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%I #16 Mar 30 2023 02:42:08
%S 3,35,462,6435,92378,1352078,20058300,300540195,4537567650,
%T 68923264410,1052049481860,16123801841550,247959266474052,
%U 3824345300380220,59132290782430712,916312070471295267,14226520737620288370
%N Bisection of A001700.
%F a(n) = binomial(4n+3, 2n+2). - _Emeric Deutsch_, Dec 09 2004
%F From _Peter Bala_, Mar 19 2023: (Start)
%F a(n) = (1/2)*Sum_{k = 0..2*n+2} binomial(2*n+2,k)^2.
%F a(n) = (1/2)*hypergeom([-2 - 2*n, -2 - 2*n], [1], 1).
%F a(n) = 2*(4*n + 1)*(4*n + 3)/((n + 1)*(2*n + 1)) * a(n-1). (End)
%F From _Peter Bala_, Mar 28 2023
%F a(n) = (1/(2*n + 2))*Sum_{k = 0..2*n+2} k*binomial(2*n+2,k)^2.
%F a(n) = 2*(n + 1)*hypergeom([-1 - 2*n, -1 - 2*n], [2], 1). (End)
%p a:=n->binomial(4*n+3,2*n+2): seq(a(n),n=0..19);
%Y Cf. A001700, A002458, A187364, A187365.
%K nonn,easy
%O 0,1
%A _N. J. A. Sloane_, Nov 20 2004
%E More terms from _Emeric Deutsch_, Dec 09 2004
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