login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A254619
a(n) = 4^n*(2*n + 1)!/n!.
4
1, 24, 960, 53760, 3870720, 340623360, 35424829440, 4250979532800, 578133216460800, 87876248902041600, 14763209815542988800, 2716430606059909939200, 543286121211981987840000, 117349802181788109373440000
OFFSET
0,2
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.
MAPLE
seq(4^n*(2*n + 1)!/n!, n = 0..13);
MATHEMATICA
Table[4^n (2n+1)!/n!, {n, 0, 20}] (* Harvey P. Dale, Oct 02 2021 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Feb 03 2015
STATUS
approved