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%I #13 Mar 15 2023 12:40:57
%S 315,280,3675,116424,5885880,399072960,33129291195,3190228041000,
%T 344161801063080,40616781150254400,5155510596280207800,
%U 695029472211496161600,98570579229528369624000,14597207555235045670540800,2243893009052293495117018875,356344642367340570239409729000
%N a(n) = 2520*(4*n)!/(n!*(n+2)!^3).
%C Row 1 of A361032.
%C The central binomial numbers A000984(n) = (2*n)!/n!^2 have the property that 6*A000984(n) is divisible by (n + 1)*(n + 2) and the result (2*n)!/(n!*(n+2)!) is the super ballot number A007054(n). Similarly, the numbers A008977(n) = (4*n)!/n!^4 appear to have the property that 2520*A008977(n) is divisible by ((n + 1)*(n + 2))^3, leading to the present sequence. Cf. A361029.
%C Conjecture: a(n) is odd iff n = 2^k - 2 for some k >= 1.
%F a(n) = 2520*A008977(n)/((n+1)*(n+2))^3.
%F a(n) = (315/2)*A008977(n+2)/((4*n+1)*(4*n+2)*(4*n+3)*(4*n+5)*(4*n+6)*(4*n+7)).
%F P-recursive: a(n) = 4*(4*n-1)*(4*n-2)*(4*n-3)/(n+2)^3 * a(n-1) with a(0) = 315.
%F The o.g.f. A(x) satisfies the differential equation
%F x^3*(1 - 256*x)*A(x)''' + x^2*(9 - 1152*x)*A(x)'' + x*(19 - 816*x)*A(x)' + (8 - 24*x)*A(x) - 2520 = 0 with A(0) = 315, A'(0) = 280 and A''(0) = 7350.
%F a(n) ~ 630*sqrt(8/Pi^3) * 2^(8*n)/n^(15/2).
%p seq(2520*(4*n)!/(n!*(n+2)!^3), n = 0..20);
%Y Cf. A007054, A008977, A361029, A361032, A361033, A361035.
%K nonn,easy
%O 0,1
%A _Peter Bala_, Mar 01 2023
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