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 A141222 Expansion of -1/(2*x)+(2*x-1)^2/(2*x*(1-4x)^(3/2)). 5
 1, 5, 22, 95, 406, 1722, 7260, 30459, 127270, 529958, 2200276, 9111830, 37650172, 155266100, 639191160, 2627302995, 10784089350, 44208873390, 181025067300, 740483276610, 3026059513620, 12355464845100 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS Apply Riordan array (1/sqrt(1-4x), xc(x)) to A131056, c(x) the g.f. of A000108. Apply Riordan array (c(x)/sqrt(1-4*x),x*c(x)^2) to A131055. Hankel transform appears to be (-1)^n*A085046(n). Central coefficients T(2n,n) of triangle A103450. [Emanuele Munarini, Jun 01 2012] LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..200 FORMULA a(n)=sum{k=0..n, (1+(k+1)2^(k-1)-0^k/2)*C(2n-k,n-k)}; a(n)=sum{k=0..n, C(2n,k)*C(n+1,2n-k)}; Equals the Narayana transform (A001263) of integer squares. - Gary W. Adamson, Jul 29 2011 Conjecture: (n+1)*a(n) +2*(-3*n-1)*a(n-1) +4*(2*n-3)*a(n-2)=0. - R. J. Mathar, Nov 24 2012 From Vaclav Kotesovec, Feb 13 2014: (Start) G.f.: -1/(2*x)+(2*x-1)^2/(2*x*(1-4x)^(3/2)). a(n) = (1+3*n+n^2) * C(2*n,n) / (n+1). Recurrence: (n+1)*(n^2 + n - 1)*a(n) = 2*(2*n-1)*(n^2 + 3*n + 1)*a(n-1). (End) MATHEMATICA Table[((1+3*n+n^2)*Binomial[2*n, n])/(n+1), {n, 0, 20}] (* Vaclav Kotesovec, Feb 13 2014 *) CoefficientList[Series[-1/(2*x)+(2*x-1)^2/(2*x*(1-4x)^(3/2)), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 13 2014 *) PROG (Maxima) a(n):=sum(binomial(2*n, k)*binomial(n+1, 2*n-k), k, 0, n); makelist(a(n), n, 0, 40); /* Emanuele Munarini, Jun 01 2012 */ CROSSREFS Sequence in context: A128746 A049675 A053154 * A127360 A116415 A026861 Adjacent sequences:  A141219 A141220 A141221 * A141223 A141224 A141225 KEYWORD easy,nonn AUTHOR Paul Barry, Jun 14 2008 EXTENSIONS Name of the sequence corrected by Vaclav Kotesovec, Feb 13 2014 STATUS approved

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Last modified September 22 00:42 EDT 2020. Contains 337276 sequences. (Running on oeis4.)