A348088 a(n) = [x^n] Product_{k=1..n} 1/(1 - (2*k-1)^2 * x).
1, 1, 91, 24970, 14057043, 13444400190, 19558289594910, 40250341173506100, 111335096965772406915, 398473840263173643939190, 1791905773077609090895008106, 9890754761467721759394797416396, 65747198205879568307026776928408110
Offset: 0
Keywords
Programs
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Mathematica
Table[SeriesCoefficient[Product[1/(1 - (2*k-1)^2*x), {k, 1, n}], {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Oct 16 2021 *) -
PARI
a(n) = polcoef(1/prod(k=1, n, 1-(2*k-1)^2*x+x*O(x^n)), n);
Formula
From Vaclav Kotesovec, Oct 16 2021, updated May 16 2025: (Start)
a(n) ~ c * d^n * n!^2 / n^(3/2), where d = 52.447924272991536496097233490380538810534457762204101802471270109895148... and c = 0.028365099209561232079163758339093959048662789595134609351298413762...
In closed form, a(n) ~ 2^(2*n) * exp(2*n) * r^(n*r + 1/2) * (1+r)^(4*n) * n^(2*n - 1/2) / (sqrt(Pi*(1 - r*(2+r))) * (2+r)^((2+r)*n - 1/2)), where r = 0.044382033760833484984013906344747760869028157215190550759633... is the root of the equation exp(4/(1+r)) = (1 + 2/r). (End)
a(n) = A381512(n,2*n-1) = (1/(2^(2*n-2)*(2*n-1)!)) * Sum_{k=0..n-1} (-1)^k * (2*n-1-2*k)^(4*n-1) * binomial(2*n-1,k) for n > 0. - Seiichi Manyama, May 16 2025