OFFSET
0,4
COMMENTS
J(n) admits a closed form J(n) = binomial(2n,n)/2^(2n+3) * Pi^2 - C(n)/D(n), with gcd(C(n),D(n)) = 1.
The sequence here gives the numerators C(n).
These integers grow quickly and show a structured factorization with many primes inherited from odd double factorials.
The Python program given in the Links validates the closed form of the integral by computing it in two independent ways: Numerical integration using high-precision quadrature and closed form formula involving binomial coefficients and the rational sequences C(n), D(n).
Proof for even powers J(n) = Integral_{0..Pi/2} x*cos(x)^(2n) dx.
1) Discrete cosine expansion of cos(x)^(2n):
Using (e^{ix}+e^{-ix})/2 raised to 2n and pairing conjugates,
one gets the standard identity
cos(x)^(2n) = 2^(-2n) * [ binomial(2n,n)
+ 2 * Sum_{m=1..n} binomial(2n, n-m) * cos(2 m x) ].
2) Insert into the integral and integrate termwise:
J(n) = Integral_{0..Pi/2} x*cos(x)^(2n) dx
= 2^(-2n) * [ binomial(2n,n) * Integral_{0..Pi/2} x dx
+ 2 * Sum_{m=1..n} binomial(2n, n-m) * Integral_{0..Pi/2} x*cos(2 m x) dx ].
The constant-term integral is
Integral_{0..Pi/2} x dx = (Pi/2)^2 / 2 = Pi^2 / 8.
For m >= 1, by integration by parts (or a table),
Integral_{0..Pi/2} x*cos(2 m x) dx
= [ x*sin(2 m x)/(2m) ]_{0}^{Pi/2} + [ cos(2 m x)/(2m)^2 ]_{0}^{Pi/2}
= (cos(m*Pi) - 1) / (4 m^2)
= ((-1)^m - 1) / (4 m^2).
3) Only odd m contribute (even m give 0), and for odd m we have ((-1)^m - 1)/(4 m^2) = -1/(2 m^2).
Therefore
J(n) = 2^(-2n) * [ binomial(2n,n) * (Pi^2/8)
+ 2 * Sum_{m=1..n} binomial(2n, n-m) * ((-1)^m - 1)/(4 m^2) ]
= 2^(-2n) * [ binomial(2n,n) * (Pi^2/8)
- Sum_{m odd <= n} binomial(2n, n-m) * (1/m^2) ].
4) Final closed form:
J(n) = binomial(2n,n) / 2^(2n+3) * Pi^2
- (1/4^n) * Sum_{1 <= m <= n, m odd} binomial(2n, n-m) / m^2.
Writing the rational sum in lowest terms as C(n)/D(n) with gcd(C(n),D(n))=1 yields
J(n) = binomial(2n,n)/2^(2n+3) * Pi^2 - C(n)/D(n).
This proves the formula used to define the sequences C(n) (numerators) and D(n) (denominators).
LINKS
Patrick Demichel, Table of n, a(n) for n = 0..1000
Patrick Demichel, Python program
FORMULA
a(n) = numerator(R(n)) where R(n) = C(n)/D(n) = (1/4^n) * Sum_{1 <= k <= n, k odd} binomial(2n, n-k) / k^2.
J(n) = binomial(2n,n)/2^(2n+3) * Pi^2 - R(n).
O.g.f. : Sum_{n>=0} J(n) t^n = (1/sqrt(1 - t)) * ( Pi^2/8 - (1/2)( Li_2(r) - Li_2(-r) ) ), where r = 2(1 - sqrt(1 - t))/t.
PROG
(Python)
from math import comb
from fractions import Fraction
def R(n: int) -> Fraction:
S = Fraction(0, 1)
for k in range(1, n+1, 2):
S += Fraction(comb(2*n, n-k), k*k)
return Fraction(S, 4**n)
def C(n: int) -> int: return R(n).numerator
if __name__ == "__main__": print([C(n) for n in range(100)])
CROSSREFS
KEYWORD
nonn,frac
AUTHOR
Patrick Demichel, Sep 30 2025
EXTENSIONS
a(21) onward corrected by Sean A. Irvine, Nov 20 2025
STATUS
approved