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A118971 a(n) = 4*binomial(5*n+3,n)/(4*n+4). 16

%I

%S 1,4,26,204,1771,16380,158224,1577532,16112057,167710664,1772645420,

%T 18974357220,205263418941,2240623268512,24648785802336,

%U 272994644359580,3041495503591365,34064252968167180,383302465665133014

%N a(n) = 4*binomial(5*n+3,n)/(4*n+4).

%C A quadrisection of A118968.

%H Michael De Vlieger, <a href="/A118971/b118971.txt">Table of n, a(n) for n = 0..924</a>

%H Paul Barry, <a href="https://arxiv.org/abs/2001.08799">Characterizations of the Borel triangle and Borel polynomials</a>, arXiv:2001.08799 [math.CO], 2020.

%H Elżbieta Liszewska, Wojciech Młotkowski, <a href="https://arxiv.org/abs/1907.10725">Some relatives of the Catalan sequence</a>, arXiv:1907.10725 [math.CO], 2019.

%F G.f.: If the inverse series of y*(1-y)^4 is G(x) then A(x)=G(x)/x.

%F 8*(4*n+1)*(2*n+1)*(4*n+3)*(n+1)*a(n) -5*(5*n+1)*(5*n+2)*(5*n+3)*(5*n-1)*a(n-1)=0. - _R. J. Mathar_, Nov 26 2012

%F a(n) = (4/5)*binomial(5*(n+1),n+1)/(5*(n+1)-1). [_Bruno Berselli_, Jan 17 2014]

%F E.g.f.: 4F4(4/5,6/5,7/5,8/5; 5/4,3/2,7/4,2; 3125*x/256). - _Ilya Gutkovskiy_, Jan 23 2018

%F G,f.: 5F4(4/5, 5/5, 6/5, 7/5, 8/5; 5/4, 6/4, 7/4, 8/4; (5^5/4^4)*x). - _Wolfdieter Lang_, Feb 06 2020

%t Table[4*Binomial[5n+3,n]/(4n+4),{n,0,30}] (* _Harvey P. Dale_, Apr 09 2012 *)

%Y Cf. A006632, A130564, A130565, A234466 (members of the same family).

%K easy,nonn

%O 0,2

%A _Paul Barry_, May 07 2006

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Last modified August 11 23:45 EDT 2020. Contains 336434 sequences. (Running on oeis4.)