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A014151
Apply partial sum operator thrice to Catalan numbers.
3
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
0,2
LINKS
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
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}]}]
PROG
(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 */
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved