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a(n) = Product_{k=1..n} (2*k*(k-1)+1).
1

%I #27 Sep 10 2023 18:09:04

%S 1,5,65,1625,66625,4064125,345450625,39035920625,5660208490625,

%T 1024497736803125,226413999833490625,59999709955875015625,

%U 18779909216188879890625,6854666863908941160078125,2885814749705664228392890625,1388076894608424493856980390625,756501907561591349152054312890625

%N a(n) = Product_{k=1..n} (2*k*(k-1)+1).

%C Fang (see link) proves that a(n) is never a square for n > 1.

%H Jin-Hui Fang, <a href="https://www.emis.de/journals/INTEGERS/papers/j16/j16.Abstract.html">Neither Product{k=1..n} (4k^2+1) nor Product{k=1..n} (2k(k-1)+1) is a perfect square</a>, Integers, A16, Volume 9 (2009).

%F a(n) ~ cosh(Pi/2) * 2^(n+1) * n^(2*n) / exp(2*n). - _Vaclav Kotesovec_, Oct 10 2016

%F a(n) = 2^n * |Gamma(1/2 + i/2 + n)|^2 * cosh(Pi/2)/Pi. - _Vladimir Reshetnikov_, Oct 11 2016

%F E.g.f.: 2F0((1-i)/2,(1+i)/2; ; 2*x). - _Benedict W. J. Irwin_, Oct 19 2016

%F a(n) = 2^(-1 + n)*Pochhammer(3/2 - i/2, -1 + n)*Pochhammer(3/2 + i/2, -1 + n), for n>=1. - _Antonio GraciĆ” Llorente_, Sep 10 2023

%t Table[Product[(2*k*(k-1)+1), {k,1,n}], {n, 1, 20}] (* _Vaclav Kotesovec_, Oct 10 2016 *)

%t Round@Table[2^n Abs[Gamma[1/2 + I/2 + n]]^2 Cosh[Pi/2]/Pi, {n, 1, 20}] (* _Vladimir Reshetnikov_, Oct 11 2016 *)

%t Rest@(CoefficientList[Series[HypergeometricPFQ[{1/2 - I/2, 1/2 + I/2}, {}, 2 x], {x, 0, 20}], x]*Range[0, 20]!) (* _Benedict W. J. Irwin_, Oct 19 2016 *)

%o (PARI) a(n) = prod(k=1, n, 2*k*(k-1)+1);

%Y Cf. A001844.

%K nonn

%O 1,2

%A _Michel Marcus_, Oct 10 2016