[math-fun] A016142
This (also) appears to be 3^(n-1) * (3^n - 1)/2. VERY WILD surmise: It's the number of integer-sided Heron triangles whose circumdiameter is the product of n distinct primes of shape 4k + 1. (3^n - 1)/2 of these are (believed to be) right triangles, so that the number of non-right ones is (3^(n-1) - 1) * (3^n - 1)/2 A003462 and a new(?) sequence 0, 8, 104, 1040, 9680, ... R.
Warning -- I believe I've overestimated the number of Heron triangles. I was assuming that if c^2 = x^2 + y^2, where c is the circumdiameter, then all pairs of (x,y)s, considered as arctans of halves of angles subtended by sides of triangles at the circumcentre, generate integer sides. But if both (x1,y1) and (x2,y2) are primitive, then you get rational sides, but not necessarily integer ones. More later, when I've thought about it. Or perhaps someone will step in with the correct formula? Best, R. On Sat, 13 Aug 2005, Richard Guy wrote:
This (also) appears to be 3^(n-1) * (3^n - 1)/2.
VERY WILD surmise: It's the number of integer-sided Heron triangles whose circumdiameter is the product of n distinct primes of shape 4k + 1.
(3^n - 1)/2 of these are (believed to be) right triangles, so that the number of non-right ones is
(3^(n-1) - 1) * (3^n - 1)/2
A003462 and a new(?) sequence 0, 8, 104, 1040, 9680, ...
R.
Alex Fink & I are now able to give the right answers. The number of incongruent integer-edged Heron triangles whose circumdiameter is the product of n distinct primes each of shape 4k + 1 is (2*7^n - 3*3^n + 1)/6 This is A016161 in OEIS, tho this fact is not mentioned there. The number of these which are right triangles is (3^n - 1)/2, as stated before, and this is A003462. The number which are not right-angled is thus (2*7^n - 6*3^n + 4)/6 or 0, 8, 88, 720, 5360, ... or eight times A016212. R + Alex Fink. On Sun, 14 Aug 2005, Richard Guy wrote:
Warning -- I believe I've overestimated the number of Heron triangles. I was assuming that if c^2 = x^2 + y^2, where c is the circumdiameter, then all pairs of (x,y)s, considered as arctans of halves of angles subtended by sides of triangles at the circumcentre, generate integer sides. But if both (x1,y1) and (x2,y2) are primitive, then you get rational sides, but not necessarily integer ones. More later, when I've thought about it. Or perhaps someone will step in with the correct formula?
Best, R.
----- Original Message ----- From: "Richard Guy" <rkg@cpsc.ucalgary.ca> To: "math-fun" <math-fun@mailman.xmission.com> Cc: "Alex Fink" <finka@math.ucalgary.ca>; <seqfan@ext.jussieu.fr> Sent: Monday, August 15, 2005 2:51 PM Subject: Re: [math-fun] Re: A016142
Alex Fink & I are now able to give the right answers. The number of incongruent integer-edged Heron triangles whose circumdiameter is the product of n distinct primes each of shape 4k + 1 is (2*7^n - 3*3^n + 1)/6
This is A016161 in OEIS.
Not true. Compute the sequence.
Sorry about that. Can't now trace how I made the mistake. Evidently 0, 1, 12, 101, 760, 5481, 38839, ... is a new sequence. Formula (2*7^n -3*3^4 + 1)/6 Characteristic polynomial (x-7)(x-3)(x-1) Manifestation: number of incongruent integer-edged Heron triangles whose circumdiameter is the product of n distinct primes of shape 4k + 1. I believe that my remarks about the number of such triangles that are not right-angled are still correct, viz. (2*7^n - 6*3^n + 4)/6 Same recurrence. 0, 0, 8, 88, 720, 5360, 38488, 272328, ... Eight times A016212. Not in OEIS per se. The number of nondegenerate right-angled such triangles is A003462, tho that fact is not noted there. Originators: Alex Fink & Richard Guy Will someone do the necessary? Thanks! On Wed, 17 Aug 2005, David Wilson wrote:
----- Original Message ----- From: "Richard Guy" <rkg@cpsc.ucalgary.ca> To: "math-fun" <math-fun@mailman.xmission.com> Cc: "Alex Fink" <finka@math.ucalgary.ca>; <seqfan@ext.jussieu.fr> Sent: Monday, August 15, 2005 2:51 PM Subject: Re: [math-fun] Re: A016142
Alex Fink & I are now able to give the right answers. The number of incongruent integer-edged Heron triangles whose circumdiameter is the product of n distinct primes each of shape 4k + 1 is (2*7^n - 3*3^n + 1)/6
This is A016161 in OEIS.
Not true. Compute the sequence.
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I recently posed a question on arranging 10 queens to make a Petersen Graph. It was solved on the 5x6 board by Daniele Degiorgi. In what is sure to become a beautiful defining example, Geoff Exoo arranged 14 queens on a 7x7 board to make the Heawood Graph. It can also be seen as a Fano Plane. For those that wish to tackle this problem, arrange seven black and seven white queens on a 7x7 board such that for every like-colored pair of queens, there is exactly one queen of the other color that attacks both. Also, no queen attacks another of the same color. Answer here: http://ginger.indstate.edu/ge/COMBIN/QUEENS/index.html A different problem: the smallest solution for N^2 queens in an NxNxN cube is N=11. A solution is at the following link: http://www.csse.monash.edu.au/~lloyd/tildeAlgDS/Recn/Queens3D/ --Ed Pegg Jr
participants (3)
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David Wilson -
ed pegg -
Richard Guy