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

%I

%S 1,9,134,2878,76966,2376934,81330523,3005537523,117938569451,

%T 4856184495787,208008478587443,9208478072445171,419215292661445171,

%U 19548493234125829171,930767164551264230296,45133682592532326893296,2224173698690413601132296,111192059034974606204132296

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

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

%C 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.

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

%H G. C. Greubel, <a href="/A167893/b167893.txt">Table of n, a(n) for n = 1..500</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CatalanNumber.html">Catalan Number</a>

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

%F 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

%F a(n) ~ 2^(6*n+6) / (63*Pi^(3/2)*n^(9/2)). - _Vaclav Kotesovec_, Jul 01 2016

%t Array[n \[Function] Sum[CatalanNumber[k]^3, {k, 1, n}], 15] (* J. Mulder, (jasper.mulder(AT)planet.nl), Jan 25 2010 *)

%t Accumulate[CatalanNumber[Range[1, 20]]^3] (* _Vincenzo Librandi_, Jul 01 2016 *)

%o (PARI) a(n)=sum(k=1,n,(binomial(k+k,k)/(k+1))^3) /* _Charles R Greathouse IV_, Jun 14 2011 */

%o (MAGMA) [&+[Catalan(i)^3: i in [1..n]]: n in [1..20]]; // _Vincenzo Librandi_, Jul 01 2016

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

%K nonn

%O 1,2

%A _Alexander Adamchuk_, Nov 15 2009

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

%E More terms from _Sean A. Irvine_, Jun 13 2011

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Last modified October 23 16:53 EDT 2018. Contains 316529 sequences. (Running on oeis4.)