A167893 a(n) = Sum_{k=1..n} Catalan(k)^3.

1, 9, 134, 2878, 76966, 2376934, 81330523, 3005537523, 117938569451, 4856184495787, 208008478587443, 9208478072445171, 419215292661445171, 19548493234125829171, 930767164551264230296, 45133682592532326893296, 2224173698690413601132296, 111192059034974606204132296

Offset

1,2

Comments

Catalan(k) = A000108(k) = (2k)!/(k!*(k+1)!) = C(2*k,k)/(k+1).

For prime p=7, p^2 divides a(p^2), and p divides all a(n) for n from (p^2-1)/2 to p^2-2.

For prime p=19 or 97, p divides all a(n) for n from (p-1)/2 to p-2.

Links

G. C. Greubel, Table of n, a(n) for n = 1..500

Eric Weisstein's World of Mathematics, Catalan Number

Formula

a(n) = Sum_{k=1..n} A033536(k).

Recurrence: (n+1)^3*a(n) = (5*n - 1)*(13*n^2 - 16*n + 7)*a(n-1) - 8*(2*n - 1)^3*a(n-2). - Vaclav Kotesovec, Jul 01 2016

a(n) ~ 2^(6*n+6) / (63*Pi^(3/2)*n^(9/2)). - Vaclav Kotesovec, Jul 01 2016

Mathematica

Array[n \[Function] Sum[CatalanNumber[k]^3, {k, 1, n}], 15] (* J. Mulder (jasper.mulder(AT)planet.nl), Jan 25 2010 *)
Accumulate[CatalanNumber[Range[1, 20]]^3] (* Vincenzo Librandi, Jul 01 2016 *)

Prog

(PARI) a(n)=sum(k=1, n, (binomial(k+k, k)/(k+1))^3) /* Charles R Greathouse IV, Jun 14 2011 */
(Magma) [&+[Catalan(i)^3: i in [1..n]]: n in [1..20]]; // Vincenzo Librandi, Jul 01 2016

Crossrefs

Cf. A000108, A014138, A167892, A167893, A001246, A033536, A014137, A094639.

Sequence in context: A163200 A279975 A296171 * A268062 A218326 A272498

Adjacent sequences: A167890 A167891 A167892 * A167894 A167895 A167896

Keyword

nonn

Author

Alexander Adamchuk, Nov 15 2009

Extensions

More terms from J. Mulder, (jasper.mulder(AT)planet.nl), Jan 25 2010

More terms from Sean A. Irvine, Jun 13 2011

Status

approved