A035324 - cp's OEIS frontend

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

Wolfdieter Lang

Replacing each '2' in the recurrence by '1' produces Pascal's triangle A007318(n-1,m-1). The columns appear as A001700, A008549, A045720, A045894, A035330, ...
Triangle T(n,k), 1 <= k <= n, given by (0, 3/1, 1/3, 5/3, 3/5, 7/5, 5/7, 9/7, 7/9, 11/9, 9/11, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Jan 28 2012
Riordan array (1, c(x)/sqrt(1-4x)) where c(x) = g.f. for Catalan numbers A000108, first column (k = 0) omitted. - Philippe Deléham, Jan 28 2012

			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.

Cf. A000108, A007318, A039598.
Row sums: A049027(n), n >= 1.
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

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

a(n+1, m) = 2*(2*n+m)*a(n, m)/(n+1) + m*a(n, m-1)/(n+1), n >= m >= 1; a(n, m) := 0, n

G.f. for column m: ((x*c(x)/sqrt(1-4*x))^m)/x, where c(x) = g.f. for Catalan numbers A000108.

a(n, m) =: s2(3; n, m).

With offset 0 (0 <= k <= n), T(n,k) = Sum_{j>=0} A039598(n,j)*binomial(j,k). - Philippe Deléham, Mar 30 2007

T(n+1,n) = 3*n = A008585(n).

T(n,k) = T(n-1,k-1) + 3*T(n-1,k) + Sum_{i>=0} T(n-1,k+1+i)*(-1)^i. - Philippe Deléham, Feb 23 2012

T(n,m) = Sum_{k=m..n} k*binomial(k-1,k-m)*2^(k-m)*binomial(2*n-k-1,n-k)/n. - Vladimir Kruchinin, Aug 07 2013

cp's OEIS Frontend

A035324 A convolution triangle of numbers, generalizing Pascal's triangle A007318.

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