A025755 10th-order Patalan numbers (generalization of Catalan numbers).

1, 1, 45, 2850, 206625, 16116750, 1316201250, 110936962500, 9568313015625, 839885253593750, 74749787569843750, 6727480881285937500, 611079513383472656250, 55937278532794804687500, 5154220664807521289062500, 477624448272163639453125000, 44478776745345238924072265625

Offset: 0

Author

Olivier Gérard

Keywords

• nonn

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• Wolfdieter Lang, On generalizations of Stirling number triangles, J. Integer Seq., Vol. 3 (2000), Article 00.2.4.
• Elżbieta Liszewska and Wojciech Młotkowski, Some relatives of the Catalan sequence, arXiv:1907.10725 [math.CO], 2019.
• Thomas M. Richardson, The Super Patalan Numbers, J. Int. Seq. 18 (2015), Article 15.3.3; arXiv preprint, arXiv:1410.5880 [math.CO], 2014.

Crossrefs

Cf. A035278, A340725.

Programs

Mathematica

CoefficientList[Series[(11 -(1 - 100*x)^(1/10))/10, {x, 0, 20}], x] (* Vincenzo Librandi, Dec 29 2012 *)
a[n_] := 100^(n-1) * Pochhammer[9/10, n-1] / n!; a[0] = 1; Array[a, 26, 0] (* Amiram Eldar, Aug 20 2025 *)

Formula

G.f.: (11-(1-100*x)^(1/10))/10.

a(n) = 10^(n-1)*9*A035278(n-1)/n!, n >= 2, where 9*A035278(n-1) = (10*n-11)(!^10) = Product_{j=2..n} (10*j - 11). - Wolfdieter Lang

Conjecture: n*a(n) + 10*(-10*n+11)*a(n-1) = 0. - R. J. Mathar, Jul 28 2014

a(n) = 100^(n-1)*Pochhammer(9/10, n-1)/n! for n >= 1. Maple confirms this satisfies Mathar's conjecture for n >= 2 (it's not true for n=1). - Robert Israel, Oct 05 2014

a(n) ~ 100^(n-1) / (Gamma(9/10) * n^(11/10)). - Amiram Eldar, Aug 20 2025