A263687 b(n) in (sqrt(2))_n = b(n) + c(n)*sqrt(2), where (x)_n is the Pochhammer symbol, b(n) and c(n) are integers.
1, 0, 2, 6, 26, 140, 896, 6636, 55804, 525168, 5468008, 62403880, 774616696, 10390122288, 149757486368, 2308301709840, 37887797229968, 659770432834688, 12148923787132832, 235858218326093664, 4814800618608693664, 103104123746671427520, 2310978427407268450048
Offset: 0
Keywords
Examples
For n = 4, (sqrt(2))_4 = sqrt(2)*(1 + sqrt(2))*(2 + sqrt(2))*(3 + sqrt(2)) = 26 + 18*sqrt(2), so a(4) = 26. G.f. = 1 + 2*x^2 + 6*x^3 + 26*x^4 + 140*x^5 + 896*x^6 + 6636*x^7 + 55804*x^8 + ...
Links
- G. C. Greubel, Table of n, a(n) for n = 0..445
- Eric Weisstein's World of Mathematics, Pochhammer Symbol.
Crossrefs
Cf. A263688.
Programs
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Mathematica
Expand@Table[(Pochhammer[Sqrt[2], n] + Pochhammer[-Sqrt[2], n])/2, {n, 0, 22}]
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PARI
{a(n) = if( n<0, 0, real(prod(k=0, n-1, quadgen(8) + k)))}; /* Michael Somos, Oct 23 2015 */
Formula
a(n) = ((sqrt(2))_n + (-sqrt(2))_n)/2.
E.g.f.: (1/(1-x)^sqrt(2)+(1-x)^sqrt(2))/2 = cosh(sqrt(2)*log(1-x)).
Recurrence: a(0) = 1, a(1) = 0, a(n+2) = (2*n+1)*a(n+1) + (2-n^2)*a(n).
a(n) ~ exp(-n)*n^(n+sqrt(2)-1/2)*sqrt(Pi/2)/Gamma(sqrt(2)).
0 = a(n)*(+7*a(n+1) - a(n+2) - 6*a(n+3) + a(n+4)) + a(n+1)*(+7*a(n+1) + 6*a(n+2) - 4*a(n+3)) + a(n+2)*(+3*a(n+2)) for all n>=0. - Michael Somos, Oct 23 2015
From Benedict W. J. Irwin, Oct 14 2016: (Start)
a(n) = (-1)^n*(binomial(-sqrt(2), n) + binomial(sqrt(2), n))*n!/2.
Conjecture: a(n) = (-1)^n * Sum_{k=0..n/2} Stirling1(n,2*k)*2^k.
(End)
Comments