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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

Table of n, a(n) for n=0..44.

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

Cf. A000079, A001147, A048993, A187075, A211402, A142963

Sequence in context: A225443 A222060 A256549 * A058175 A112906 A137375

Adjacent sequences:  A211605 A211606 A211607 * A211609 A211610 A211611

KEYWORD

easy,nonn,tabl

AUTHOR

Philippe Deléham, Feb 10 2013

STATUS

approved

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Last modified February 21 12:59 EST 2018. Contains 299411 sequences. (Running on oeis4.)