanned/39-3/nyblom.pdf
Cf. A001227, A057716, A098033.
The Tripathi paper is a later work from the same journal containing similar theorems, but no back reference?
The proof of Theorem 2.1 is too fast for me in that it does not explain why the square is completed, or what happens otherwise. It turns out that this step is not necessary. I found a mistake in my implementation of Rich's solution, see instead:
RichTriples[TNn_] := Sort[Select[ {TNn, k, m} /. Solve[MapThread[ #1 == #2 &, {{TNn (TNn + 1)/#, #}, {(m + k + 1), (m - k)}}]][[1]] & /@ Complement[Divisors[TNn (TNn + 1)], {TNn}], And[And @@ (IntegerQ /@ #), And @@ (# > 0 & /@ #)] &]] Length@RichTriples[#] & /@ Range[2, 10] Out[] = {1, 1, 1, 3, 3, 1, 2, 5, 3}
This is the correct count when including the so-called "trivial" solutions (at first, I misunderstood that these should be excluded for no reason). I think it also needs to be said and proven that (x-y)*(x+y) = x^2 - y^2 is a unique factorization, but perhaps Fibonacci authors think that this is an obvious fact?
Wrong again! --Brad
On Fri, Feb 28, 2020 at 3:51 PM Victor Miller <victorsmiller@gmail.com> wrote:
This one's also relevant:
MR2550124 (2010k:11056) Reviewed Ulas, Maciej(PL-JAGL) A note on Sierpiński's problem related to triangular numbers. (English summary) Colloq. Math. 117 (2009), no. 2, 165–173. 11D72 (11D25 11D41)
The author considers the system of equations tx+ty=tp,ty+tz=tq,tx+tz=tr,t_x+t_y=t_p,\quad t_y+t_z=t_q,\quad
t_x+t_z=t_r,
where tn:=n(n+1)/2t_n\coloneq n(n+1)/2 denotes a triangular number. In fact the problem was suggested by W. Sierpiński, who asked the following question: Is it possible to find three different triangular numbers such that the sum of any pair of them is a triangular number? The author answers the above question by giving all integer solutions of the above system which satisfy the condition x<y<z<1000x<y<z<1000. Further, he also shows that there are infinitely many integer solutions to the system in question by constructing two one-parameter polynomial solutions of that system. Considering the rational case, the author proves that this system has a rational three-parameter solution. Finally, the above result on the rational case is used to construct a rational two-parameter solution of a related system. Reviewed by Attila Bérczes
On Fri, Feb 28, 2020 at 4:26 PM Victor Miller <victorsmiller@gmail.com> wrote:
Neil, I'm sorry to hear that Richard is not doing well. I hope that he pulls through. In the meantime, it appears that Sierpinski has looked at the problem:
MR0182602 (32 #85) Reviewed Sierpiński, W. On triangular numbers which are sums of two smaller triangular numbers. (Polish) Wiadom. Mat. (2) 7 (1963), 27–28. 10.05
The author proves the following theorem: The necessary and sufficient condition that a triangular number tn=12n(n+1)t_n=\frac 1{2}n(n+1) be the sum of two smaller triangular numbers is that n2+(n+1)2n^2+(n+1)^2 be a composite number. From the hypothesis H of A. Schinzel [Schinzel and the author, Acta Arith. 4 (1958), 185–208; erratum, ibid. 5 (1959), 259; MR0106202], it follows that there exist infinitely many triangular numbers which are not sums of two smaller triangular numbers. Reviewed by J. W. Andrushkiw
and Vaidya:
MR0379350 (52 #255) Reviewed Vaidya, A. M. On representing an integer as a sum of two triangular numbers. Vidya B 15 (1972), no. 2, 104–105. 10A40
The author proves that a positive integer kk is the sum of two triangular numbers if and only if 4k+1=u2+v24k+1=u^2+v^2, where u−v≥3u-v\geq 3. He also observes that for a positive integer nn we have n2+(n+1)2=u2+v2n^2+(n+1)^2=u^2+v^2 with u−v≥3u-v\geq 3 if and only if n2+(n+1)2n^2+(n+1)^2 is not a prime number. Combining these two results the author obtains the following theorem of W. Sierpiński [Elem. Math. 17 (1962), 63–65; MR0150088]: The triangular number n(n+1)/2n(n+1)/2 is the sum of two nonzero triangular numbers if and only if n2+(n+1)2n^2+(n+1)^2 is not a prime number. Reviewed by K. Szymiczek
and Haggard proves something close to what we want:
MR1467589 (99f:11022) Reviewed Haggard, Paul W.(1-ENC) Pythagorean triples and sums of triangular numbers. (English summary) Internat. J. Math. Ed. Sci. Tech. 28 (1997), no. 1, 109–116. 11B65 (11D09)
An integer is said to be a triangular number if it is of the form tn=n(n+1)/2t_n=n(n+1)/2. It is known [see, e.g., D. M. Burton, Elementary number theory, fourth edition, McGraw-Hill, New York, 1998 (p.16)] that tx=ty+tzt_x=t_y+t_z, where x=n(n+3)/2+1x=n(n+3)/2+1, y=n+1y=n+1, z=n(n+3)/2z=n(n+3)/2 and n≥1n\geq 1. In this paper the author shows that for each primitive Pythagorean triple (A,B,C)(A,B,C), of positive integers with AA odd such that A2+B2=C2A^2+B^2=C^2 and C−B=1C-B=1, one can find a triple (a,b,c)(a,b,c) of integers such that ta=tb+tct_a=t_b+t_c. Reviewed by Jau-Shyong Shiue
On Fri, Feb 28, 2020 at 3:21 PM Neil Sloane <njasloane@gmail.com> wrote:
Concerning the question about the general solution to the Trythagorean Triples problem, I happened to be speaking to Richard Guy this morning and I thought the question might cheer him up (he is in the hospital, and not doing well, I'm sorry to say. He is 104.) A colleague kindly put me on the speaker phone in his room. But he was not well enough to answer.
I have all ten volumes of his Reviews in Number Theory, but I was hoping not to have to search through them.
On Fri, Feb 28, 2020 at 1:19 AM Brad Klee <bradklee@gmail.com> wrote:
Hi Neil, your parameterization is more or less the same as what I came up with a week ago on Thurs. Feb 20. The purpose of the last email was to say that a similar parameterization (ranges over divisors) can be written for Pythagorean triples. The (t,u,v) form you gave does not make an entirely strong analogy. My head is hurting from recent wave of attacks and local problems, (despite having fun at arthropod museum) so I can not write too much more, but probably will anyways. To reiterate:
Pythagorean is to Trithagorean as
{n, d/2 - n^2/(2 d), d/2 + n^2/(2 d)}
is to
{n, n*(n + 1)/d - (d + 2)/4, n*(n + 1)/d + (d - 2)/4}.
— Brad
On Feb 27, 2020, at 11:27 PM, Neil Sloane <njasloane@gmail.com> wrote:
PS Brad, I was not trying to "diss"you, I simply did not understand what you were saying. I proposed a formula that seemed to give all the trythagorean triples. I'm quite willing to believe either of the following: You had already given that formula two days ago My formula is wrong, and you had given the correct formula two days ago! Best regards Neil
Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com
>> On Thu, Feb 27, 2020 at 11:47 PM Neil Sloane < njasloane@gmail.com> wrote: > Dear Grad, > Trouble is, I don't speak Mathematica. Could you possibly translate what > you said in that last email into human? > Best regards > Neil > Neil J. A. Sloane, President, OEIS Foundation. > 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. > Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. > Phone: 732 828 6098; home page: http://NeilSloane.com > Email: njasloane@gmail.com >> On Thu, Feb 27, 2020 at 10:46 PM Brad Klee <bradklee@gmail.com> wrote: >> Hi Neil, >> I think you may have missed this little piece of code: >> PythagoreanTriples[sqNn_] := Select[{sqNn, k, m} /. Solve[ >> MapThread[#1 == #2 &, {{(sqNn^2)/#, #}, {(m - k), (m + >> k)}}]][[1]] & /@ Divisors[sqNn^2][[1 ;; -1]], >> And[And @@ (IntegerQ /@ #), And @@ (# > 0 & /@ #)] &] >> (*https://oeis.org/A046079*) >> Length[PythagoreanTriples[#]] & /@ Range[2, 20] >> Out[] = {0, 1, 1, 1, 1, 1, 2, 2, 1, 1, 4, 1, 1, 4, 3, 1, 2, 1, 4} >> If you hack out the part that says Solve[...] and maybe put a "d" in >> for the "#" sign, then you can get that [n,d/2-n^2/(2*d),d/2+n^2/(2*d)], >> with "d" for divisor. Just to double-check for possible typo, let's try: >> Expand[{n, d/2 - n^2/(2 d), d/2 + n^2/(2 d)} /. {x_, y_, z_} :>x^2 + y^2 >> - z^2] >> Out[]=0 >> Then, we can refine the idea of an "analog", >> Pythagorean : Trithagorean :: >> {n, d/2 - n^2/(2 d), d/2 + n^2/(2 d)} : {n, n*(n + 1)/d - (d + 2)/4, >> n*(n + 1)/d + (d - 2)/4}. >> The other relevant sequence is https://oeis.org/A309507, see >> my message from Sat. Feb 22. --Brad >> On Thu, Feb 27, 2020 at 3:54 PM Neil Sloane < njasloane@gmail.com> wrote: >>> I mentioned this on Feb 19 2020 and several people sent very helpful >>> replies. >>> For squares, we have the theorem that all solutions to x^2+y^2=z^2 in >>> integers are given by Pythagorean triples, t*[u^2-v^2, 2uv, u^2+v^2]. >>> Expanding on the replies from Michael Collins and Rich, it looks like >> the >>> following is the analog for triangular numbers T_n = n(n+1)/2. >>> ALL solutions to T_n+T_k=T_m are given by the list of what one might >> call >>> Trithagorean triples: these are the triples >>> [n,k,m] = [n, T/Q-(Q+1)/2,T/Q+(Q-1)/2] where n >= 2, T=n(n+1)/2, and Q >> is >>> any odd divisor of T less than n, >>> plus these triples with the first two coordinates swapped. >>> I haven't found this in the literature, but it can hardly be new. >>> _______________________________________________ >>> math-fun mailing list >>> math-fun@mailman.xmission.com >>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> _______________________________________________ >> math-fun mailing list >> math-fun@mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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The link was broken, but I think you mean this: Günhan Caglayan, "Covering a Triangular Number with Pentagonal Numbers", Mathematical Intelligencer 42, Dec. 2020, p. 55. https://link.springer.com/article/10.1007/s00283-019-09953-0 A nice construction! It probably would seem more related if we were to try and find geometric interpretation à la Euclid's proof of the Pythagorean theorem. As Area = sqrt(3)/4*n^2, for an equilateral triangle of edge-length n, we can immediately replace the squares off each edge, with equilateral triangles off each edge, or in fact, with any shape regular polygon, including pentagons. --Brad On Sat, Feb 29, 2020 at 11:59 PM Olivier Gerard <olivier.gerard@gmail.com> wrote:
Slightly related, there is this nice demonstration of T(6n-1) = 12 P(n) + 3 n where P(n) is the n-th pentagonal number in the new issue of the Intelligencer
On Sat, Feb 29, 2020 at 8:14 PM Brad Klee <bradklee@gmail.com> wrote:
M. A. Nyblom, "On the representation of the integers as a difference of nonconsecutive triangular numbers", Fibonacci Quarterly 39:3 (2001), pp. 256-263.
https://www.fq.math.ca/Scanned/39-3/nyblom.pdf
Cf. A001227, A057716, A098033.
The Tripathi paper is a later work from the same journal containing similar theorems, but no back reference?
The proof of Theorem 2.1 is too fast for me in that it does not explain why the square is completed, or what happens otherwise. It turns out that this step is not necessary. I found a mistake in my implementation of Rich's solution, see instead:
RichTriples[TNn_] := Sort[Select[ {TNn, k, m} /. Solve[MapThread[ #1 == #2 &, {{TNn (TNn + 1)/#, #}, {(m + k + 1), (m - k)}}]][[1]] & /@ Complement[Divisors[TNn (TNn + 1)], {TNn}], And[And @@ (IntegerQ /@ #), And @@ (# > 0 & /@ #)] &]] Length@RichTriples[#] & /@ Range[2, 10] Out[] = {1, 1, 1, 3, 3, 1, 2, 5, 3}
This is the correct count when including the so-called "trivial" solutions (at first, I misunderstood that these should be excluded for no reason). I think it also needs to be said and proven that (x-y)*(x+y) = x^2 - y^2 is a unique factorization, but perhaps Fibonacci authors think that this is an obvious fact?
Wrong again! --Brad
On Fri, Feb 28, 2020 at 3:51 PM Victor Miller <victorsmiller@gmail.com> wrote:
This one's also relevant:
MR2550124 (2010k:11056) Reviewed Ulas, Maciej(PL-JAGL) A note on Sierpiński's problem related to triangular numbers. (English summary) Colloq. Math. 117 (2009), no. 2, 165–173. 11D72 (11D25 11D41)
The author considers the system of equations tx+ty=tp,ty+tz=tq,tx+tz=tr,t_x+t_y=t_p,\quad t_y+t_z=t_q,\quad
t_x+t_z=t_r,
where tn:=n(n+1)/2t_n\coloneq n(n+1)/2 denotes a triangular number. In fact the problem was suggested by W. Sierpiński, who asked the following question: Is it possible to find three different triangular numbers such that the sum of any pair of them is a triangular number? The author answers the above question by giving all integer solutions of the above system which satisfy the condition x<y<z<1000x<y<z<1000. Further, he also shows that there are infinitely many integer solutions to the system in question by constructing two one-parameter polynomial solutions of that system. Considering the rational case, the author proves that this system has a rational three-parameter solution. Finally, the above result on the rational case is used to construct a rational two-parameter solution of a related system. Reviewed by Attila Bérczes
On Fri, Feb 28, 2020 at 4:26 PM Victor Miller <victorsmiller@gmail.com> wrote:
Neil, I'm sorry to hear that Richard is not doing well. I hope that he pulls through. In the meantime, it appears that Sierpinski has looked at the problem:
MR0182602 (32 #85) Reviewed Sierpiński, W. On triangular numbers which are sums of two smaller triangular numbers. (Polish) Wiadom. Mat. (2) 7 (1963), 27–28. 10.05
The author proves the following theorem: The necessary and sufficient condition that a triangular number tn=12n(n+1)t_n=\frac 1{2}n(n+1) be the sum of two smaller triangular numbers is that n2+(n+1)2n^2+(n+1)^2 be a composite number. From the hypothesis H of A. Schinzel [Schinzel and the author, Acta Arith. 4 (1958), 185–208; erratum, ibid. 5 (1959), 259; MR0106202], it follows that there exist infinitely many triangular numbers which are not sums of two smaller triangular numbers. Reviewed by J. W. Andrushkiw
and Vaidya:
MR0379350 (52 #255) Reviewed Vaidya, A. M. On representing an integer as a sum of two triangular numbers. Vidya B 15 (1972), no. 2, 104–105. 10A40
The author proves that a positive integer kk is the sum of two triangular numbers if and only if 4k+1=u2+v24k+1=u^2+v^2, where u−v≥3u-v\geq 3. He also observes that for a positive integer nn we have n2+(n+1)2=u2+v2n^2+(n+1)^2=u^2+v^2 with u−v≥3u-v\geq 3 if and only if n2+(n+1)2n^2+(n+1)^2 is not a prime number. Combining these two results the author obtains the following theorem of W. Sierpiński [Elem. Math. 17 (1962), 63–65; MR0150088]: The triangular number n(n+1)/2n(n+1)/2 is the sum of two nonzero triangular numbers if and only if n2+(n+1)2n^2+(n+1)^2 is not a prime number. Reviewed by K. Szymiczek
and Haggard proves something close to what we want:
MR1467589 (99f:11022) Reviewed Haggard, Paul W.(1-ENC) Pythagorean triples and sums of triangular numbers. (English summary) Internat. J. Math. Ed. Sci. Tech. 28 (1997), no. 1, 109–116. 11B65 (11D09)
An integer is said to be a triangular number if it is of the form tn=n(n+1)/2t_n=n(n+1)/2. It is known [see, e.g., D. M. Burton, Elementary number theory, fourth edition, McGraw-Hill, New York, 1998 (p.16)] that tx=ty+tzt_x=t_y+t_z, where x=n(n+3)/2+1x=n(n+3)/2+1, y=n+1y=n+1, z=n(n+3)/2z=n(n+3)/2 and n≥1n\geq 1. In this paper the author shows that for each primitive Pythagorean triple (A,B,C)(A,B,C), of positive integers with AA odd such that A2+B2=C2A^2+B^2=C^2 and C−B=1C-B=1, one can find a triple (a,b,c)(a,b,c) of integers such that ta=tb+tct_a=t_b+t_c. Reviewed by Jau-Shyong Shiue
On Fri, Feb 28, 2020 at 3:21 PM Neil Sloane <njasloane@gmail.com> wrote:
Concerning the question about the general solution to the Trythagorean Triples problem, I happened to be speaking to Richard Guy this morning and I thought the question might cheer him up (he is in the hospital, and not doing well, I'm sorry to say. He is 104.) A colleague kindly put me on the speaker phone in his room. But he was not well enough to answer.
I have all ten volumes of his Reviews in Number Theory, but I was hoping not to have to search through them.
On Fri, Feb 28, 2020 at 1:19 AM Brad Klee <bradklee@gmail.com> wrote:
Hi Neil, your parameterization is more or less the same as what I came up with a week ago on Thurs. Feb 20. The purpose of the last email was to say that a similar parameterization (ranges over divisors) can be written for Pythagorean triples. The (t,u,v) form you gave does not make an entirely strong analogy. My head is hurting from recent wave of attacks and local problems, (despite having fun at arthropod museum) so I can not write too much more, but probably will anyways. To reiterate:
Pythagorean is to Trithagorean as
{n, d/2 - n^2/(2 d), d/2 + n^2/(2 d)}
is to
{n, n*(n + 1)/d - (d + 2)/4, n*(n + 1)/d + (d - 2)/4}.
— Brad
> On Feb 27, 2020, at 11:27 PM, Neil Sloane <njasloane@gmail.com> wrote: > > PS Brad, I was not trying to "diss"you, I simply did not understand what > you were saying. > I proposed a formula that seemed to give all the trythagorean triples. > I'm quite willing to believe either of the following: > You had already given that formula two days ago > My formula is wrong, and you had given the correct formula two days ago! > Best regards > Neil > > Neil J. A. Sloane, President, OEIS Foundation. > 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. > Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. > Phone: 732 828 6098; home page: http://NeilSloane.com > Email: njasloane@gmail.com > > > >>> On Thu, Feb 27, 2020 at 11:47 PM Neil Sloane < njasloane@gmail.com> wrote: >> Dear Grad, >> Trouble is, I don't speak Mathematica. Could you possibly translate what >> you said in that last email into human? >> Best regards >> Neil >> Neil J. A. Sloane, President, OEIS Foundation. >> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. >> Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. >> Phone: 732 828 6098; home page: http://NeilSloane.com >> Email: njasloane@gmail.com >>> On Thu, Feb 27, 2020 at 10:46 PM Brad Klee <bradklee@gmail.com> wrote: >>> Hi Neil, >>> I think you may have missed this little piece of code: >>> PythagoreanTriples[sqNn_] := Select[{sqNn, k, m} /. Solve[ >>> MapThread[#1 == #2 &, {{(sqNn^2)/#, #}, {(m - k), (m + >>> k)}}]][[1]] & /@ Divisors[sqNn^2][[1 ;; -1]], >>> And[And @@ (IntegerQ /@ #), And @@ (# > 0 & /@ #)] &] >>> (*https://oeis.org/A046079*) >>> Length[PythagoreanTriples[#]] & /@ Range[2, 20] >>> Out[] = {0, 1, 1, 1, 1, 1, 2, 2, 1, 1, 4, 1, 1, 4, 3, 1, 2, 1, 4} >>> If you hack out the part that says Solve[...] and maybe put a "d" in >>> for the "#" sign, then you can get that [n,d/2-n^2/(2*d),d/2+n^2/(2*d)], >>> with "d" for divisor. Just to double-check for possible typo, let's try: >>> Expand[{n, d/2 - n^2/(2 d), d/2 + n^2/(2 d)} /. {x_, y_, z_} :>x^2 + y^2 >>> - z^2] >>> Out[]=0 >>> Then, we can refine the idea of an "analog", >>> Pythagorean : Trithagorean :: >>> {n, d/2 - n^2/(2 d), d/2 + n^2/(2 d)} : {n, n*(n + 1)/d - (d + 2)/4, >>> n*(n + 1)/d + (d - 2)/4}. >>> The other relevant sequence is https://oeis.org/A309507, see >>> my message from Sat. Feb 22. --Brad >>> On Thu, Feb 27, 2020 at 3:54 PM Neil Sloane < njasloane@gmail.com> wrote: >>>> I mentioned this on Feb 19 2020 and several people sent very helpful >>>> replies. >>>> For squares, we have the theorem that all solutions to x^2+y^2=z^2 in >>>> integers are given by Pythagorean triples, t*[u^2-v^2, 2uv, u^2+v^2]. >>>> Expanding on the replies from Michael Collins and Rich, it looks like >>> the >>>> following is the analog for triangular numbers T_n = n(n+1)/2. >>>> ALL solutions to T_n+T_k=T_m are given by the list of what one might >>> call >>>> Trithagorean triples: these are the triples >>>> [n,k,m] = [n, T/Q-(Q+1)/2,T/Q+(Q-1)/2] where n >= 2, T=n(n+1)/2, and Q >>> is >>>> any odd divisor of T less than n, >>>> plus these triples with the first two coordinates swapped. >>>> I haven't found this in the literature, but it can hardly be new. >>>> _______________________________________________ >>>> math-fun mailing list >>>> math-fun@mailman.xmission.com >>>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>> _______________________________________________ >>> math-fun mailing list >>> math-fun@mailman.xmission.com >>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > _______________________________________________ > math-fun mailing list > math-fun@mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Hello Math-Fun, I would like to transmit two _composite_ numbers A and B in the form of single number C. How can I do that? For instance, if Alice concatenates A et B, Bob will have a problem in reconstructing those numbers out of C: A = 865445 B = 12377007 C = 86544512377007 Is there a clever, economic way to say where C must be split? Or do you know another technique (dealing perhaps with prime numbers)? This has nothing to do with cryptography (I guess) -- just a (silly) question. Best, É.
Interleave bits. On Tue, Mar 10, 2020 at 10:30 AM Éric Angelini <bk263401@skynet.be> wrote:
Hello Math-Fun, I would like to transmit two _composite_ numbers A and B in the form of single number C. How can I do that?
For instance, if Alice concatenates A et B, Bob will have a problem in reconstructing those numbers out of C: A = 865445 B = 12377007 C = 86544512377007
Is there a clever, economic way to say where C must be split? Or do you know another technique (dealing perhaps with prime numbers)? This has nothing to do with cryptography (I guess) -- just a (silly) question. Best, É.
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-- -- http://cube20.org/ -- http://golly.sf.net/ --
Nice -- I've just googled yr answer, Tom, and seen how it works, thank you!
Interleave bits.
-----------------
Hello Math-Fun, I would like to transmit two _composite_ numbers A and B in the form of single number C. How can I do that?
For instance, if Alice concatenates A et B, Bob will have a problem in reconstructing those numbers out of C: A = 865445 B = 12377007 C = 86544512377007
Is there a clever, economic way to say where C must be split? Or do you know another technique (dealing perhaps with prime numbers)? This has nothing to do with cryptography (I guess) -- just a (silly) question. Best, É.
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Ah, I remember having read somewhere that I could interleave the digits of A and B, filling the missing spaces with 0's: A = 865445 B = 12377007 A = 00865445 B = 12377007 I'll start the interleave with the longest term (in this case B) to avoid a leading 0 in C: C = 1020387875040475 Correct? Best, É
You'll probably want the Cantor pairing function: https://en.wikipedia.org/wiki/Pairing_function -- APG.
Sent: Tuesday, March 10, 2020 at 6:30 PM From: "Éric Angelini" <bk263401@skynet.be> To: math-fun <math-fun@mailman.xmission.com> Subject: [math-fun] A simple naive question
Hello Math-Fun, I would like to transmit two _composite_ numbers A and B in the form of single number C. How can I do that?
For instance, if Alice concatenates A et B, Bob will have a problem in reconstructing those numbers out of C: A = 865445 B = 12377007 C = 86544512377007
Is there a clever, economic way to say where C must be split? Or do you know another technique (dealing perhaps with prime numbers)? This has nothing to do with cryptography (I guess) -- just a (silly) question. Best, É.
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If you don't mind big numbers, encode A,B as C = (2^A)*(3^B). On Tue, Mar 10, 2020 at 1:30 PM Éric Angelini <bk263401@skynet.be> wrote:
Hello Math-Fun, I would like to transmit two _composite_ numbers A and B in the form of single number C. How can I do that?
For instance, if Alice concatenates A et B, Bob will have a problem in reconstructing those numbers out of C: A = 865445 B = 12377007 C = 86544512377007
Is there a clever, economic way to say where C must be split? Or do you know another technique (dealing perhaps with prime numbers)? This has nothing to do with cryptography (I guess) -- just a (silly) question. Best, É.
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participants (8)
-
Adam P. Goucher -
Brad Klee -
Neil Sloane -
Olivier Gerard -
Tomas Rokicki -
Victor Miller -
W. Edwin Clark -
Éric Angelini