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A254619 a(n) = 4^n*(2*n + 1)!/n!.

%I A254619 #11 Oct 02 2021 17:29:01
%S A254619 1,24,960,53760,3870720,340623360,35424829440,4250979532800,
%T A254619 578133216460800,87876248902041600,14763209815542988800,
%U A254619 2716430606059909939200,543286121211981987840000,117349802181788109373440000
%N A254619 a(n) = 4^n*(2*n + 1)!/n!.
%F A254619 E.g.f.: 1/(1 - 16*x)^(3/2) = 1 + 24*x + 960*x^2/2! + 53760*x^3/3! + ....
%F A254619 Recurrence equation: a(n) = 8*(2*n + 1)*a(n-1) with a(0) = 1.
%F A254619 2nd order recurrence equation: a(n) = (20*n + 6)*a(n-1) - 16*(2*n - 1)^2*a(n-2) with a(0) = 1, a(1) = 24.
%F A254619 Define a sequence b(n) := a(n)*sum {k = 0..n} 1/((2*k + 1)*4^k) beginning [1, 26, 1052, 59032, 4251984, 374204832, 38917967808, ...]. It is not difficult to check that b(n) also satisfies the previous 2nd order recurrence equation (and so is an integer sequence). From this observation we can obtain the continued fraction expansion
%F A254619 log(3) = Sum {k >= 0} 1/((2*k + 1)*4^k) = 1 + 2/(24 - 16*3^2/(46 - 16*5^2/(66 - ... - 16*(2*n - 1)^2/((20*n + 6) - ... )))).
%F A254619 Alternative 2nd order recurrence equation: a(n) = (12*n + 10)*a(n-1) + 16*(2*n - 1)^2*a(n-2) with a(0) = 1, a(1) = 24.
%F A254619 Define now a sequence c(n) := a(n)*sum {k = 0..n} (-1)^k/((2*k + 1)*4^k) beginning [1, 22, 892, 49832, 3589584, 315853152, 32849393088, ...], which, along with a(n), satisfies the alternative 2nd order recurrence equation. From this observation we find the continued fraction expansion 2*arctan(1/2) = Sum {k >= 0} (-1)^k/((2*k + 1)*4^k) = 1 - 2/(24 + 16*3^2/(34 + 16*5^2/(46 + ... + 16*(2*n - 1)^2/((12*n + 10) + ... )))). Cf. A254381 and A254620.
%p A254619 seq(4^n*(2*n + 1)!/n!, n = 0..13);
%t A254619 Table[4^n (2n+1)!/n!,{n,0,20}] (* _Harvey P. Dale_, Oct 02 2021 *)
%Y A254619 Cf. A002391, A034910, A073000, A254381, A254620.
%K A254619 nonn,easy
%O A254619 0,2
%A A254619 _Peter Bala_, Feb 03 2015