A000556 Expansion of exp(-x) / (1 - exp(x) + exp(-x)).

1, 1, 5, 31, 257, 2671, 33305, 484471, 8054177, 150635551, 3130337705, 71556251911, 1784401334897, 48205833997231, 1402462784186105, 43716593539939351, 1453550100421124417, 51350258701767067711, 1920785418183176050505, 75839622064482770570791

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

References

• Anthony G. Shannon and Richard L. Ollerton. "A note on Ledin's summation problem." The Fibonacci Quarterly 59:1 (2021), 47-56.
• N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..401
• Gregory Dresden, On the Brousseau sums Sum_{i=1..n} i^p*Fibonacci(i), arxiv.org:2206.00115 [math.NT], 2022.
• Paul Kinlaw, Michael Morris, and Samanthak Thiagarajan, Sums related to the Fibonacci sequence, Husson University (2021).
• G. Ledin, Jr., On a certain kind of Fibonacci sums, Fib. Quart., 5 (1967), 45-58.
• R. L. Ollerton and A. G. Shannon, A Note on Brousseau's Summation Problem, Fibonacci Quart. 58 (2020), no. 5, 190-199.
• Eric Weisstein's MathWorld, Polylogarithm.
• Eric Weisstein's MathWorld, Golden Ratio.
• Eric Weisstein's MathWorld, Lucas Number.

Crossrefs

Cf. A005923, A216794.

Programs

Maple

a:= proc(n) option remember; `if`(n=0, 1, add(
      a(n-j)*binomial(n, j)*(2^j-1), j=1..n))
    end:
seq(a(n), n=0..20);  # Alois P. Heinz, Oct 05 2019

Mathematica

CoefficientList[Series[E^(-x)/(1-E^x+E^(-x)), {x, 0, 20}], x] * Range[0, 20]! (* Vaclav Kotesovec, May 04 2015 *)
Round@Table[(-1)^(n+1) (PolyLog[-n, 1-GoldenRatio] GoldenRatio + PolyLog[-n, GoldenRatio]/GoldenRatio)/Sqrt[5], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 30 2015 *)

PARI

a(n) = sum(k=0, n, k!*fibonacci(k+1)*stirling(n, k, 2)); \\ Michel Marcus, Oct 30 2015

Formula

a(n) = Sum_{k=0..n} k!*Fibonacci(k+1)*Stirling2(n,k).

E.g.f.: 1/(1 + U(0)) where U(k) = 1 - 2^k/(1 - x/(x - (k+1)*2^k/U(k+1) )); (continued fraction 3rd kind, 3-step ). - Sergei N. Gladkovskii, Dec 05 2012

a(n) ~ 2*n! / ((5+sqrt(5)) * log((1+sqrt(5))/2)^(n+1)). - Vaclav Kotesovec, May 04 2015

a(n) = (-1)^(n+1)*(Li_{-n}(1-phi)*phi + Li_{-n}(phi)/phi)/sqrt(5), where Li_n(x) is the polylogarithm, phi=(1+sqrt(5))/2 is the golden ratio. - Vladimir Reshetnikov, Oct 30 2015

John W. Layman observes that this is also Sum (-2)^k*binomial(n, k)*b(n-k), where b() = A005923.

From Greg Dresden, May 13 2022 (Start):

For n > 0, a(n) = 1 + 2*Sum_{k=0..floor(n/2-1)} binomial(n,2*k+1) * a(n-2*k-1).

For n > 0, a(n) = Sum_{k=0..n-1} binomial(n,k)*A000557(k).

(End)