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%I M3966 N1638 #87 Jun 02 2022 09:58:20
%S 1,1,5,31,257,2671,33305,484471,8054177,150635551,3130337705,
%T 71556251911,1784401334897,48205833997231,1402462784186105,
%U 43716593539939351,1453550100421124417,51350258701767067711,1920785418183176050505,75839622064482770570791
%N Expansion of exp(-x) / (1 - exp(x) + exp(-x)).
%D Anthony G. Shannon and Richard L. Ollerton. "A note on Ledin's summation problem." The Fibonacci Quarterly 59:1 (2021), 47-56.
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H Alois P. Heinz, <a href="/A000556/b000556.txt">Table of n, a(n) for n = 0..401</a>
%H Gregory Dresden, <a href="https://arxiv.org/abs/2206.00115">On the Brousseau sums Sum_{i=1..n} i^p*Fibonacci(i)</a>, arxiv.org:2206.00115 [math.NT], 2022.
%H Paul Kinlaw, Michael Morris, and Samanthak Thiagarajan, <a href="https://www.researchgate.net/publication/350886459_SUMS_RELATED_TO_THE_FIBONACCI_SEQUENCE">Sums related to the Fibonacci sequence</a>, Husson University (2021).
%H G. Ledin, Jr., <a href="http://www.fq.math.ca/Scanned/5-1/ledin.pdf">On a certain kind of Fibonacci sums</a>, Fib. Quart., 5 (1967), 45-58.
%H R. L. Ollerton and A. G. Shannon, <a href="https://www.fq.math.ca/Papers1/58-5/ollerton.pdf">A Note on Brousseau's Summation Problem</a>, Fibonacci Quart. 58 (2020), no. 5, 190-199.
%H Eric Weisstein's MathWorld, <a href="http://mathworld.wolfram.com/Polylogarithm.html">Polylogarithm</a>.
%H Eric Weisstein's MathWorld, <a href="http://mathworld.wolfram.com/GoldenRatio.html">Golden Ratio</a>.
%H Eric Weisstein's MathWorld, <a href="http://mathworld.wolfram.com/LucasNumber.html">Lucas Number</a>.
%F a(n) = Sum_{k=0..n} k!*Fibonacci(k+1)*Stirling2(n,k).
%F 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
%F a(n) ~ 2*n! / ((5+sqrt(5)) * log((1+sqrt(5))/2)^(n+1)). - _Vaclav Kotesovec_, May 04 2015
%F 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
%F _John W. Layman_ observes that this is also Sum (-2)^k*binomial(n, k)*b(n-k), where b() = A005923.
%F From _Greg Dresden_, May 13 2022 (Start):
%F For n > 0, a(n) = 1 + 2*Sum_{k=0..floor(n/2-1)} binomial(n,2*k+1) * a(n-2*k-1).
%F For n > 0, a(n) = Sum_{k=0..n-1} binomial(n,k)*A000557(k).
%F (End)
%p a:= proc(n) option remember; `if`(n=0, 1, add(
%p a(n-j)*binomial(n, j)*(2^j-1), j=1..n))
%p end:
%p seq(a(n), n=0..20); # _Alois P. Heinz_, Oct 05 2019
%t CoefficientList[Series[E^(-x)/(1-E^x+E^(-x)), {x, 0, 20}], x] * Range[0, 20]! (* _Vaclav Kotesovec_, May 04 2015 *)
%t Round@Table[(-1)^(n+1) (PolyLog[-n, 1-GoldenRatio] GoldenRatio + PolyLog[-n, GoldenRatio]/GoldenRatio)/Sqrt[5], {n, 0, 20}] (* _Vladimir Reshetnikov_, Oct 30 2015 *)
%o (PARI) a(n) = sum(k=0, n, k!*fibonacci(k+1)*stirling(n, k, 2)); \\ _Michel Marcus_, Oct 30 2015
%Y Cf. A005923, A216794.
%K nonn,easy
%O 0,3
%A _N. J. A. Sloane_
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