Original entry on oeis.org
0, 0, 0, 0, 1, 45, 1470, 43890, 1291815, 38710035, 1199167200, 38692476900, 1304976397725, 46070080281225, 1702810398539250, 65862570279255750, 2663551451057371875, 112503209942059311375, 4957166849516125744500
Offset: 1
a(6)=45 increasing ternary 5-forest with n=6 vertices: there are three such 5-forests (four one vertex trees together with any of the three different 2-vertex trees) each with binomial(6,2)= 15 increasing labelings. W. Lang, Sep 14 2007.
A035324
A convolution triangle of numbers, generalizing Pascal's triangle A007318.
Original entry on oeis.org
1, 3, 1, 10, 6, 1, 35, 29, 9, 1, 126, 130, 57, 12, 1, 462, 562, 312, 94, 15, 1, 1716, 2380, 1578, 608, 140, 18, 1, 6435, 9949, 7599, 3525, 1045, 195, 21, 1, 24310, 41226, 35401, 19044, 6835, 1650, 259, 24, 1, 92378, 169766, 161052, 97954, 40963, 12021, 2450
Offset: 1
Triangle begins:
1;
3, 1;
10, 6, 1;
35, 29, 9, 1;
126, 130, 57, 12, 1;
462, 562, 312, 94, 15, 1;
Triangle (0, 3, 1/3, 5/3, 3/5, ...) DELTA (1,0,0,0,0,0, ...) has an additional first column (1,0,0,...).
- Reinhard Zumkeller, Rows n = 1..120 of triangle, flattened
- Milan Janjić, Pascal Matrices and Restricted Words, J. Int. Seq., Vol. 21 (2018), Article 18.5.2.
- Wolfdieter Lang, On generalizations of Stirling number triangles, J. Integer Seqs., Vol. 3 (2000), #00.2.4.
- Wolfdieter Lang, First 10 rows.
Alternating row sums give
A000108 (Catalan numbers).
If offset 0 (n >= m >= 0): convolution triangle based on
A001700 (central binomial coeffs. of odd order).
-
a035324 n k = a035324_tabl !! (n-1) !! (k-1)
a035324_row n = a035324_tabl !! (n-1)
a035324_tabl = map snd $ iterate f (1, [1]) where
f (i, xs) = (i + 1, map (`div` (i + 1)) $
zipWith (+) ((map (* 2) $ zipWith (*) [2 * i + 1 ..] xs) ++ [0])
([0] ++ zipWith (*) [2 ..] xs))
-- Reinhard Zumkeller, Jun 30 2013
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a[n_, m_] /; n >= m >= 1 := a[n, m] = 2*(2*(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; Flatten[ Table[ a[n, m], {n, 1, 10}, {m, 1, n}]] (* Jean-François Alcover, Feb 21 2012, from first formula *)
-
@cached_function
def T(n, k):
if n == 0: return n^k
return sum(binomial(2*i-1, i)*T(n-1, k-i) for i in (1..k-n+1))
A035324 = lambda n,k: T(k, n)
for n in (1..8): print([A035324(n, k) for k in (1..n)]) # Peter Luschny, Aug 16 2016
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