A048966 A convolution triangle of numbers obtained from A025748.
1, 3, 1, 15, 6, 1, 90, 39, 9, 1, 594, 270, 72, 12, 1, 4158, 1953, 567, 114, 15, 1, 30294, 14580, 4482, 1008, 165, 18, 1, 227205, 111456, 35721, 8667, 1620, 225, 21, 1, 1741905, 867834, 287199, 73656, 15075, 2430, 294, 24, 1, 13586859, 6857136, 2328183, 623106, 136323, 24354, 3465, 372, 27, 1
Offset: 1
Examples
Triangle begins:
1;
3, 1;
15, 6, 1;
90, 39, 9, 1;
594, 270, 72, 12, 1;
4158, 1953, 567, 114, 15, 1;
Links
- Reinhard Zumkeller, Rows n = 1..125 of triangle, flattened
- W. Lang, On generalizations of Stirling number triangles, J. Integer Seqs., Vol. 3 (2000), #00.2.4.
Crossrefs
Programs
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Haskell
a048966 n k = a048966_tabl !! (n-1) !! (k-1) a048966_row n = a048966_tabl !! (n-1) a048966_tabl = [1] : f 2 [1] where f x xs = ys : f (x + 1) ys where ys = map (flip div x) $ zipWith (+) (map (* 3) $ zipWith (*) (map (3 * (x - 1) -) [1..]) (xs ++ [0])) (zipWith (*) [1..] ([0] ++ xs)) -- Reinhard Zumkeller, Feb 19 2014 -
Mathematica
a[n_, m_] /; n >= m >= 1 := a[n, m] = 3*(3*(n-1) - m)*a[n-1, m]/n + m*a[n-1, m-1]/n; a[n_, m_] /; n < m := 0; a[n_, 0] = 0; a[1, 1] = 1; Table[a[n, m], {n, 1, 10}, {m, 1, n}] // Flatten (* Jean-François Alcover, Apr 26 2011, after given formula *)
Formula
a(n, m) = 3*(3*(n-1)-m)*a(n-1, m)/n + m*a(n-1, m-1)/n, n >= m >= 1; a(n, m) := 0, n
G.f. for m-th column: ((1-(1-9*x)^(1/3))/3)^m.
a(n,m) = m/n * sum(k=0..n-m, binomial(k,n-m-k) * 3^k*(-1)^(n-m-k) * binomial(n+k-1,n-1)). - Vladimir Kruchinin, Feb 08 2011
Comments