Wikipedia:Reference desk/Archives/Mathematics/2024 March 16

= March 16 =

Smallest triangular number with prime signature the same as A025487(n)
Smallest triangular number with prime signature the same as (n), or 0 if no such number exists. (there is a similar sequence in OEIS)

For n = 1 through n = 29, this sequence is: (I have not confirmed the n = 15 term is 0, but it seems that it is 0, i.e. it seems that there is no triangular number of the form p^3*q^2 with p, q both primes)

1, 3, 0, 6, 0, 28, 0, 136, 66, 0, 36, 496, 276, 0, 0, 118341, 120, 0, 1631432881, 300, 8128, 210, 0, 528, 0, 29403, 1176, 32896, 630

Is it possible to extend this sequence to n = 100? 1.165.194.85 (talk) 00:51, 16 March 2024 (UTC)


 * This paper by Cohn implies that there are no prime solutions $$(p,q)$$ to $$p^{3} + 1 = 2q^{2}$$ and $$2p^{2} + 1 = q^{3}$$. If anyone here can prove that the elliptic curves $$y^{2} = x^{3} \pm 4$$ have no nontrivial integral points (for $$-4$$ that would be $$(2, \pm 2)$$ and $$(5, \pm 11)$$, while for $$4$$ that would be $$(0, \pm 2)$$), then that implies that there are no triangular numbers of the form $$p^{3}q^{2}$$ at all. GalacticShoe (talk) 06:22, 16 March 2024 (UTC)
 * I'm not sure what you mean by "at all". If you drop the requirement that q is prime then there are many solutions, for example 242⋅243/2 = 33332, 12167⋅12168/2 = 233782. --RDBury (talk) 08:42, 16 March 2024 (UTC)
 * Sorry, shoulda clarified; no triangular numbers of the form $$p^{3}q^{2}$$ at all where $$p, q$$ are prime. Naturally, the two cases mentioned earlier (the latter example of which I hastily posted before remembering that I completely forgot about the prime signature part) are either not covered by Cohn's paper, with the $$242$$ case being invalid since we spread powers of $$3$$ among the cube and the square, or they are one of Cohn's special case, as with $$12167$$. GalacticShoe (talk) 09:51, 16 March 2024 (UTC)
 * Thanks for the clarification. I don't have access to the Cohn paper, but it seems to me that the p3+1=2q2 is trivial to exclude, and similarly for p3-1=2q2. That leaves 2p3±1=q2. Using brute force I found 2⋅233+2=1562 but none, even when p is only required to be odd and q can be anything, where the difference is one. Note that this generates the 12167 example, but that's not the way I found it. I'm not sure how y2=x3±4 is related, but I'm no expert on elliptic curves so I may have to take your word on that. In any case, given that the p3q2 is proving so difficult, it seems unlikely that getting to n=100 is feasible. Finding entries where there is a solution should only require a computer search, but if none are found then proving that there are none to be found can be difficult. It seems ironic that while there is apparently no p3q2 solution, there is a p5q2, though this seems less likely at first glance. RDBury (talk) 18:24, 16 March 2024 (UTC)
 * The $$y^{2} = x^{3} \pm 4$$ condition arises because, if we write $$x = 2p$$ and $$y = 2q$$, then an integer solution to $$y^{2} = x^{3} \pm 4$$ implies $$(2q)^{2} = (2p)^{3} \pm 4 \Rightarrow 4q^{2} = 8p^{3} \pm 4 \Rightarrow q^{2} = 2p^{3} \pm 1$$ is a rational, possibly integer solution to the equalities we are looking at. All integer solutions to $$q^{2} = 2p^{3} \pm 1$$ can be generated this way, so the lack of a nontrivial integer solution to $$y^{2} = x^{3} \pm 4$$ (or the existence only of solutions where $$x, y$$ are odd) would rule out any further triangular numbers of the form $$p^{2}q^{3}$$ with $$p, q$$ prime. GalacticShoe (talk) 21:33, 16 March 2024 (UTC)
 * Right, I should have seen that. The 2p3+1=q2 case may be easier than I thought; it reduces to 2p3=(q+1)(q-1). You can rule out p=2, q=2, giving q odd, but then by unique factorization q±1 divides p, which is impossible. I thought of applying a similar trick to 2p3-1=q2, or 2p3=(q+i)(q-i), and using unique factorization over Gaussian integers. But I quickly got bogged down with this. --RDBury (talk) 02:10, 17 March 2024 (UTC)
 * A similar disproof is that after proving $$p = 2$$ and $$q = 2$$ impossible by simple casework, $$q$$ odd implies that $$(q+1)(q-1)$$ is a multiple of $$4$$, which $$2p^{3}$$ cannot be. I imagine that $$2p^{3} - 1 = q^{2}$$ is indeed not as simple to pin down. GalacticShoe (talk) 02:45, 17 March 2024 (UTC)
 * So if $$y^2=x^3 \pm 4$$ has no integer solution other than $$(0,\pm 2)$$, $$(2,\pm 2)$$, $$(5,\pm 11)$$, than a(15) in the above sequence is 0? How about a(30), a(31), a(32), …? Do you have the status for a(n) for all $$n \le 100$$? has the proof for which some prime signatures do not exists for triangular numbers such as $$p^6q^2$$ (i.e. a(33) = 0), see the comment of Jinyuan Wang in Aug 22 2020. (You can also check whether a(1) ~ a(29) which I have listed above are correct) (a(n) is the smallest triangular number with prime signature the same as (n), or a(n) is 0 if no such number exists) 61.224.130.152 (talk) 03:21, 18 March 2024 (UTC)
 * I mentioned this a few paragraphs back. Given that finding the value of a single entry is turning out to be a project in itself, and that it seems unlikely that computations will get any easier going on, I don't see getting to 100 as feasible. A081978 is asking for less information, and even so it took some high powered number theory to get it's values. --RDBury (talk) 15:36, 18 March 2024 (UTC)
 * So which a(n) for $$n\le 100$$ are known to be 0? And which a(n) for $$n\le 100$$ have known nonzero values? I want the status for $$n \le 100$$. 203.73.106.200 (talk) 08:37, 25 March 2024 (UTC)
 * Now I use a PARI/GP program to compute, after the PARI/GP programs in and :
 * a(n)=my(f=vecsort(factor(n)[, 2],, 4), p); prod(i=1, #f, (p=nextprime(p+1))^f[i])
 * isA025487(n)=my(k=valuation(n, 2), t); n>>=k; forprime(p=3, default(primelimit), t=valuation(n, p); if(t>k, return(0), k=t); if(k, n/=p^k, return(n==1)))
 * b(n)=for(k=1,2^24,if(a(k*(k+1)/2)==n,return(k*(k+1)/2)))
 * for(k=1,2^12,if(isA025487(k),print(k, ",",b(k))))
 * The result is (∞ loop for a(n)=0, but some zeros a(n) are not confirmed, and instead only conjectured):
 * 1,1
 * 2,3
 * 4,0 (there is no triangular number which is a square of a prime)
 * 6,6
 * 8,0 (triangular number cannot be nth power if n>2)
 * 12,28
 * 16,0 (triangular number cannot be nth power if n>2)
 * 24,136
 * 30,66
 * 32,0 (triangular number cannot be nth power if n>2)
 * 36,36
 * 48,496
 * 60,276
 * 64,0 (triangular number cannot be nth power if n>2)
 * 72,0 (not confirmed, but must be > 10^15 if nonzero, see https://www.numbersaplenty.com/both_Achilles_and_triangular.html)
 * 96,118341
 * 120,120
 * 128,0 (triangular number cannot be nth power if n>2)
 * 144,1631432881
 * 180,300
 * 192,8128
 * 210,210
 * 216,0 (triangular number cannot be nth power if n>2)
 * 240,528
 * 256,0 (triangular number cannot be nth power if n>2)
 * 288,29403
 * 360,1176
 * 384,32896
 * 420,630
 * 432,0 (not confirmed, but must be > 10^15 if nonzero, see https://www.numbersaplenty.com/both_Achilles_and_triangular.html)
 * 480,2080
 * 512,0 (triangular number cannot be nth power if n>2)
 * 576,0 (see Jinyuan Wang proof on Aug 22 2020 in )
 * 720,209628
 * 768,86086881
 * 840,6216
 * 864,0 (not confirmed, but must be > 10^15 if nonzero, see https://www.numbersaplenty.com/both_Achilles_and_triangular.html)
 * 900,2172602007770041 (after )
 * 960,8256
 * 1024,0 (triangular number cannot be nth power if n>2)
 * 1080,93096
 * 1152,0 (not confirmed, but must be > 10^15 if nonzero, see https://www.numbersaplenty.com/both_Achilles_and_triangular.html)
 * 1260,4950
 * 1296,0 (triangular number cannot be nth power if n>2)
 * 1440,2016
 * 1536,774860661
 * 1680,4560
 * 1728,0 (triangular number cannot be nth power if n>2)
 * 1800,0 (not confirmed, but must be > 10^15 if nonzero, see https://www.numbersaplenty.com/both_Achilles_and_triangular.html)
 * 1920,295296
 * 2048,0 (triangular number cannot be nth power if n>2)
 * 2160,3240
 * 2304,0 (it seems that we can use the Jinyuan Wang proof on Aug 22 2020 for 576 in to prove that it is 0)
 * 2310,3570
 * 2520,9180
 * 2592,0 (not confirmed, but must be > 10^15 if nonzero, see https://www.numbersaplenty.com/both_Achilles_and_triangular.html)
 * 2880,662976
 * 3072,38009927549623740385753 (after )
 * 3360,51040
 * 3456,0 (not confirmed, but must be > 10^15 if nonzero, see https://www.numbersaplenty.com/both_Achilles_and_triangular.html)
 * 3600,41616
 * 3840,130816
 * 4096,0 (triangular number cannot be nth power if n>2)


 * 220.132.230.56 (talk) 14:12, 26 March 2024 (UTC)