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%I #42 Aug 12 2023 21:12:37
%S 1,1,6,56,616,7392,93632,1230592,16612992,228890112,3204461568,
%T 45445091328,651379642368,9419951751168,137262154088448,
%U 2013178259963904,29694379334467584,440175505428578304
%N 4th-order Patalan numbers (generalization of Catalan numbers).
%H Vincenzo Librandi, <a href="/A025749/b025749.txt">Table of n, a(n) for n = 0..200</a>
%H W. Lang, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL3/LANG/lang.html">On generalizations of Stirling number triangles</a>, J. Integer Seqs., Vol. 3 (2000), #00.2.4.
%H Elżbieta Liszewska, Wojciech Młotkowski, <a href="https://arxiv.org/abs/1907.10725">Some relatives of the Catalan sequence</a>, arXiv:1907.10725 [math.CO], 2019.
%H T. M. Richardson, <a href="http://arxiv.org/abs/1410.5880">The Super Patalan Numbers</a>, arXiv preprint arXiv:1410.5880, 2014 and <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Richardson/rich2.html">J. Int. Seq. 18 (2015) # 15.3.3</a>
%F From _Wolfdieter Lang_: (Start)
%F G.f.: (5-(1-16*x)^(1/4))/4.
%F a(n) = 4^(n-1)*3*A034176(n-1)/n!, n >= 2;
%F 3*A034176(n-1) = (4*n-5)(!^4) := Product_{j=2..n} (4*j - 5). (End)
%F 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
%F n*a(n) +4*(-4*n+5)*a(n-1)=0. - _R. J. Mathar_, Apr 05 2018
%t 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_ *)
%o (Maxima)
%o 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 */
%Y Equals 2^n * A048779(n), n > 1.
%K nonn
%O 0,3
%A _Olivier Gérard_
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