OFFSET
0,2
COMMENTS
Convolution of A000984(n) and 2^n*A000984(n). Convolution of A000984(n) and A059304. 4th binomial transform of A000984.
Largest coefficient of (1 + 6*x + x^2)^n. - Philippe Deléham, Oct 02 2007
Also number of paths from (0,0) to (n,0) using steps U=(1,1), H=(1,0) and D=(1,-1), the H steps can have 6 colors. - N-E. Fahssi, Mar 31 2008
Self-convolution of a(n)/4^n gives A126646. - Vladimir Reshetnikov, Oct 10 2016
Diagonal of rational function 1/(1 - (x^2 + 6*x*y + y^2)). - Gheorghe Coserea, Aug 03 2018
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..1000 (terms 0..200 from Vincenzo Librandi)
Hacène Belbachir, Abdelghani Mehdaoui and László Szalay, Diagonal Sums in the Pascal Pyramid, II: Applications, J. Int. Seq., Vol. 22 (2019), Article 19.3.5.
Francesc Fite, Kiran S. Kedlaya, Victor Rotger and Andrew V. Sutherland, Sato-Tate distributions and Galois endomorphism modules in genus 2, arXiv preprint arXiv:1110.6638 [math.NT], 2011 (the sequence b_{6,n}).
Rigoberto Flórez, Leandro Junes and José L. Ramírez, Further Results on Paths in an n-Dimensional Cubic Lattice, Journal of Integer Sequences, Vol. 21 (2018), Article 18.1.2.
Tony D. Noe, On the Divisibility of Generalized Central Trinomial Coefficients, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7.
FORMULA
G.f.: 1/sqrt(1 - 12*x + 32*x^2).
E.g.f.: exp(6*x)*BesselI(0, 2*x).
a(n) = Sum_{k=0..n} 2^k*binomial(2*k, k)*binomial(2*(n-k), n-k).
a(n) = Sum_{k=0..n} 4^(n-k)*binomial(n,k)*binomial(2k,k). - Paul Barry, Mar 08 2005
D-finite with recurrence: n*a(n) = 6*(2*n-1)*a(n-1) - 32*(n-1)*a(n-2). - Vaclav Kotesovec, Oct 15 2012
a(n) ~ 2^(3*n+1/2)/sqrt(Pi*n). - Vaclav Kotesovec, Oct 15 2012
a(n) = 4^n*hypergeometric([-n, 1/2], [1], -1). - Peter Luschny, May 19 2015
a(n) = Sum_{k=0..n} 8^(n-k) * (-1)^k * binomial(n,k) * binomial(2*k,k). - Seiichi Manyama, Apr 22 2019
a(n) = Sum_{k=0..floor(n/2)} 6^(n-2*k) * binomial(n,2*k) * binomial(2*k,k). - Seiichi Manyama, May 04 2019
From Peter Bala, Jan 10 2022: (Start)
3*x + x^2*exp(Sum_{n >= 1} a(n)*x^n/n) = 3*x + x^2 + 6*x^3 + 37*x^4 + 234*x^5 + 1514*x^6 + ... is the o.g.f. of A025230.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for prime p and positive integers n and k.
a(n) = (1/Pi) * Integral_{x = -1..1} (4 + 4*x^2)^n/sqrt(1 - x^2) dx = (1/Pi) * Integral_{x = -1..1} (8 - 4*x^2)^n/sqrt(1 - x^2) dx. (End)
EXAMPLE
G.f. = 1 + 6*x + 38*x^2 + 252*x^3 + 1734*x^4 + 12276*x^5 + 88796*x^6 + ...
MATHEMATICA
Table[SeriesCoefficient[1/(Sqrt[1-4*x]*Sqrt[1-8*x]), {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Oct 15 2012 *)
a[ n_] := If[n < 0, 0, 4^n Hypergeometric2F1[-n, 1/2, 1, -1]]; (* Michael Somos, May 06 2017 *)
a[ n_] := SeriesCoefficient[ D[ InverseJacobiSD[2 x, -1] / 2, x], {x, 0, 2 n}]; (* Michael Somos, May 06 2017 *)
PROG
(PARI) x='x+O('x^66); Vec(1/sqrt(1-12*x+32*x^2)) \\ Joerg Arndt, May 11 2013
(PARI) {a(n) = sum(k=0, n, 8^(n-k)*(-1)^k*binomial(n, k)*binomial(2*k, k))} \\ Seiichi Manyama, Apr 22 2019
(PARI) {a(n) = sum(k=0, n\2, 6^(n-2*k)*binomial(n, 2*k)*binomial(2*k, k))} \\ Seiichi Manyama, May 04 2019
(Sage)
a = lambda n: 4^n*hypergeometric([-n, 1/2], [1], -1)
[simplify(a(n)) for n in range(23)] # Peter Luschny, May 19 2015
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Sep 07 2004
STATUS
approved