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a(n) = Sum of binomial(n,k)*binomial(2n+k,k) over all k.
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%I #29 Jan 27 2017 04:32:10

%S 1,4,26,190,1462,11584,93536,765314,6323270,52638760,440815036,

%T 3709445084,31340292076,265683004240,2258793820988,19251776923210,

%U 164440378882630,1407266585304760,12063701803046300,103571977632247076

%N a(n) = Sum of binomial(n,k)*binomial(2n+k,k) over all k.

%C Modification of A001850 inspired by the Apéry numbers A005259.

%C From _Paul Barry_, Feb 17 2009: (Start)

%C Central coefficient of (1 + 4x + 5x^2 + 2x^3)^n. The coefficients are the 4th row of A029635.

%C The third row of A029635 corresponds to the central Delannoy numbers A001850. (End)

%H G. C. Greubel, <a href="/A114496/b114496.txt">Table of n, a(n) for n = 0..1000</a>

%H P. Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL16/Barry2/barry231.html">A Note on a Family of Generalized Pascal Matrices Defined by Riordan Arrays</a>, Journal of Integer Sequences, 16 (2013), #13.5.4.

%F a(n) = Sum_{k=0..n} (binomial(n,k)*binomial(2n+k,k)).

%F Recurrence: 20*n*(2*n - 1)*a(n) = (371*n^2 - 411*n + 120)*a(n-1) -2*(81*n^2 - 299*n + 278)*a(n-2) + 4*(n-2)*(2*n-5)*a(n-3). - _Vaclav Kotesovec_, Oct 19 2012

%F a(n) ~ sqrt(1734 + 442*sqrt(17))*((71 + 17*sqrt(17))/16)^n/(68*sqrt(Pi*n)). - _Vaclav Kotesovec_, Oct 19 2012

%F From _Peter Bala_, Oct 05 2015: (Start)

%F a(n) = Sum_{i = 0..n} 2^(n-i)*binomial(2*n,i)*binomial(n,i).

%F 4*n*(2*n - 1)*(17*n - 23)*a(n) = (1207*n^3 - 2840*n^2 + 1897*n - 360)*a(n-1) - 2*(n - 1)*(17*n - 6)*(2*n - 3)*a(n-2) with a(0) = 1 and a(1) = 4.

%F 1 + x*exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 4*x^2 + 21*x^3 + 126*x^4 + ... is the o.g.f. for A003168. (End)

%t Table[Sum[Binomial[n, k]*Binomial[2n+k, k], {k, 0, n}], {n,0,25}]

%o (PARI) a(n) = sum(k=0, n, 2^(n-k)*binomial(2*n,k)*binomial(n,k));

%o vector(50, n, a(n-1)) \\ _Altug Alkan_, Oct 05 2015

%Y Cf. A114497, A114498, A003168, A156894.

%Y Cf. A156886. - _Paul Barry_, Feb 17 2009

%K nonn,easy

%O 0,2

%A _Eric Rowland_, Dec 01 2005