[go: up one dir, main page]

login
A025749
4th-order Patalan numbers (generalization of Catalan numbers).
8
1, 1, 6, 56, 616, 7392, 93632, 1230592, 16612992, 228890112, 3204461568, 45445091328, 651379642368, 9419951751168, 137262154088448, 2013178259963904, 29694379334467584, 440175505428578304
OFFSET
0,3
LINKS
W. Lang, On generalizations of Stirling number triangles, J. Integer Seqs., Vol. 3 (2000), #00.2.4.
Elżbieta Liszewska, Wojciech Młotkowski, Some relatives of the Catalan sequence, arXiv:1907.10725 [math.CO], 2019.
T. M. Richardson, The Super Patalan Numbers, arXiv preprint arXiv:1410.5880, 2014 and J. Int. Seq. 18 (2015) # 15.3.3
FORMULA
From Wolfdieter Lang: (Start)
G.f.: (5-(1-16*x)^(1/4))/4.
a(n) = 4^(n-1)*3*A034176(n-1)/n!, n >= 2;
3*A034176(n-1) = (4*n-5)(!^4) := Product_{j=2..n} (4*j - 5). (End)
a(n) := (4^(n-1)*Sum_{k=1..n-1} binomial(n+k-1,n-1) * Sum_{j=0..k} binomial(j,n-3*k+2*j-1)*4^(j-k)*binomial(k,j)*3^(-n+3*k-j+1)*2^(n-3*k+j-1)*(-1)^(n-3*k+2*j-1))/n. - Vladimir Kruchinin, Apr 01 2011
n*a(n) +4*(-4*n+5)*a(n-1)=0. - R. J. Mathar, Apr 05 2018
MATHEMATICA
a[n_] := (4^(n-1)*Sum[ Binomial[n+k-1, n-1]*Sum[ Binomial[j, n-3*k+2*j-1] * 4^(j-k) * Binomial[k, j] * 3^(-n+3*k-j+1) * 2^(n-3*k+j-1) * (-1)^(n-3*k+2*j-1), {j, 0, k}], {k, 1, n-1}])/n; a[0] = a[1] = 1; Table[a[n], {n, 0, 17}] (* Jean-François Alcover, Mar 05 2013, after Vladimir Kruchinin *)
PROG
(Maxima)
a(n):=(4^(n-1)*sum(binomial(n+k-1, n-1)*sum(binomial(j, n-3*k+2*j-1)*4^(j-k)*binomial(k, j)*3^(-n+3*k-j+1)*2^(n-3*k+j-1)*(-1)^(n-3*k+2*j-1), j, 0, k), k, 1, n-1))/n; /* Vladimir Kruchinin, Apr 01 2011 */
CROSSREFS
Equals 2^n * A048779(n), n > 1.
Sequence in context: A048348 A227384 A199755 * A365766 A297705 A231690
KEYWORD
nonn
STATUS
approved