OFFSET
0,2
COMMENTS
This is not the sequence A025547(n+1)_{n>=0}, because a(32) = 1420993851085122917681925 but A025547(33) = 18472920064106597929865025. Hence it is also not the sequence A350670.
The alternating sum Sum_{k=0..n} (-1)^k/(2*k+1) = (Psi(n + 3/2) - Psi((2*n - (-1)^n)/4 + 1) - log(2) + Pi/2)/2, with the Digamma function Psi(z).
Proof by subtracting twice the negative fractions from Sum_{k=0..n} 1/(2*k+1) = A350669(n)/A350670(n) (Abramowitz-Stegun, p. 258, eq. 6.3.4.), using Sum_{j=0..k} 1/(4*j + 3) = A074637((k+1)/4)/A074638(k+1) (Abramowitz-Stegun, p. 258, eqs. 6.3.6. with 6.3.5.) and, finally, replacing in the results for the even and odd n cases the formula for Psi(3/4) = -A200134.
LINKS
Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions. p.258, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy]. p. 258.
FORMULA
a(n) = denominator( (Psi(n + 3/2) - Psi((2*n - (-1)^n)/4 + 1) - log(2) + Pi/2)/2), for n >= 0, with the Digamma function. See the above comment.
a(n) = denominator(Pi/4 + (-1)^n * (Psi((n + 5/2)/2) - Psi((n + 3/2)/2))/4). - Vaclav Kotesovec, May 16 2022
MATHEMATICA
Denominator @ Accumulate @ Table[(-1)^k/(2*k + 1), {k, 0, 25}] (* Amiram Eldar, Apr 08 2022 *)
PROG
(PARI) a(n) = denominator(sum(k=0, n, (-1)^k / (2*k+1))); \\ Michel Marcus, Apr 07 2022
(Python)
from fractions import Fraction
def A352395(n): return sum(Fraction(-1 if k % 2 else 1, 2*k+1) for k in range(n+1)).denominator # Chai Wah Wu, May 18 2022
CROSSREFS
KEYWORD
nonn,frac,easy
AUTHOR
Wolfdieter Lang, Apr 06 2022
STATUS
approved