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%I #36 Feb 16 2017 11:43:06
%S 1,12,176,2688,41728,653312,10280960,162332672,2569207808,40732459008,
%T 646621167616,10275491151872,163421593010176,2600786039144448,
%U 41413155926048768,659738836733001728,10514182345513238528,167619477382960775168
%N a(n) = 2^(2*n-1)*binomial(2*n,n) + 2^(4*n-1).
%C 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
%D The right-hand side of a binomial coefficient identity in H. W. Gould, Combinatorial Identities, Morgantown, 1972, (3.97), page 33.
%H G. C. Greubel, <a href="/A045952/b045952.txt">Table of n, a(n) for n = 0..825</a>
%H H. W. Gould, ed. J. Quaintance, <a href="http://www.math.wvu.edu/~gould/Vol.4.PDF">Combinatorial Identities</a>, May 2010, (7.5) p. 34.
%F a(n) + A055787(n) = 2^(4n). - _David Scambler_, Aug 13 2012
%F 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]
%F 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
%F a(n) = 2^(4*n-1)*((Gamma(n+1/2))/(sqrt(Pi)*Gamma(n+1))+1). - _Alexander R. Povolotsky_, Aug 13 2012
%F 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).
%F a(n) = Sum_{k=0..n}(2^(2*n) * C(2*n, k)). - _David Scambler_, Aug 13 2012
%t Table[2^(2n-1) Binomial[2n,n]+2^(4n-1),{n,0,30}] (* _Harvey P. Dale_, Mar 14 2013 *)
%o (PARI) for(n=0,25, print1(2^(2*n-1)*binomial(2*n,n) + 2^(4*n-1), ", ")) \\ _G. C. Greubel_, Feb 16 2017
%Y Cf. A055787.
%K nonn
%O 0,2
%A _N. J. A. Sloane_, Jul 15 2000
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