A290000 a(n) = Product_{k=1..n-1} (3^k + 1).

1, 1, 4, 40, 1120, 91840, 22408960, 16358540800, 35792487270400, 234870301468364800, 4623187014103292723200, 272999193182799435304960000, 48361261073946554365403054080000, 25701205307660304745058529866383360000, 40976048450930207702360695570691784048640000

Offset: 0

Author

Ilya Gutkovskiy, Jun 06 2020

Keywords

• nonn

Links

• G. C. Greubel, Table of n, a(n) for n = 0..50

Crossrefs

Cf. A027871, A034472, A047656, A119600, A132324, A323716.

Sequences of the form Product_{j=1..n-1} (m^j + 1): A000012 (m=0), A011782 (m=1), A028362 (m=2), this sequence (m=3), A309327 (m=4).

Programs

Magma

[n lt 3 select 1 else (&*[3^j +1: j in [1..n-1]]): n in [1..20]]; // G. C. Greubel, Feb 21 2021

Mathematica

Table[Product[3^k + 1, {k, 1, n - 1}], {n, 0, 14}]

PARI

a(n) = prod(k=1, n-1, 3^k + 1); \\ Michel Marcus, Jun 06 2020

Sage

from sage.combinat.q_analogues import q_pochhammer
[1]+[3^(binomial(n,2))*q_pochhammer(n-1, -1/3, 1/3) for n in (1..20)] # G. C. Greubel, Feb 21 2021

Formula

G.f. A(x) satisfies: A(x) = 1 + x * A(3*x) / (1 - x).

G.f.: Sum_{k>=0} 3^(k*(k - 1)/2) * x^k / Product_{j=0..k-1} (1 - 3^j*x).

a(0) = 1; a(n) = Sum_{k=0..n-1} 3^k * a(k).

a(n) ~ c * 3^(n*(n - 1)/2), where c = Product_{k>=1} (1 + 1/3^k) = 1.564934018567011537938849... = A132324.

a(n) = 3^(binomial(n+1,2))*(-1/3;1/3){n}, where (a;q){n} is the q-Pochhammer symbol. - G. C. Greubel, Feb 21 2021