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A180875
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Sum_{j>=1} j^n*2^j/binomial(2*j,j) = r_n*Pi/2 + s_n with integer r_n and s_n; sequence gives s_n.
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9
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1, 3, 11, 55, 355, 2807, 26259, 283623, 3473315, 47552791, 719718067, 11932268231, 215053088835, 4186305575415, 87534887434835, 1956680617267879, 46561960552921315, 1175204650272267479, 31357650670190565363, 881958890078887314567, 26078499305918584929155, 808742391638178302137783
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OFFSET
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0,2
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COMMENTS
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LINKS
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F. J. Dyson, N. E. Frankel and M. L. Glasser, Lehmer's Interesting Series, arXiv:1009.4274 [math-ph], 2010-2011; see Table IV on p. 14.
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FORMULA
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a(0)=1; if n>=1, then a(n) = a(n-1) + 1 + Sum_{m=1..n} binomial(n,m)*a(n-m). - Detlef Meya, Jan 22 2018
E.g.f.: 2*(arcsin(exp(x/2)/sqrt(2)) - Pi/4) * sqrt(exp(x)/(2-exp(x))^3) + exp(x)/(2-exp(x)). - Seiichi Manyama, Oct 21 2019
a(n) ~ Pi * n^(n+1) / (sqrt(2) * exp(n) * (log(2))^(n + 3/2)). - Vaclav Kotesovec, Oct 22 2019
E.g.f.: d/dx (f(x) * Integral f(x) dx), where f(x) = sqrt(exp(x)/(2-exp(x))), cf. A014307. - Seiichi Manyama, Oct 22 2019
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MAPLE
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f := n -> sum(j^n*(j!)^2*2^j/(2*j)!, j = 1..infinity):
seq(f(n), n = 0..5); # gives
# [1+(1/2)*Pi, 3+Pi, 11+(7/2)*Pi, 55+(35/2)*Pi, 355+113*Pi, 2807+(1787/2)*Pi].
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MATHEMATICA
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Table[Expand[FunctionExpand[FullSimplify[Sum[j^n*2^j/Binomial[2*j, j], {j, 1, Infinity}]]]][[1]], {n, 0, 20}] (* Vaclav Kotesovec, May 14 2020 *)
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PROG
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(PARI) N=20; x='x+O('x^N); f=sqrt(exp(x)/(2-exp(x))); Vec(serlaplace(deriv(f*intformal(f)))) \\ Seiichi Manyama, Oct 22 2019
(Python) # An alternative version of the sequence starts (for n >= 0):
# 0, 1, 3, 11, ..., or in terms of the approximation: [(1/2)*Pi, 1+(1/2)*Pi,
# 3+Pi, 11+(7/2)*Pi, ...]. Similar to the formula of Detlef Meya above, the
# sequence then can be computed (without a special initial case) as:
from functools import cache
from math import comb as binomial
@cache
def a(n): return n + sum((binomial(n, j) - 1) * a(n - j) for j in range(1, n))
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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