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a(n) = Sum_{k=0..n} binomial(n,k) * binomial(n+1,k+1) * 3^k.
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%I #30 Sep 02 2024 13:04:51

%S 1,5,30,193,1286,8754,60460,421985,2968902,21019510,149572292,

%T 1068795930,7664092060,55121602436,397464604440,2872406652001,

%U 20799171328070,150869330458830,1096046132412628,7973709600124958,58081342410990516,423551998861478140

%N a(n) = Sum_{k=0..n} binomial(n,k) * binomial(n+1,k+1) * 3^k.

%H Vincenzo Librandi, <a href="/A098663/b098663.txt">Table of n, a(n) for n = 0..200</a>

%F G.f.: ((1+2*x) - sqrt(1-8*x+4*x^2))/(6*x*sqrt(1-8*x+4*x^2)).

%F E.g.f.: exp(4x)*(BesselI(0, 2*sqrt(3)*x) + BesselI(1, 2*sqrt(3)*x)/sqrt(3)).

%F Recurrence: (n+1)*(2*n-1)*a(n) = 2*(8*n^2-3)*a(n-1) - 4*(n-1)*(2*n+1)*a(n-2). - _Vaclav Kotesovec_, Oct 15 2012

%F a(n) ~ sqrt(12+7*sqrt(3))*(4+2*sqrt(3))^n/(3*sqrt(Pi*n)). - _Vaclav Kotesovec_, Oct 15 2012

%F a(n) = 3^n*hypergeom([-n, -n - 1], [1], 1/3). - _Detlef Meya_, May 21 2024

%F From _Peter Bala_, Sep 02 2024: (Start)

%F The following formulas assume an offset of 1 (i.e., a(1) = 1, a(2) = 5, etc.):

%F a(n) = (1/3) * [x^n] ((2*x - 1)/(1 + x))^n = (1/3) * A255688(n).

%F a(n) = (1/3) * Sum_{k = 0..n} binomial(n, k)*binomial(n+k-1, k)*2^(n-k).

%F a(n) = (1/3) * 2^n * hypergeom([n, -n], [1], -1/2).

%F The Gauss congruences a(n*p^r) == a(n*p^(r-1)) (mod p^r) hold for all primes p >= 5 and all positive integers n and r. (End)

%p seq(simplify(3^n*hypergeom([-n, -n-1], [1], 1/3)), n = 0..20); # _Peter Bala_, Sep 02 2024

%t Table[Sum[Binomial[n,k]Binomial[n+1,k+1]3^k,{k,0,n}],{n,0,20}] (* _Harvey P. Dale_, Nov 08 2011 *)

%t a[n_] := 3^n*HypergeometricPFQ[{-n, -n - 1}, {1}, 1/3]; Flatten[Table[a[n], {n,0,21}]] (* _Detlef Meya_, May 21 2024 *)

%o (PARI) my(x='x+O('x^66)); Vec(((1+2*x)-sqrt(1-8*x+4*x^2))/(6*x*sqrt(1-8*x+4*x^2))) \\ _Joerg Arndt_, May 12 2013

%Y Fourth binomial transform of A098662.

%Y Cf. A255688.

%K easy,nonn

%O 0,2

%A _Paul Barry_, Sep 20 2004