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%I #21 Oct 08 2017 17:39:53
%S 1,6,1,56,12,1,616,148,18,1,7392,1904,276,24,1,93632,25312,4080,440,
%T 30,1,1230592,344960,59808,7360,640,36,1,16612992,4792128,876960,
%U 118224,11960,876,42,1,228890112,67586816,12900416,1860992,209200,18096,1148
%N A convolution triangle of numbers obtained from A025749.
%C a(n,1) = A025749(n); a(n,1)= 4^(n-1)*3*A034176(n-1)/n!, n >= 2.
%C G.f. for m-th column: ((1-(1-16*x)^(1/4))/4)^m.
%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.
%F a(n, m) = 4*(4*(n-1)-m)*a(n-1, m)/n + m*a(n-1, m-1)/n, n >= m >= 1; a(n, m) := 0, n < m; a(n, 0) := 0; a(1, 1)=1.
%F a(n,m) = (m/n) * 4^(n-m) * Sum_{k=1..n-m} binomial(n+k-1, n-1) * Sum_{j=0..k} binomial(j, n-m-3*k+2*j) * 4^(j-k) * binomial(k,j) * 3^(-n+m+3*k-j) * 2^(n-m-3*k+j) * (-1)^(n-m-3*k+2*j), n > m; a(n,n)=1. - _Vladimir Kruchinin_, Feb 08 2011
%t a[n_, n_] = 1; a[n_, m_] := m/n * 4^(n-m) * Sum[ Binomial[n+k-1, n-1] * Sum[ Binomial[j, n-m-3*k+2*j] * 4^(j-k) * Binomial[k, j] * 3^(-n+m+3*k-j) * 2^(n-m-3*k+j) * (-1)^(n-m-3*k+2*j), {j, 0, k}], {k, 1, n-m}]; Table[a[n, m], {n, 1, 9}, {m, 1, n}] // Flatten (* _Jean-François Alcover_, Jul 05 2013, after _Vladimir Kruchinin_ *)
%Y Cf. A048966. Row sums = A025757.
%K easy,nonn,tabl
%O 1,2
%A _Wolfdieter Lang_
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