A156894 a(n) = Sum_{k=0..n} binomial(n,k)*binomial(2*n+k-1,k).
1, 3, 19, 138, 1059, 8378, 67582, 552576, 4563235, 37972290, 317894394, 2674398268, 22590697614, 191475925332, 1627653567916, 13870754053388, 118464647799075, 1013709715774130, 8689197042438274, 74594573994750972, 641252293546113434, 5519339268476249676, 47558930664216470628
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Peter Bala, Notes on A156894
Programs
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Magma
A156894:= func< n | (&+[ Binomial(n,k)*Binomial(2*n+k-1,k): k in [0..n]]) >; [A156894(n): n in [0..30]]; // G. C. Greubel, Jan 06 2022
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Maple
a := n -> hypergeom([-n, 2*n], [1], -1); seq(round(evalf(a(n),32)), n=0..19); # Peter Luschny, Aug 02 2014
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Mathematica
Table[Sum[Binomial[n,k]Binomial[2n+k-1,k],{k,0,n}],{n,0,20}] (* Harvey P. Dale, Nov 12 2014 *) -
PARI
a(n) = if (n < 1, 1, sum(k=0, n, binomial(n,k)*binomial(2*n+k-1,k))); vector(50, n, a(n-1)) \\ Altug Alkan, Oct 05 2015
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Sage
[round( hypergeometric([-n, 2*n], [1], -1) ) for n in (0..30)] # G. C. Greubel, Jan 06 2022
Formula
a(n) = [x^n] ((1+x)/(1-x)^2)^n.
a(n) = (4*(n+1)*(2*n+1)*A003169(n+1) - (5*n+1)*(2*n-1)*A003169(n))/(17*n + 5) for n>0. - Mark van Hoeij, Jul 14 2010
a(n) = Hypergeometric2F1([-n, 2*n], [1], -1). - Peter Luschny, Aug 02 2014
Conjecture: 64*n*(2*n-1)*a(n) -16*(89*n^2 -134*n +63)*a(n-1) +4*(661*n^2 -2619*n +2576)*a(n-2) -3*(119*n^2 -713*n +1092)*a(n-3) +6*(2*n-7)*(n-4)*a(n-4) = 0. - R. J. Mathar, Feb 05 2015
Conjecture: 16*n*(782*n +5365)*(2*n-1)*a(n) +8*(3128*n^3 -362053*n^2 +593930*n -290328)*a(n-1) -3*(726869*n^3 -5105981*n^2 +11667946*n -8715544)*a(n-2) +158*(2*n-5)*(n-3)*(391*n -764)*a(n-3) = 0. - R. J. Mathar, Feb 05 2015
Conjecture: 4*n*(2*n-1)*(17*n^2 -52*n +39)*a(n) -(1207*n^4 -4899*n^3 +6692*n^2 -3504*n +576)*a(n-1) +2*(n-2)*(2*n-3)*(17*n^2 -18*n +4)*a(n-2) = 0. - R. J. Mathar, Feb 05 2015 [the Maple command sumrecursion (binomial(n,k) * binomial(2*n+k-1,k), k, a(n)) verifies this recurrence. - Peter Bala, Oct 05 2015 ]
a(n) ~ sqrt(578 + 306*sqrt(17)) * (71 + 17*sqrt(17))^n / (17 * sqrt(Pi*n) * 2^(4*n+2)). - Vaclav Kotesovec, Feb 05 2015
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 3*x + 14*x^2 + 79*x^3 + ... is the o.g.f. of A003169 (taken with offset 0). - Peter Bala, Oct 05 2015
From Peter Bala, Mar 20 2020: (Start)
More generally, we conjecture that a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. (End)