A025750 5th-order Patalan numbers (generalization of Catalan numbers).

1, 1, 10, 150, 2625, 49875, 997500, 20662500, 439078125, 9513359375, 209293906250, 4661546093750, 104884787109375, 2380077861328125, 54401779687500000, 1251240932812500000, 28934946571289062500, 672311993862304687500, 15687279856787109375000, 367412607172119140625000

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. A034687, A049393, A340722.

Programs

Mathematica

CoefficientList[Series[(6-(1-25x)^(1/5))/5,{x,0,20}],x] (* Harvey P. Dale, Dec 06 2012 *)
a[0] = 1; a[n_] := ((-5)^(n - 1)*Sum[5^(n - k)*StirlingS1[n, k], {k, 1, n}])/n!; Table[a[n], {n, 0, 16}] (* Jean-François Alcover, Mar 19 2013, after Vladimir Kruchinin *)
a[n_] := 25^(n-1) * Pochhammer[4/5, n-1]/n!; a[0] = 1; Array[a, 20, 0] (* Amiram Eldar, Aug 20 2025 *)

Maxima

a(n):=if n=0 then 1 else (sum((-1)^(n-k-1)*binomial(n+k-1,n-1)*sum(2^j*binomial(k,j)*sum(binomial(j,i-j)*binomial(k-j,n-3*(k-j)-i-1)*5^(3*(k-j)+i),i,j,n-k+j-1),j,0,k),k,0,n-1))/(n); /* Vladimir Kruchinin, Dec 10 2011 */

Maxima

a(n):=if n=0 then 1 else -binomial(1/5,n)*(-25)^n/5; /* Tani Akinari, Sep 17 2015 */

Formula

G.f.: (6-(1-25*x)^(1/5))/5.

a(n) = 5^(n-1)*4*A034301(n-1)/n!, n >= 2, where 4*A034301(n-1) = (5*n-6)(!^5) = Product_{j=2..n} (5*j-6). - Wolfdieter Lang

a(n) = Sum_{k=0..n-1} (-1)^(n-k-1)*binomial(n+k-1,n-1) * Sum_{j=0..k} 2^j*binomial(k,j) * Sum_{i=j..n-k+j-1} binomial(j,i-j)*binomial(k-j,n-3*(k-j)-i-1)*5^(3*(k-j)+i)/n, n > 0, a(0) = 1. - Vladimir Kruchinin, Dec 10 2011

a(n) = ((-5)^(n-1)*Sum_{k=1..n} (5)^(n-k)*stirling1(n,k))/n!, n>0, a(0) = 1. - Vladimir Kruchinin, Mar 19 2013

From Karol A. Penson, Feb 05 2025: (Start)

a(n) without the initial 1 (i.e., a(n) for n >= 1) is given by

a(n+1) = 5^(2*n)*gamma(n + 4/5)/(gamma(4/5)*(n + 1)!), n >= 0.

a(n+1) = Integral_{x=0..25} x^n*W(x) dx, n >= 0,

where W(x) = sin(Pi/5)*5^(2/5)*(1 - x/25)^(1/5)/(5*Pi*x^(1/5)). W(x) is a positive function on x = (0, 25), is singular at x = 0 with the singularity (x)^(-1/5), and it goes to zero at x = 25. (End)

a(n) ~ 25^(n-1) / (Gamma(4/5) * n^(6/5)). - Amiram Eldar, Aug 20 2025