A014151 Apply partial sum operator thrice to Catalan numbers.
1, 4, 11, 27, 66, 170, 471, 1398, 4381, 14282, 47897, 164012, 570639, 2010678, 7158569, 25709157, 93020112, 338736224, 1240496193, 4565563209, 16878057692, 62644246662, 233346693759, 872045012633, 3268643350608, 12285088109136, 46288732360369, 174813127020311, 661606223839322
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Programs
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Mathematica
Flatten[{1, RecurrenceTable[{n*(n+1)*a[n] == 2*n*(3*n+1)*a[n-1] - (9*n^2+7*n-4)*a[n-2] + 2*(n+1)*(2*n+1)*a[n-3],a[1]==4,a[2]==11,a[3]==27},a,{n,100}]}] -
PARI
sm(v)={my(s=vector(#v));s[1]=v[1];for(n=2,#v,s[n]=v[n]+s[n-1]);s;} C(n)=binomial(2*n,n)/(n+1); sm(sm(sm(vector(66,n,C(n-1))))) /* Joerg Arndt, May 04 2013 */
Formula
D-finite with recurrence: n*(n+1)*a(n) = 2*n*(3*n+1)*a(n-1) - (9*n^2+7*n-4)*a(n-2) + 2*(n+1)*(2*n+1)*a(n-3). - Vaclav Kotesovec, Oct 07 2012
a(n) ~ 2^(2*n+6)/(27*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 07 2012
G.f.: C(x)/(1-x)^3, where C(x) is g.f. of Catalan numbers. - Vladimir Kruchinin, Oct 18 2016
a(n) = Sum_{k=0..n} binomial(n+3,k+3) * r(k), r(k) = A005043(k). - Vladimir Kruchinin, Oct 18 2016