login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

Expansion of 1/((1-10x)(1-11x)(1-12x)).
0

%I #19 Aug 31 2018 16:52:26

%S 1,33,727,13365,221431,3428733,50631967,721942485,10021257511,

%T 136192514733,1819621847407,23973890545605,312209398691191,

%U 4026262617877533,51492399583946047,653858524870924725

%N Expansion of 1/((1-10x)(1-11x)(1-12x)).

%H <a href="/index/Rec">Index entries for linear recurrences with constant coefficients</a>, signature (33,-362,1320)

%F If we define f(m,j,x) = Sum_{k=j..m} binomial(m,k)*Stirling2(k,j)*x^(m-k) then a(n-2) = f(n,2,10) for n >= 2. - _Milan Janjic_, Apr 26 2009

%F a(n) = 33*a(n-1) - 362*a(n-2) + 1320*a(n-3), n >= 3. - _Vincenzo Librandi_, Mar 18 2011

%F a(n) = 23*a(n-1) - 132*a(n-2) + 10^n, n >= 2. - _Vincenzo Librandi_, Mar 18 2011

%F a(n) = 6*12^(n+1) - 11^(n+2) + 5*10^(n+1). - _R. J. Mathar_, Mar 18 2011

%t CoefficientList[Series[1/((1-10x)(1-11x)(1-12x)),{x,0,30}],x] (* or *) LinearRecurrence[{33,-362,1320},{1,33,727},30] (* _Harvey P. Dale_, Apr 27 2012 *)

%K nonn

%O 0,2

%A _N. J. A. Sloane_