A045952 a(n) = 2^(2*n-1)*binomial(2*n,n) + 2^(4*n-1).

1, 12, 176, 2688, 41728, 653312, 10280960, 162332672, 2569207808, 40732459008, 646621167616, 10275491151872, 163421593010176, 2600786039144448, 41413155926048768, 659738836733001728, 10514182345513238528, 167619477382960775168

Offset

0,2

Comments

Number of 4-ary words of length 2n in which the combined count of 0's and 1's is not greater than n. - David Scambler, Aug 13 2012

References

The right-hand side of a binomial coefficient identity in H. W. Gould, Combinatorial Identities, Morgantown, 1972, (3.97), page 33.

Links

G. C. Greubel, Table of n, a(n) for n = 0..825

H. W. Gould, ed. J. Quaintance, Combinatorial Identities, May 2010, (7.5) p. 34.

Formula

a(n) + A055787(n) = 2^(4n). - David Scambler, Aug 13 2012

n*a(n) +8*(3-4*n)*a(n-1) +128*(2*n-3)*a(n-2)=0. - R. J. Mathar, Aug 13 2012 [Proof by inserting the binomial expression at all three places, converting everything to Gamma-functions, and let Maple 16 simplify to zero. Robert Israel, Aug 13 2012]

a(n) = 8/n*((2*n-1)*a(n-1)+2^(4*n-5)). Using this formula we can express a(n) and a(n-1) via a(n-2) and easily obtain the hypergeometric recurrence above. - Vladimir Shevelev, Aug 13 2012

a(n) = 2^(4*n-1)*((Gamma(n+1/2))/(sqrt(Pi)*Gamma(n+1))+1). - Alexander R. Povolotsky, Aug 13 2012

a(n) = Sum_{k=0..n}(C(4*k, 2*k) * C(4*n-4*k, 2*n-2*k)). (LHS of Gould's identity).

a(n) = Sum_{k=0..n}(2^(2*n) * C(2*n, k)). - David Scambler, Aug 13 2012

Mathematica

Table[2^(2n-1) Binomial[2n, n]+2^(4n-1), {n, 0, 30}] (* Harvey P. Dale, Mar 14 2013 *)

Prog

(PARI) for(n=0, 25, print1(2^(2*n-1)*binomial(2*n, n) + 2^(4*n-1), ", ")) \\ G. C. Greubel, Feb 16 2017

Crossrefs

Cf. A055787.

Sequence in context: A307549 A052208 A234531 * A227714 A358951 A363152

Adjacent sequences: A045949 A045950 A045951 * A045953 A045954 A045955

Keyword

nonn

Author

N. J. A. Sloane, Jul 15 2000

Status

approved