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

0,3

Comments

The Pochhammer symbol (sqrt(2))_n = Gamma(n + sqrt(2))/Gamma(sqrt(2)) = sqrt(2)*(1 + sqrt(2))*(2 + sqrt(2))*...*(n - 1 + sqrt(2)).

(sqrt(2))_n = a(n) + A263688(n)*sqrt(2).

Links

G. C. Greubel, Table of n, a(n) for n = 0..445

Eric Weisstein's World of Mathematics, Pochhammer Symbol.

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)

Example

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 + ...

Mathematica

Expand@Table[(Pochhammer[Sqrt[2], n] + Pochhammer[-Sqrt[2], n])/2, {n, 0, 22}]

Prog

(PARI) {a(n) = if( n<0, 0, real(prod(k=0, n-1, quadgen(8) + k)))}; /* Michael Somos, Oct 23 2015 */

Crossrefs

Cf. A263688.

Sequence in context: A332796 A134094 A009575 * A180891 A224529 A171151

Adjacent sequences: A263684 A263685 A263686 * A263688 A263689 A263690

Keyword

nonn

Author

Vladimir Reshetnikov, Oct 23 2015

Status

approved