A025211 Expansion of 1/((1-2x)(1-3x)(1-4x)(1-5x)).
1, 14, 125, 910, 5901, 35574, 204205, 1132670, 6129101, 32566534, 170691885, 885423630, 4556561101, 23305343894, 118631189165, 601616805790, 3042056477901, 15346559343654, 77279066272045, 388583895311150, 1951684190615501, 9793511186181814, 49108010998116525
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..500
- Index entries for linear recurrences with constant coefficients, signature (14,-71,154,-120)
Programs
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Magma
[3^(n+3)/2 -2*4^(n+2)-2^(n+2)/3+5^(n+3)/6: n in [0..30]]; // Vincenzo Librandi, Jun 21 2011
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Mathematica
CoefficientList[Series[1/((1 - 2 x) (1 - 3 x) (1 - 4 x) (1 - 5 x)), {x, 0, 25}], x] (* Vladimir Joseph Stephan Orlovsky, Jun 20 2011 *) LinearRecurrence[{14,-71,154,-120},{1,14,125,910},30] (* Harvey P. Dale, Feb 05 2020 *)
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PARI
a(n)=n-=2;3^n*3/2-2*4^n-2^n/3+5^n*5/6 \\ Charles R Greathouse IV, Jun 21 2011
Formula
If we define f(m,j,x) = Sum_{k=j..m} binomial(m,k)*stirling2(k,j)*x^(m-k) then a(n-3) = (-1)^(n-1)*f(n,3,-5), (n >= 3). - Milan Janjic, Apr 26 2009
If we define f(m,j,x) = Sum_{k=j..m} binomial(m,k)*stirling2(k,j)*x^(m-k) then a(n-3) = f(n,3,2), (n >= 3). - Milan Janjic, Apr 26 2009
a(n) = 3^(n+3)/2 - 2*4^(n+2) - 2^(n+2)/3 + 5^(n+3)/6. - R. J. Mathar, Mar 22 2011
O.g.f.: 1/((1-2*x)*(1-3*x)*(1-4*x)*(1-5*x)).
E.g.f.: (d^3/dx^3) (exp(2*x)*((exp(x)-1)^3)/3!). See the Sheffer comment given above. - Wolfdieter Lang, Oct 08 2011
Comments