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A211608 Triangle T(n,k), 0 <= k <= n, given by (0, 1, 0, 2, 0, 3, 0, 4, 0, 5, ...) DELTA (1, 2, 3, 4, 5, 6, 7, 8, 9, ...) where DELTA is the operator defined in A084938. 1
1, 0, 1, 0, 1, 3, 0, 1, 9, 15, 0, 1, 21, 90, 105, 0, 1, 45, 375, 1050, 945, 0, 1, 93, 1350, 6825, 14175, 10395, 0, 1, 189, 4515, 36750, 132300, 218295, 135135, 0, 1, 381, 14490, 178605, 992250, 2765070, 3783780, 2027025 (list; table; graph; refs; listen; history; text; internal format)
OFFSET
0,6
LINKS
K. N. Boyadzhiev, Series with central binomial coefficients, Catalan numbers, and harmonic numbers, J. Integer Seq. 15 (2012), Article 12.1.7.
FORMULA
T(n,k) = A048993(n,k)*A001147(k).
T(n,k) = A211402(n,k)/(2^(n-k)).
T(n,k) = k*T(n-1,k) + (2*k-1)*T(n-1,k-1), T(0,0) = 1, T(n,k) = 0 if k<0 or if k>n.
G.f.: F(x,t) = 1 + x*t + (x+3*x^2)*t^2/2! + (x+9*x^2+15*x^3)*t^3/3! + ... = Sum_{n = 0..inf} R(n,x)* t^n/n!.
The row polynomials R(n,x) satisfy the recursion R(n+1,x) = (x+2*x^2)*R'(n,x) + x*R(n,x) where ' indicates differentiation with respect to x.
R(n,x) = 1/sqrt(1 + 2*x)*Sum_{k >= 0} binomial(2*k,k)/2^k*k^n * x^k/(1 + 2*x)^k (see Boyadzhiev, eqn. 19). - Peter Bala, Jan 18 2018
EXAMPLE
Triangle begins :
1
0, 1
0, 1, 3
0, 1, 9, 15
0, 1, 21, 90, 105
0, 1, 45, 375, 1050, 945
0, 1, 93, 1350, 6825, 14175, 10395
CROSSREFS
Sequence in context: A225443 A222060 A256549 * A058175 A353010 A112906
KEYWORD
easy,nonn,tabl
AUTHOR
Philippe Deléham, Feb 10 2013
STATUS
approved

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Last modified June 12 22:58 EDT 2024. Contains 373360 sequences. (Running on oeis4.)