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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 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 August 12 09:17 EDT 2020. Contains 336438 sequences. (Running on oeis4.)