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

0,2

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

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

Cf. A000108, A014137, A005043

Sequence in context: A301874 A027439 A108985 * A176759 A266009 A096124

Adjacent sequences: A014148 A014149 A014150 * A014152 A014153 A014154

Keyword

nonn,easy

Author

N. J. A. Sloane.

Status

approved