A016075 Expansion of 1/((1-8*x)*(1-9*x)*(1-10*x)*(1-11*x)).
1, 38, 905, 17290, 289821, 4453638, 64331905, 887339330, 11810819141, 152832918238, 1933092302505, 23997027406170, 293289532268461, 3537885908902838, 42204462297434705, 498697803478957810, 5844588402226277781, 68011678300853991438, 786547256602640400505
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
- Index entries for linear recurrences with constant coefficients, signature (38,-539,3382,-7920)
Programs
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Magma
m:=20; R
:=PowerSeriesRing(Integers(), m); Coefficients(R!(1/((1-8*x)*(1-9*x)*(1-10*x)*(1-11*x)))); // Vincenzo Librandi, Jun 24 2013 -
Magma
I:=[1, 38, 905, 17290]; [n le 4 select I[n] else 38*Self(n-1)-539*Self(n-2)+3382*Self(n-3)-7920*Self(n-4): n in [1..20]]; // Vincenzo Librandi, Jun 24 2013
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Mathematica
CoefficientList[Series[1/((1-8*x)*(1-9*x)*(1-10*x)*(1-11*x)), {x,0,20}], x] (* Vincenzo Librandi, Jun 23 2013 *) -
PARI
x='x+O('x^30); Vec(1/((1-8*x)*(1-9*x)*(1-10*x)*(1-11*x))) \\ G. C. Greubel, Feb 07 2018
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) = f(n,3,8), (n>=3). - Milan Janjic, Apr 26 2009
a(n) = 38*a(n-1) - 539*a(n-2) + 3382*a(n-3) - 7920*a(n-4), n>=4. - Vincenzo Librandi, Mar 17 2011
a(n) = 21*a(n-1) - 110*a(n-2) + 9^(n+1) - 8^(n+1), n>=2. - Vincenzo Librandi, Mar 17 2011
a(n) = 11^(n+3)/6 -5*10^(n+2) -4*8^(n+2)/3 + 9^(n+3)/2. - R. J. Mathar, Mar 18 2011