Re: [math-fun] 4-5-6
Man, that Hirschhorn is dangerous! billgosper$ python rattri.py 5 3 | head 2098817 2016840 220597 In[524]:= ratang[823543, 1024303, 220597] Out[524]= {(\[Pi] - ArcCos[1461083/1647086])/(\[Pi] - ArcCos[-(11472481/11529602)]), (\[Pi] - ArcCos[-(11472481/11529602)])/(\[Pi] - ArcCos[-(13/14)]), (\[Pi] - ArcCos[-(13/14)])/(\[Pi] - ArcCos[1461083/1647086])} In[525]:= N@% Out[525]= {26.7290445719883, 0.261887400469822, 0.142857142857143} In[526]:= FI /@ {823543, 1024303, 220597} Out[526]= {7^7,7 41 43 83,13 71 239} Anecdote: The only time I spoke with Erdös was at the Ramanujan Centenary at UIUC. I told him my conjecture that any sum of four distinct odd squares was the sum of four distinct even squares. After about 30 sec, he said "This is not my problem." Later I repeated the conjecture to Hirschhorn, who handed me a handwritten algebraic proof the next day. --rwg On 2018-07-16 10:03, Tomas Rokicki wrote:
Here's a python script that generates integer triangles with angles in any rational ratio (pass the numerator and denominator on the command line).
I'm sure someone skilled in Python can write this much more simply.
Given this, the only thing we need to show is that these are the only such integer triangles.
Sample usage:
abacus:inttri rokicki$ python rattri.py 1 6 | head
1771561 2271060 566839
46656 72930 30421
4826809 8761896 4632263
117649 240240 146329
11390625 25591020 17151679
262144 637560 458081
# # Given two integers h and k, find integer triangles with angles in the # ratio h:k. Based on the generic parameterization given here: # #
https://en.wikipedia.org/wiki/Integer_triangle#Integer_triangles_with_one_an...
# import sys from fractions import gcd import math # h = int(sys.argv[1]) ; k = int(sys.argv[2]) ; q = 1 coscrit = math.cos(math.pi/(h+k)) def choose(a, b): if b < 0 or b > a: return 0 if b + b > a: b = a - b r = 1 for i in range(1, b+1): r = r * (a - i + 1) / i return r ; while True: for p in range(1, q): if q*coscrit >= p: continue if gcd(p, q) > 1: continue a = 0 b = 0 c = 0 for i in range(0, (h+1)/2): a += ((-1)**i)*choose(h, 2*i+1)*(p**(h-2*i-1))*((q*q-p*p)**i) for i in range(0, (k+1)/2): b += ((-1)**i)*choose(k, 2*i+1)*(p**(k-2*i-1))*((q*q-p*p)**i) for i in range(0, (h+k+1)/2): c += ((-1)**i)*choose(h+k, 2*i+1)*(p**(h+k-2*i-1))*((q*q-p*p)**i) a *= q**k b *= q**h g = gcd(gcd(a, b), c) a /= g b /= g c /= g print a, b, c q = q + 1
On Sun, Jul 15, 2018 at 2:55 PM Bill Gosper <billgosper@gmail.com> wrote:
Wow. I haven't even digested WDS's thing. --Bill Maybe it's time for MathOverflow, e.g.?
On Sun, Jul 15, 2018 at 1:36 PM Tomas Rokicki <rokicki@gmail.com> wrote:
Any more progress on this?
I have a large number of additional examples if you are interested, such as this triple for 6:
117649 146329 240240
and these for 5:
231 1024 1220
3024 3125 5555
7776 12155 17214
16807 34320 42833
36400 59049 89001
32768 79695 92168
59049 162656 178695
100000 302499 320210
161051 206460 323851
I have not yet tried to find recurrences for these . . .
It's abundantly clear empirically that your hypothesis works both ways (with suitable caveats). There should be a way to prove this.
-tom
On Sat, Jul 14, 2018 at 4:14 PM Bill Gosper <billgosper@gmail.com> wrote:
OMG! Nick Baxter somehow found 1024 | 1220 | 231 -> {23.09066960680706, 0.2165376788608164, 1/5}, 46656 | 72930 | 30421 -> {9.357241092137794, 0.6412146423203051, 1/6}
23.1--a toothpick! Even more toothpicular is 357 | 343 | 20 -> {18.99091913441869, 0.01755224857596323, 3}, because it's 1/19 of a smaller angle.
[Massive chop!]
On my few encounters with Erdos, I got the sense that his engagement with other people’s questions had a much more precipitous drop-off than most mathematicians’. He either found a problem interesting or he didn’t, whereas most people admit subtler gradations of interest. Also, if I recall correctly, he was bored by nearly all of enumerative combinatorics. I suspect that hypergeometric functions bored him too; I’m not surprised that, upon hearing Bill’s question, he quickly convinced himself it required tools outside his wheelhouse and decided to leave the problem to other people who were more adept in using those tools. I don’t really blame him. In a world with finite mental resources, isn’t it in some sense a waste for person A to tackle a problem that person B could solve faster, especially if at the same time person B is tackling a problem that person A could solve faster? Of course I don’t actually look at math research as a thousand races to a thousand finish lines, but it is one aspect of the enterprise. Part of success in research is figuring out what you’re good at. Come to think of it, this relates to my earlier facetious post about finding the thing, or a combination of things, that you’re the best at. I once showed Bill Thurston a (Euclidean) geometry problem that he seemed interested in until he was unable to solve it using hyperbolic geometry, at which point he lost interest. He wasn’t interested in finding just any solution to the problem; he wanted to find a Thurston-style solution. And why not? Why shouldn’t he focus on creating the kinds of beauty that he was uniquely equipped to create, and let others do the same in their own metiers? If there is a Supreme Fascist, and Erdos meets Him, the S.F. won’t say “Why weren’t you Euler?”; he’ll say “Why weren’t you Erdos? ... Oh wait, never mind. You WERE Erdos. In fact, you were EXTREMELY Erdos.” Jim Propp On Thursday, July 19, 2018, Bill Gosper <billgosper@gmail.com> wrote:
Man, that Hirschhorn is dangerous!
billgosper$ python rattri.py 5 3 | head
2098817 2016840 220597
In[524]:= ratang[823543, 1024303, 220597]
Out[524]= {(\[Pi] - ArcCos[1461083/1647086])/(\[Pi] -
ArcCos[-(11472481/11529602)]), (\[Pi] -
ArcCos[-(11472481/11529602)])/(\[Pi] - ArcCos[-(13/14)]), (\[Pi] -
ArcCos[-(13/14)])/(\[Pi] - ArcCos[1461083/1647086])}
In[525]:= N@%
Out[525]= {26.7290445719883, 0.261887400469822, 0.142857142857143}
In[526]:= FI /@ {823543, 1024303, 220597}
Out[526]= {7^7,7 41 43 83,13 71 239}
Anecdote: The only time I spoke with Erdös was at the Ramanujan Centenary
at UIUC. I told him my conjecture that any sum of four distinct odd squares was
the sum of four distinct even squares. After about 30 sec, he said "This is not
my problem." Later I repeated the conjecture to Hirschhorn, who handed me
a handwritten algebraic proof the next day.
--rwg
On 2018-07-16 10:03, Tomas Rokicki wrote:
Here's a python script that generates integer triangles with angles in any rational ratio (pass the numerator and denominator on the command line).
I'm sure someone skilled in Python can write this much more simply.
Given this, the only thing we need to show is that these are the only such integer triangles.
Sample usage:
abacus:inttri rokicki$ python rattri.py 1 6 | head
1771561 2271060 566839
46656 72930 30421
4826809 8761896 4632263
117649 240240 146329
11390625 25591020 17151679
262144 637560 458081
# # Given two integers h and k, find integer triangles with angles in the # ratio h:k. Based on the generic parameterization given here: # #
https://en.wikipedia.org/wiki/Integer_triangle#Integer_ triangles_with_one_angle_equal_to_an_arbitrary_ rational_number_times_another_angle
# import sys from fractions import gcd import math # h = int(sys.argv[1]) ; k = int(sys.argv[2]) ; q = 1 coscrit = math.cos(math.pi/(h+k)) def choose(a, b): if b < 0 or b > a: return 0 if b + b > a: b = a - b r = 1 for i in range(1, b+1): r = r * (a - i + 1) / i return r ; while True: for p in range(1, q): if q*coscrit >= p: continue if gcd(p, q) > 1: continue a = 0 b = 0 c = 0 for i in range(0, (h+1)/2): a += ((-1)**i)*choose(h, 2*i+1)*(p**(h-2*i-1))*((q*q-p*p)**i) for i in range(0, (k+1)/2): b += ((-1)**i)*choose(k, 2*i+1)*(p**(k-2*i-1))*((q*q-p*p)**i) for i in range(0, (h+k+1)/2): c += ((-1)**i)*choose(h+k, 2*i+1)*(p**(h+k-2*i-1))*((q*q- p*p)**i) a *= q**k b *= q**h g = gcd(gcd(a, b), c) a /= g b /= g c /= g print a, b, c q = q + 1
On Sun, Jul 15, 2018 at 2:55 PM Bill Gosper <billgosper@gmail.com> wrote:
Wow. I haven't even digested WDS's thing. --Bill Maybe it's time for MathOverflow, e.g.?
On Sun, Jul 15, 2018 at 1:36 PM Tomas Rokicki <rokicki@gmail.com> wrote:
Any more progress on this?
I have a large number of additional examples if you are interested, such as this triple for 6:
117649 146329 240240
and these for 5:
231 1024 1220
3024 3125 5555
7776 12155 17214
16807 34320 42833
36400 59049 89001
32768 79695 92168
59049 162656 178695
100000 302499 320210
161051 206460 323851
I have not yet tried to find recurrences for these . . .
It's abundantly clear empirically that your hypothesis works both ways (with suitable caveats). There should be a way to prove this.
-tom
On Sat, Jul 14, 2018 at 4:14 PM Bill Gosper <billgosper@gmail.com> wrote:
OMG! Nick Baxter somehow found 1024 | 1220 | 231 -> {23.09066960680706, 0.2165376788608164, 1/5}, 46656 | 72930 | 30421 -> {9.357241092137794, 0.6412146423203051, 1/6}
23.1--a toothpick! Even more toothpicular is 357 | 343 | 20 -> {18.99091913441869, 0.01755224857596323, 3}, because it's 1/19 of a smaller angle.
[Massive chop!]
math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Bill, "Sums of four distinct odd squares" seemed to be missing from OEIS so I created A316833 - could you check? Was Mike H.'s proof ever published? 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, Jul 19, 2018 at 6:55 AM, James Propp <jamespropp@gmail.com> wrote:
On my few encounters with Erdos, I got the sense that his engagement with other people’s questions had a much more precipitous drop-off than most mathematicians’. He either found a problem interesting or he didn’t, whereas most people admit subtler gradations of interest. Also, if I recall correctly, he was bored by nearly all of enumerative combinatorics. I suspect that hypergeometric functions bored him too; I’m not surprised that, upon hearing Bill’s question, he quickly convinced himself it required tools outside his wheelhouse and decided to leave the problem to other people who were more adept in using those tools.
I don’t really blame him. In a world with finite mental resources, isn’t it in some sense a waste for person A to tackle a problem that person B could solve faster, especially if at the same time person B is tackling a problem that person A could solve faster? Of course I don’t actually look at math research as a thousand races to a thousand finish lines, but it is one aspect of the enterprise. Part of success in research is figuring out what you’re good at.
Come to think of it, this relates to my earlier facetious post about finding the thing, or a combination of things, that you’re the best at. I once showed Bill Thurston a (Euclidean) geometry problem that he seemed interested in until he was unable to solve it using hyperbolic geometry, at which point he lost interest. He wasn’t interested in finding just any solution to the problem; he wanted to find a Thurston-style solution. And why not? Why shouldn’t he focus on creating the kinds of beauty that he was uniquely equipped to create, and let others do the same in their own metiers?
If there is a Supreme Fascist, and Erdos meets Him, the S.F. won’t say “Why weren’t you Euler?”; he’ll say “Why weren’t you Erdos? ... Oh wait, never mind. You WERE Erdos. In fact, you were EXTREMELY Erdos.”
Jim Propp
On Thursday, July 19, 2018, Bill Gosper <billgosper@gmail.com> wrote:
Man, that Hirschhorn is dangerous!
billgosper$ python rattri.py 5 3 | head
2098817 2016840 220597
In[524]:= ratang[823543, 1024303, 220597]
Out[524]= {(\[Pi] - ArcCos[1461083/1647086])/(\[Pi] -
ArcCos[-(11472481/11529602)]), (\[Pi] -
ArcCos[-(11472481/11529602)])/(\[Pi] - ArcCos[-(13/14)]), (\[Pi] -
ArcCos[-(13/14)])/(\[Pi] - ArcCos[1461083/1647086])}
In[525]:= N@%
Out[525]= {26.7290445719883, 0.261887400469822, 0.142857142857143}
In[526]:= FI /@ {823543, 1024303, 220597}
Out[526]= {7^7,7 41 43 83,13 71 239}
Anecdote: The only time I spoke with Erdös was at the Ramanujan Centenary
at UIUC. I told him my conjecture that any sum of four distinct odd squares was
the sum of four distinct even squares. After about 30 sec, he said "This is not
my problem." Later I repeated the conjecture to Hirschhorn, who handed me
a handwritten algebraic proof the next day.
--rwg
On 2018-07-16 10:03, Tomas Rokicki wrote:
Here's a python script that generates integer triangles with angles in any rational ratio (pass the numerator and denominator on the command line).
I'm sure someone skilled in Python can write this much more simply.
Given this, the only thing we need to show is that these are the only such integer triangles.
Sample usage:
abacus:inttri rokicki$ python rattri.py 1 6 | head
1771561 2271060 566839
46656 72930 30421
4826809 8761896 4632263
117649 240240 146329
11390625 25591020 17151679
262144 637560 458081
# # Given two integers h and k, find integer triangles with angles in the # ratio h:k. Based on the generic parameterization given here: # #
https://en.wikipedia.org/wiki/Integer_triangle#Integer_ triangles_with_one_angle_equal_to_an_arbitrary_ rational_number_times_another_angle
# import sys from fractions import gcd import math # h = int(sys.argv[1]) ; k = int(sys.argv[2]) ; q = 1 coscrit = math.cos(math.pi/(h+k)) def choose(a, b): if b < 0 or b > a: return 0 if b + b > a: b = a - b r = 1 for i in range(1, b+1): r = r * (a - i + 1) / i return r ; while True: for p in range(1, q): if q*coscrit >= p: continue if gcd(p, q) > 1: continue a = 0 b = 0 c = 0 for i in range(0, (h+1)/2): a += ((-1)**i)*choose(h, 2*i+1)*(p**(h-2*i-1))*((q*q-p*p)**i) for i in range(0, (k+1)/2): b += ((-1)**i)*choose(k, 2*i+1)*(p**(k-2*i-1))*((q*q-p*p)**i) for i in range(0, (h+k+1)/2): c += ((-1)**i)*choose(h+k, 2*i+1)*(p**(h+k-2*i-1))*((q*q- p*p)**i) a *= q**k b *= q**h g = gcd(gcd(a, b), c) a /= g b /= g c /= g print a, b, c q = q + 1
On Sun, Jul 15, 2018 at 2:55 PM Bill Gosper <billgosper@gmail.com> wrote:
Wow. I haven't even digested WDS's thing. --Bill Maybe it's time for MathOverflow, e.g.?
On Sun, Jul 15, 2018 at 1:36 PM Tomas Rokicki <rokicki@gmail.com> wrote:
Any more progress on this?
I have a large number of additional examples if you are interested, such as this triple for 6:
117649 146329 240240
and these for 5:
231 1024 1220
3024 3125 5555
7776 12155 17214
16807 34320 42833
36400 59049 89001
32768 79695 92168
59049 162656 178695
100000 302499 320210
161051 206460 323851
I have not yet tried to find recurrences for these . . .
It's abundantly clear empirically that your hypothesis works both ways (with suitable caveats). There should be a way to prove this.
-tom
On Sat, Jul 14, 2018 at 4:14 PM Bill Gosper <billgosper@gmail.com> wrote:
OMG! Nick Baxter somehow found 1024 | 1220 | 231 -> {23.09066960680706, 0.2165376788608164, 1/5}, 46656 | 72930 | 30421 -> {9.357241092137794, 0.6412146423203051, 1/6}
23.1--a toothpick! Even more toothpicular is 357 | 343 | 20 -> {18.99091913441869, 0.01755224857596323, 3}, because it's 1/19 of a smaller angle.
[Massive chop!]
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
I don’t know about formally published, but the proof is discussed on the Mathworld site on square numbers, about halfway down, and references Michael Hirschhorn. It gives the identity (4a+1)^2+(4b+1)^2+(4c+1)^2+(4d+1)^2 = 4[ (a+b+c+d+1)^2 + (a-b-c+d)^2 + (a-b+c-d)^2 + (a+b-c-d)^2 ]. You can find this mentioned in his own book at https://books.google.com/books?id=60QwDwAAQBAJ&pg=PA295&lpg=PA295&dq=hirschh... Steve On Jul 19, 2018, at 10:57 AM, Neil Sloane <njasloane@gmail.com<mailto:njasloane@gmail.com>> wrote: Bill, "Sums of four distinct odd squares" seemed to be missing from OEIS so I created A316833 - could you check? Was Mike H.'s proof ever published? 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: https://urldefense.proofpoint.com/v2/url?u=http-3A__NeilSloane.com&d=DwIGaQ&... Email: njasloane@gmail.com<mailto:njasloane@gmail.com> On Thu, Jul 19, 2018 at 6:55 AM, James Propp <jamespropp@gmail.com> wrote: On my few encounters with Erdos, I got the sense that his engagement with other people’s questions had a much more precipitous drop-off than most mathematicians’. He either found a problem interesting or he didn’t, whereas most people admit subtler gradations of interest. Also, if I recall correctly, he was bored by nearly all of enumerative combinatorics. I suspect that hypergeometric functions bored him too; I’m not surprised that, upon hearing Bill’s question, he quickly convinced himself it required tools outside his wheelhouse and decided to leave the problem to other people who were more adept in using those tools. I don’t really blame him. In a world with finite mental resources, isn’t it in some sense a waste for person A to tackle a problem that person B could solve faster, especially if at the same time person B is tackling a problem that person A could solve faster? Of course I don’t actually look at math research as a thousand races to a thousand finish lines, but it is one aspect of the enterprise. Part of success in research is figuring out what you’re good at. Come to think of it, this relates to my earlier facetious post about finding the thing, or a combination of things, that you’re the best at. I once showed Bill Thurston a (Euclidean) geometry problem that he seemed interested in until he was unable to solve it using hyperbolic geometry, at which point he lost interest. He wasn’t interested in finding just any solution to the problem; he wanted to find a Thurston-style solution. And why not? Why shouldn’t he focus on creating the kinds of beauty that he was uniquely equipped to create, and let others do the same in their own metiers? If there is a Supreme Fascist, and Erdos meets Him, the S.F. won’t say “Why weren’t you Euler?”; he’ll say “Why weren’t you Erdos? ... Oh wait, never mind. You WERE Erdos. In fact, you were EXTREMELY Erdos.” Jim Propp On Thursday, July 19, 2018, Bill Gosper <billgosper@gmail.com> wrote: Man, that Hirschhorn is dangerous! billgosper$ python rattri.py 5 3 | head 2098817 2016840 220597 In[524]:= ratang[823543, 1024303, 220597] Out[524]= {(\[Pi] - ArcCos[1461083/1647086])/(\[Pi] - ArcCos[-(11472481/11529602)]), (\[Pi] - ArcCos[-(11472481/11529602)])/(\[Pi] - ArcCos[-(13/14)]), (\[Pi] - ArcCos[-(13/14)])/(\[Pi] - ArcCos[1461083/1647086])} In[525]:= N@% Out[525]= {26.7290445719883, 0.261887400469822, 0.142857142857143} In[526]:= FI /@ {823543, 1024303, 220597} Out[526]= {7^7,7 41 43 83,13 71 239} Anecdote: The only time I spoke with Erdös was at the Ramanujan Centenary at UIUC. I told him my conjecture that any sum of four distinct odd squares was the sum of four distinct even squares. After about 30 sec, he said "This is not my problem." Later I repeated the conjecture to Hirschhorn, who handed me a handwritten algebraic proof the next day. --rwg On 2018-07-16 10:03, Tomas Rokicki wrote: Here's a python script that generates integer triangles with angles in any rational ratio (pass the numerator and denominator on the command line). I'm sure someone skilled in Python can write this much more simply. Given this, the only thing we need to show is that these are the only such integer triangles. Sample usage: abacus:inttri rokicki$ python rattri.py 1 6 | head 1771561 2271060 566839 46656 72930 30421 4826809 8761896 4632263 117649 240240 146329 11390625 25591020 17151679 262144 637560 458081 # # Given two integers h and k, find integer triangles with angles in the # ratio h:k. Based on the generic parameterization given here: # # https://urldefense.proofpoint.com/v2/url?u=https-3A__en.wikipedia.org_wiki_I... triangles_with_one_angle_equal_to_an_arbitrary_ rational_number_times_another_angle # import sys from fractions import gcd import math # h = int(sys.argv[1]) ; k = int(sys.argv[2]) ; q = 1 coscrit = math.cos(math.pi/(h+k)) def choose(a, b): if b < 0 or b > a: return 0 if b + b > a: b = a - b r = 1 for i in range(1, b+1): r = r * (a - i + 1) / i return r ; while True: for p in range(1, q): if q*coscrit >= p: continue if gcd(p, q) > 1: continue a = 0 b = 0 c = 0 for i in range(0, (h+1)/2): a += ((-1)**i)*choose(h, 2*i+1)*(p**(h-2*i-1))*((q*q-p*p)**i) for i in range(0, (k+1)/2): b += ((-1)**i)*choose(k, 2*i+1)*(p**(k-2*i-1))*((q*q-p*p)**i) for i in range(0, (h+k+1)/2): c += ((-1)**i)*choose(h+k, 2*i+1)*(p**(h+k-2*i-1))*((q*q- p*p)**i) a *= q**k b *= q**h g = gcd(gcd(a, b), c) a /= g b /= g c /= g print a, b, c q = q + 1 On Sun, Jul 15, 2018 at 2:55 PM Bill Gosper <billgosper@gmail.com> wrote: Wow. I haven't even digested WDS's thing. --Bill Maybe it's time for MathOverflow, e.g.? On Sun, Jul 15, 2018 at 1:36 PM Tomas Rokicki <rokicki@gmail.com> wrote: Any more progress on this? I have a large number of additional examples if you are interested, such as this triple for 6: 117649 146329 240240 and these for 5: 231 1024 1220 3024 3125 5555 7776 12155 17214 16807 34320 42833 36400 59049 89001 32768 79695 92168 59049 162656 178695 100000 302499 320210 161051 206460 323851 I have not yet tried to find recurrences for these . . . It's abundantly clear empirically that your hypothesis works both ways (with suitable caveats). There should be a way to prove this. -tom On Sat, Jul 14, 2018 at 4:14 PM Bill Gosper <billgosper@gmail.com> wrote: OMG! Nick Baxter somehow found 1024 | 1220 | 231 -> {23.09066960680706, 0.2165376788608164, 1/5}, 46656 | 72930 | 30421 -> {9.357241092137794, 0.6412146423203051, 1/6} 23.1--a toothpick! Even more toothpicular is 357 | 343 | 20 -> {18.99091913441869, 0.01755224857596323, 3}, because it's 1/19 of a smaller angle. [Massive chop!] _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://urldefense.proofpoint.com/v2/url?u=https-3A__mailman.xmission.com_cg... _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://urldefense.proofpoint.com/v2/url?u=https-3A__mailman.xmission.com_cg... _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://urldefense.proofpoint.com/v2/url?u=https-3A__mailman.xmission.com_cg...
Steve, Thank you, that was very helpful, and I updated A316833. Michael Hirschhorn's book looks very interesting and judging by the excerpts that your link showed (look at page 295!) could lead to many other new sequences. If anyone would like to help entering them into the OEIS, please do! 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, Jul 19, 2018 at 11:26 AM, Lucas, Stephen K - lucassk < lucassk@jmu.edu> wrote:
I don’t know about formally published, but the proof is discussed on the Mathworld site on square numbers, about halfway down, and references Michael Hirschhorn. It gives the identity (4a+1)^2+(4b+1)^2+(4c+1)^2+(4d+1)^2 = 4[ (a+b+c+d+1)^2 + (a-b-c+d)^2 + (a-b+c-d)^2 + (a+b-c-d)^2 ].
You can find this mentioned in his own book at https://books.google.com/books?id=60QwDwAAQBAJ&pg= PA295&lpg=PA295&dq=hirschhorn+sum+four+distinct+odd+squares& source=bl&ots=P0bLaMkf63&sig=B1tCdmh6jDWeYBfza_arOJwvtgE&hl=en&sa=X&ved= 2ahUKEwiUrtmOuKvcAhWhct8KHTzfCe8Q6AEwB3oECAEQAQ#v=onepage&q= hirschhorn%20sum%20four%20distinct%20odd%20squares&f=false
Steve
On Jul 19, 2018, at 10:57 AM, Neil Sloane <njasloane@gmail.com<mailto:nj asloane@gmail.com>> wrote:
Bill, "Sums of four distinct odd squares" seemed to be missing from OEIS so I created A316833 - could you check? Was Mike H.'s proof ever published?
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: https://urldefense.proofpoint. com/v2/url?u=http-3A__NeilSloane.com&d=DwIGaQ&c= eLbWYnpnzycBCgmb7vCI4uqNEB9RSjOdn_5nBEmmeq0&r=vge6KOo90zMf7Wx14WFtiQ&m= KlvEAtFe4zKCr3Aql2E0mi4T73AzYJ8kLGVSN9IC010&s=zft- IZ1e4jFVTGQ3CXQOdJ4AHPEdQ1E2Vr39GfbCVSc&e= Email: njasloane@gmail.com<mailto:njasloane@gmail.com>
On Thu, Jul 19, 2018 at 6:55 AM, James Propp <jamespropp@gmail.com> wrote:
On my few encounters with Erdos, I got the sense that his engagement with other people’s questions had a much more precipitous drop-off than most mathematicians’. He either found a problem interesting or he didn’t, whereas most people admit subtler gradations of interest. Also, if I recall correctly, he was bored by nearly all of enumerative combinatorics. I suspect that hypergeometric functions bored him too; I’m not surprised that, upon hearing Bill’s question, he quickly convinced himself it required tools outside his wheelhouse and decided to leave the problem to other people who were more adept in using those tools.
I don’t really blame him. In a world with finite mental resources, isn’t it in some sense a waste for person A to tackle a problem that person B could solve faster, especially if at the same time person B is tackling a problem that person A could solve faster? Of course I don’t actually look at math research as a thousand races to a thousand finish lines, but it is one aspect of the enterprise. Part of success in research is figuring out what you’re good at.
Come to think of it, this relates to my earlier facetious post about finding the thing, or a combination of things, that you’re the best at. I once showed Bill Thurston a (Euclidean) geometry problem that he seemed interested in until he was unable to solve it using hyperbolic geometry, at which point he lost interest. He wasn’t interested in finding just any solution to the problem; he wanted to find a Thurston-style solution. And why not? Why shouldn’t he focus on creating the kinds of beauty that he was uniquely equipped to create, and let others do the same in their own metiers?
If there is a Supreme Fascist, and Erdos meets Him, the S.F. won’t say “Why weren’t you Euler?”; he’ll say “Why weren’t you Erdos? ... Oh wait, never mind. You WERE Erdos. In fact, you were EXTREMELY Erdos.”
Jim Propp
On Thursday, July 19, 2018, Bill Gosper <billgosper@gmail.com> wrote:
Man, that Hirschhorn is dangerous!
billgosper$ python rattri.py 5 3 | head
2098817 2016840 220597
In[524]:= ratang[823543, 1024303, 220597]
Out[524]= {(\[Pi] - ArcCos[1461083/1647086])/(\[Pi] -
ArcCos[-(11472481/11529602)]), (\[Pi] -
ArcCos[-(11472481/11529602)])/(\[Pi] - ArcCos[-(13/14)]), (\[Pi] -
ArcCos[-(13/14)])/(\[Pi] - ArcCos[1461083/1647086])}
In[525]:= N@%
Out[525]= {26.7290445719883, 0.261887400469822, 0.142857142857143}
In[526]:= FI /@ {823543, 1024303, 220597}
Out[526]= {7^7,7 41 43 83,13 71 239}
Anecdote: The only time I spoke with Erdös was at the Ramanujan Centenary
at UIUC. I told him my conjecture that any sum of four distinct odd squares was
the sum of four distinct even squares. After about 30 sec, he said "This is not
my problem." Later I repeated the conjecture to Hirschhorn, who handed me
a handwritten algebraic proof the next day.
--rwg
On 2018-07-16 10:03, Tomas Rokicki wrote: Here's a python script that generates integer triangles with angles in any rational ratio (pass the numerator and denominator on the command line).
I'm sure someone skilled in Python can write this much more simply.
Given this, the only thing we need to show is that these are the only such integer triangles.
Sample usage:
abacus:inttri rokicki$ python rattri.py 1 6 | head
1771561 2271060 566839
46656 72930 30421
4826809 8761896 4632263
117649 240240 146329
11390625 25591020 17151679
262144 637560 458081
# # Given two integers h and k, find integer triangles with angles in the # ratio h:k. Based on the generic parameterization given here: # #
https://urldefense.proofpoint.com/v2/url?u=https-3A__en. wikipedia.org_wiki_Integer-5Ftriangle-23Integer-5F&d=DwIGaQ&c= eLbWYnpnzycBCgmb7vCI4uqNEB9RSjOdn_5nBEmmeq0&r=vge6KOo90zMf7Wx14WFtiQ&m= KlvEAtFe4zKCr3Aql2E0mi4T73AzYJ8kLGVSN9IC010&s=hw_ IL80wLiOSox14LBwnUt5aSicU-M0_hUyosICK79s&e= triangles_with_one_angle_equal_to_an_arbitrary_ rational_number_times_another_angle # import sys from fractions import gcd import math # h = int(sys.argv[1]) ; k = int(sys.argv[2]) ; q = 1 coscrit = math.cos(math.pi/(h+k)) def choose(a, b): if b < 0 or b > a: return 0 if b + b > a: b = a - b r = 1 for i in range(1, b+1): r = r * (a - i + 1) / i return r ; while True: for p in range(1, q): if q*coscrit >= p: continue if gcd(p, q) > 1: continue a = 0 b = 0 c = 0 for i in range(0, (h+1)/2): a += ((-1)**i)*choose(h, 2*i+1)*(p**(h-2*i-1))*((q*q-p*p)**i) for i in range(0, (k+1)/2): b += ((-1)**i)*choose(k, 2*i+1)*(p**(k-2*i-1))*((q*q-p*p)**i) for i in range(0, (h+k+1)/2): c += ((-1)**i)*choose(h+k, 2*i+1)*(p**(h+k-2*i-1))*((q*q- p*p)**i) a *= q**k b *= q**h g = gcd(gcd(a, b), c) a /= g b /= g c /= g print a, b, c q = q + 1
On Sun, Jul 15, 2018 at 2:55 PM Bill Gosper <billgosper@gmail.com> wrote:
Wow. I haven't even digested WDS's thing. --Bill Maybe it's time for MathOverflow, e.g.?
On Sun, Jul 15, 2018 at 1:36 PM Tomas Rokicki <rokicki@gmail.com> wrote:
Any more progress on this?
I have a large number of additional examples if you are interested, such as this triple for 6:
117649 146329 240240
and these for 5:
231 1024 1220
3024 3125 5555
7776 12155 17214
16807 34320 42833
36400 59049 89001
32768 79695 92168
59049 162656 178695
100000 302499 320210
161051 206460 323851
I have not yet tried to find recurrences for these . . .
It's abundantly clear empirically that your hypothesis works both ways (with suitable caveats). There should be a way to prove this.
-tom
On Sat, Jul 14, 2018 at 4:14 PM Bill Gosper <billgosper@gmail.com> wrote:
OMG! Nick Baxter somehow found 1024 | 1220 | 231 -> {23.09066960680706, 0.2165376788608164, 1/5}, 46656 | 72930 | 30421 -> {9.357241092137794, 0.6412146423203051, 1/6}
23.1--a toothpick! Even more toothpicular is 357 | 343 | 20 -> {18.99091913441869, 0.01755224857596323, 3}, because it's 1/19 of a smaller angle.
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participants (4)
-
Bill Gosper -
James Propp -
Lucas, Stephen K - lucassk -
Neil Sloane