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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.
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%I #14 Jan 18 2018 13:22:45

%S 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,

%T 6825,14175,10395,0,1,189,4515,36750,132300,218295,135135,0,1,381,

%U 14490,178605,992250,2765070,3783780,2027025

%N 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.

%H K. N. Boyadzhiev, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL15/Boyadzhiev/boyadzhiev6.html">Series with central binomial coefficients, Catalan numbers, and harmonic numbers</a>, J. Integer Seq. 15 (2012), Article 12.1.7.

%F T(n,k) = A048993(n,k)*A001147(k).

%F T(n,k) = A211402(n,k)/(2^(n-k)).

%F 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.

%F 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!.

%F 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.

%F 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

%e Triangle begins :

%e 1

%e 0, 1

%e 0, 1, 3

%e 0, 1, 9, 15

%e 0, 1, 21, 90, 105

%e 0, 1, 45, 375, 1050, 945

%e 0, 1, 93, 1350, 6825, 14175, 10395

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

%K easy,nonn,tabl

%O 0,6

%A _Philippe Deléham_, Feb 10 2013