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%I #23 Jul 07 2017 13:27:47
%S 1,2,18,504,32760,4127760,895723920,308129028480,158070191610240,
%T 115391239875475200,115506631115350675200,153854832645647099366400,
%U 266015005644323834804505600,584700982406223788900303308800,1605004196705084300531332582656000
%N a(n) = Product_{k=0..n} (k^3+1).
%H G. C. Greubel, <a href="/A255433/b255433.txt">Table of n, a(n) for n = 0..150</a>
%H Erhan Gürela, Ali Ulas Özgür Kisisel, <a href="http://dx.doi.org/10.1016/j.jnt.2009.07.014">A note on the products (1^mu + 1)(2^mu + 1)···(n^mu + 1)</a>, Journal of Number Theory, Volume 130, Issue 1, January 2010, Pages 187-191.
%H Chuan Ze Niu, <a href="https://arxiv.org/abs/1612.08158">(1^3+1)(2^3+1)...(n^3+1) is not a cube</a>, arXiv:1612.08158 [math.NT], 2016.
%F a(n) ~ 2*sqrt(2*Pi) * cosh(sqrt(3)*Pi/2) * n^(3*n + 3/2) / exp(3*n).
%F a(n) = 2*A158621(n). - _Vaclav Kotesovec_, Jul 11 2015
%t Table[Product[k^3 + 1, {k, 0, n}], {n, 0, 20}]
%t FullSimplify[Table[(Cosh[Sqrt[3]*Pi/2] * Gamma[2+n] * Gamma[1/2 - I*Sqrt[3]/2 + n] * Gamma[1/2 + I*Sqrt[3]/2 + n])/Pi, {n, 0, 20}]]
%t FoldList[Times,Range[0,20]^3+1] (* _Harvey P. Dale_, Jul 07 2017 *)
%o (PARI) a(n) = prod(k=1, n, 1+k^3); \\ _Michel Marcus_, Jan 25 2016
%Y Cf. A101686, A158621, A255434, A255435.
%K nonn,easy
%O 0,2
%A _Vaclav Kotesovec_, Feb 23 2015