A156894 - OEIS

%I #45 Sep 08 2022 08:45:41

%S 1,3,19,138,1059,8378,67582,552576,4563235,37972290,317894394,

%T 2674398268,22590697614,191475925332,1627653567916,13870754053388,

%U 118464647799075,1013709715774130,8689197042438274,74594573994750972,641252293546113434,5519339268476249676,47558930664216470628

%N a(n) = Sum_{k=0..n} binomial(n,k)*binomial(2*n+k-1,k).

%H G. C. Greubel, <a href="/A156894/b156894.txt">Table of n, a(n) for n = 0..1000</a>

%H Peter Bala, <a href="/A156894/a156894.pdf">Notes on A156894</a>

%F a(n) = [x^n] ((1+x)/(1-x)^2)^n.

%F 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

%F a(n) = Hypergeometric2F1([-n, 2*n], [1], -1). - _Peter Luschny_, Aug 02 2014

%F 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

%F 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

%F 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 ]

%F 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

%F 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

%F From _Peter Bala_, Mar 20 2020: (Start)

%F a(p) == 3 ( mod p^3 ) for prime p >= 5. Cf. A002003, A103885 and A119259.

%F 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)

%p a := n -> hypergeom([-n, 2*n], [1], -1);

%p seq(round(evalf(a(n),32)), n=0..19); # _Peter Luschny_, Aug 02 2014

%t Table[Sum[Binomial[n,k]Binomial[2n+k-1,k],{k,0,n}],{n,0,20}] (* _Harvey P. Dale_, Nov 12 2014 *)

%o (PARI) a(n) = if (n < 1, 1, sum(k=0, n, binomial(n,k)*binomial(2*n+k-1,k)));

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

%o (Magma)

%o A156894:= func< n | (&+[ Binomial(n,k)*Binomial(2*n+k-1,k): k in [0..n]]) >;

%o [A156894(n): n in [0..30]]; // _G. C. Greubel_, Jan 06 2022

%o (Sage) [round( hypergeometric([-n, 2*n], [1], -1) ) for n in (0..30)] # _G. C. Greubel_, Jan 06 2022

%Y Cf. A002003, A003169, A103885, A114496, A123164.

%K easy,nonn

%O 0,2

%A _Paul Barry_, Feb 17 2009