A201365 Expansion of e.g.f. exp(x) / (5 - 4*exp(x)).
1, 5, 45, 605, 10845, 243005, 6534045, 204972605, 7348546845, 296387331005, 13282361478045, 654762261324605, 35211177242722845, 2051349014835939005, 128701394409842982045, 8651475271312083756605, 620334325261670875138845, 47259638324026516284867005
Offset: 0
Examples
E.g.f.: E(x) = 1 + 5*x + 45*x^2/2! + 605*x^3/3! + 10845*x^4/4! + 243005*x^5/5! + ... O.g.f.: A(x) = 1 + 5*x + 45*x^2 + 605*x^3 + 10845*x^4 + 243005*x^5 + ... where A(x) = 1 + 5*x/(1+x) + 2!*5^2*x^2/((1+x)*(1+2*x)) + 3!*5^3*x^3/((1+x)*(1+2*x)*(1+3*x)) + 4!*5^4*x^4/((1+x)*(1+2*x)*(1+3*x)*(1+4*x)) + ...
Links
- G. C. Greubel, Table of n, a(n) for n = 0..358
Crossrefs
Programs
-
Magma
R
:=PowerSeriesRing(Rationals(), 20); Coefficients(R!(Laplace( 1/(5*Exp(-x) -4) ))); // G. C. Greubel, Jun 08 2020 -
Maple
seq(coeff(series(1/(5*exp(-x) - 4), x, n+1)*n!, x, n), n = 0..20); # G. C. Greubel, Jun 08 2020
-
Mathematica
Table[Sum[(-1)^(n-k)*5^k*StirlingS2[n,k]*k!,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Jun 13 2013 *) With[{nn=20},CoefficientList[Series[Exp[x]/(5-4Exp[x]),{x,0,nn}],x] Range[ 0,nn]!] (* Harvey P. Dale, Jul 09 2015 *) a[n_]:= If[n<0, 0, PolyLog[ -n, 4/5]/4]; (* Michael Somos, Apr 27 2019 *) -
PARI
{a(n)=n!*polcoeff(exp(x+x*O(x^n))/(5 - 4*exp(x+x*O(x^n))), n)} -
PARI
{a(n)=polcoeff(sum(m=0, n, 5^m*m!*x^m/prod(k=1, m, 1+k*x+x*O(x^n))), n)} -
PARI
{a(n)=sum(k=0, n, (-1)^(n-k)*5^k*stirling(n, k, 2)*k!)} -
Sage
[sum( (-1)^(n-j)*5^j*factorial(j)*stirling_number2(n,j) for j in (0..n)) for n in (0..20)] # G. C. Greubel, Jun 08 2020
Formula
O.g.f.: A(x) = Sum_{n>=0} n! * 5^n*x^n / Product_{k=0..n} (1+k*x).
O.g.f.: A(x) = 1/(1 - 5*x/(1-4*x/(1 - 10*x/(1-8*x/(1 - 15*x/(1-12*x/(1 - 20*x/(1-16*x/(1 - 25*x/(1-20*x/(1 - ...))))))))))), a continued fraction.
a(n) = Sum_{k=0..n} (-1)^(n-k) * 5^k * Stirling2(n,k) * k!.
a(n) = Sum_{k=0..n} A123125(n,k)*5^k*4^(n-k). - Philippe Deléham, Nov 30 2011
a(n) ~ n! / (4*(log(5/4))^(n+1)) . - Vaclav Kotesovec, Jun 13 2013
a(n) = log(5/4) * Integral_{x = 0..oo} (ceiling(x))^n * (5/4)^(-x) dx. - Peter Bala, Feb 14 2015
a(n) = (1/4) Sum_{k>=1} (4/5)^k * n^k. - Michael Somos, Apr 27 2019
a(n) = 1 + 4 * Sum_{k=0..n-1} binomial(n,k) * a(k). - Ilya Gutkovskiy, Jun 08 2020
From Seiichi Manyama, Nov 15 2023: (Start)
a(0) = 1; a(n) = -5*Sum_{k=1..n} (-1)^k * binomial(n,k) * a(n-k).
a(0) = 1; a(n) = 5*a(n-1) + 4*Sum_{k=1..n-1} binomial(n-1,k) * a(n-k). (End)
a(n) = (5/4)*A094417(n) - (1/4)*0^n. - Seiichi Manyama, Dec 21 2023