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

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

1,2

Comments

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

Links

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.

Formula

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<m; a(n, 0) := 0, a(1, 1)=1;

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

Example

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,...).

Mathematica

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 *)

Prog

(Haskell)
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
(Sage)
@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

Crossrefs

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).

Sequence in context: A171509 A171505 A134283 * A171814 A091965 A171568

Adjacent sequences: A035321 A035322 A035323 * A035325 A035326 A035327

Keyword

easy,nice,nonn,tabl

Author

Wolfdieter Lang

Status

approved