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A110255
Numerators in the fractional coefficients that form the partial quotients of the continued fraction representation of the inverse tangent of 1/x.
7
1, 3, 5, 28, 81, 704, 325, 768, 20825, 311296, 83349, 1507328, 1334025, 3145728, 5337189, 130023424, 1366504425, 7516192768, 5466528925, 12884901888, 87470372561, 2954937499648, 349899121845, 12919261626368, 22394407746529
OFFSET
1,2
COMMENTS
Limit a(2*n-1)/A110256(2*n-1) = limit A110257(n)/A110258(n) = 4/Pi.
Limit a(2*n)/A110256(2*n) = limit A110259(n)/A110260(n) = Pi.
LINKS
EXAMPLE
arctan(1/x) = 1/x - 1/(3*x^3) + 1/(5*x^5) - 1/(7*x^7) +-...
= [0; x, 3*x, 5/4*x, 28/9*x, 81/64*x, 704/225*x, 325/256*x,
768/245*x, 20825/16384*x, 311296/99225*x, 83349/65536*x,
1507328/480249*x, 1334025/1048576*x, 3145728/1002001*x,...]
= 1/(x + 1/(3*x + 1/(5/4*x + 1/(28/9*x + 1/(81/64*x +...))))).
The coefficients of x in the even-indexed partial quotients converge to Pi:
{3, 28/9, 704/225, 768/245, 311296/99225, ...}.
The coefficients of x in the odd-indexed partial quotients converge to 4/Pi:
{1, 5/4, 81/64, 325/256, 20825/16384, ...}.
PROG
(PARI) {a(n)=numerator(subst((contfrac( sum(k=0, n, (-1)^k/x^(2*k+1)/(2*k+1)), n+1))[n+1], x, 1))}
CROSSREFS
Cf. A110256 (denominators), A110257/A110258 (odd-indexed), A110259/A110260 (even-indexed).
Sequence in context: A163783 A103526 A054388 * A290330 A300676 A016552
KEYWORD
cofr,frac,nonn
AUTHOR
Paul D. Hanna, Jul 18 2005
STATUS
approved