login
A339411
Product of partial sums of odd squares.
0
1, 1, 10, 350, 29400, 4851000, 1387386000, 631260630000, 429257228400000, 415950254319600000, 553213838245068000000, 979741707532015428000000, 2253405927323635484400000000, 6591212337421633791870000000000, 24084289880938649875492980000000000, 108258883014819231190340945100000000000
OFFSET
0,3
FORMULA
a(n) = Product_{i=1..n} (Sum_{j=1..i} (2*j - 1)^2).
a(n) = Product_{i=1..n} binomial(2*i + 1, 3).
a(n) = Product_{i=1..n} A000447(i).
a(n) = ((2*n)! * (2*n+1)!) / (n! * 12^n).
a(n) / A135438(n) = A000108(n).
a(n) = (Gamma(2*n + 2)*Gamma(n + 1/2))/(3^n*sqrt(Pi)). - Peter Luschny, Dec 11 2020
D-finite with recurrence 3*a(n) -n*(2*n-1)*(2*n+1)*a(n-1)=0. - R. J. Mathar, Jan 25 2023
EXAMPLE
a(4) = (1^2) * (1^2 + 3^2) * (1^2 + 3^2 + 5^2) * (1^2 + 3^2 + 5^2 + 7^2) = 1 * 10 * 35 * 84 = 29400.
MAPLE
a:= proc(n) option remember;
`if`(n=0, 1, a(n-1)*(4*n^3-n)/3)
end:
seq(a(n), n=0..15); # Alois P. Heinz, Dec 03 2020
MATHEMATICA
Array[((2 #)!*(2 # + 1)!)/(#!*12^#) &, 16, 0] (* Michael De Vlieger, Dec 10 2020 *)
PROG
(PARI) for(n=0, 9, print((2*n)!*(2*n+1)!/(n!*12^n)))
(PARI) for(n=0, 9, print(prod(i=1, n, sum(j=1, i, (2*j-1)^2))))
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Werner Schulte, Dec 03 2020
STATUS
approved