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%I #20 Mar 21 2024 14:24:43
%S 1,4,41,488,8881,176556,4622745,128838480,4403082465,157917434580,
%T 6659489632905,292097166060600,14653855170875025,759940716395000700,
%U 44202442040567948025,2645857155729629066400,175060715455871850866625
%N a(n) = s(1)*s(2)*...*s(n)*(1/s(1) - 1/s(2) + ... + c/s(n)), where c = (-1)^(n+1) and s(k) = 4*k - 3 for k = 1, 2, 3, ....
%F a(n) ~ (Pi^(3/2) + 2*sqrt(Pi)*log(1 + sqrt(2))) * 2^(2*n - 2) * n^(n - 1/4) / (Gamma(1/4) * exp(n)). - _Vaclav Kotesovec_, Jan 02 2020
%F From _Peter Bala_, Mar 21 2024: (Start)
%F a(n) = Product_{k = 0..n} (4*k + 1) * Sum_{k = 0..n} (-1)^k/(4*k + 1).
%F a(n) = 4*a(n-1) + (4*n - 3)^2*a(n-2) with a(0) = 1 and a(1) = 4.
%F b(n) := Product_{k = 0..n} (4*k + 1) = A007696(n+1) satisfies the same 3-term recurrence with b(0) = 1 and b(1) = 5, leading to the continued fraction expansion for the constant A181048 = Sum_{k >= 0} (-1)^k/(4*k + 1) = 1/(1 + 1^2/(4 + 5^2/(4 + 9^2/(4 + 13^2/(4 + ... ))))) due to Euler. (End)
%p a := proc(n) option remember; if n = 0 then 1 elif n = 1 then 4 else 4*a(n-1) + (4*n - 3)^2*a(n-2) end if; end:
%p seq(a(n), n = 0..20);
%t Table[Product[4*k - 3, {k, 1, n}] * Sum[(-1)^(k+1)/(4*k - 3), {k, 1, n}], {n, 1, 20}] (* _Vaclav Kotesovec_, Jan 02 2020 *)
%o (PARI) a(n) = prod(k=1, n, 4*k-3)*sum(k=1, n, (-1)^(k+1)/(4*k-3)); \\ _Michel Marcus_, Jul 06 2019
%Y Cf. A024199, A024217, A024396, A007696, A181048.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_
%E More terms from _Sean A. Irvine_, Jul 06 2019
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