A254619 a(n) = 4^n*(2*n + 1)!/n!.
1, 24, 960, 53760, 3870720, 340623360, 35424829440, 4250979532800, 578133216460800, 87876248902041600, 14763209815542988800, 2716430606059909939200, 543286121211981987840000, 117349802181788109373440000
Offset: 0
Programs
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Maple
seq(4^n*(2*n + 1)!/n!, n = 0..13);
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Mathematica
Table[4^n (2n+1)!/n!,{n,0,20}] (* Harvey P. Dale, Oct 02 2021 *)
Formula
E.g.f.: 1/(1 - 16*x)^(3/2) = 1 + 24*x + 960*x^2/2! + 53760*x^3/3! + ....
Recurrence equation: a(n) = 8*(2*n + 1)*a(n-1) with a(0) = 1.
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.
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
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) - ... )))).
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.
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.