cp's OEIS Frontend

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

%I A167893 #25 Feb 16 2025 08:33:11
%S A167893 1,9,134,2878,76966,2376934,81330523,3005537523,117938569451,
%T A167893 4856184495787,208008478587443,9208478072445171,419215292661445171,
%U A167893 19548493234125829171,930767164551264230296,45133682592532326893296,2224173698690413601132296,111192059034974606204132296
%N A167893 a(n) = Sum_{k=1..n} Catalan(k)^3.
%C A167893 Catalan(k) = A000108(k) = (2k)!/(k!*(k+1)!) = C(2*k,k)/(k+1).
%C A167893 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 A167893 For prime p=19 or 97, p divides all a(n) for n from (p-1)/2 to p-2.
%H A167893 G. C. Greubel, <a href="/A167893/b167893.txt">Table of n, a(n) for n = 1..500</a>
%H A167893 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/CatalanNumber.html">Catalan Number</a>
%F A167893 a(n) = Sum_{k=1..n} A033536(k).
%F A167893 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 A167893 a(n) ~ 2^(6*n+6) / (63*Pi^(3/2)*n^(9/2)). - _Vaclav Kotesovec_, Jul 01 2016
%t A167893 Array[n \[Function] Sum[CatalanNumber[k]^3, {k, 1, n}], 15] (* J. Mulder (jasper.mulder(AT)planet.nl), Jan 25 2010 *)
%t A167893 Accumulate[CatalanNumber[Range[1, 20]]^3] (* _Vincenzo Librandi_, Jul 01 2016 *)
%o A167893 (PARI) a(n)=sum(k=1,n,(binomial(k+k,k)/(k+1))^3) /* _Charles R Greathouse IV_, Jun 14 2011 */
%o A167893 (Magma) [&+[Catalan(i)^3: i in [1..n]]: n in [1..20]]; // _Vincenzo Librandi_, Jul 01 2016
%Y A167893 Cf. A000108, A014138, A167892, A167893, A001246, A033536, A014137, A094639.
%K A167893 nonn
%O A167893 1,2
%A A167893 _Alexander Adamchuk_, Nov 15 2009
%E A167893 More terms from J. Mulder, (jasper.mulder(AT)planet.nl), Jan 25 2010
%E A167893 More terms from _Sean A. Irvine_, Jun 13 2011