A110257 Numerators in the coefficients that form the odd-indexed partial quotients of the continued fraction representation of the inverse tangent of 1/x.

1, 5, 81, 325, 20825, 83349, 1334025, 5337189, 1366504425, 5466528925, 87470372561, 349899121845, 22394407746529, 89580335298125, 1433319858545625, 5733391194015525, 5871086572691471625

Offset

1,2

Comments

Limit a(n)/A110258(n) = limit A110255(2*n-1)/A110256(2*n-1) = 4/Pi.

Links

Paul D. Hanna, Table of n, a(n) for n = 1..200

Formula

a(n) = A110255(2*n-1).

a(n) = (4*n+1)*A002894(n)/4^A000120(n). - Peter Luschny, Mar 23 2014

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, ...}.

Maple

a := n -> (4*n+1)*binomial(2*n, n)^2/4^(add(i, i=convert(n, base, 2)));
seq(a(n), n=0..16);  # Peter Luschny, Mar 23 2014

Mathematica

a[n_] := (4n+1) Binomial[2n, n]^2 / 4^DigitCount[n, 2, 1];
Array[a, 16] (* Jean-François Alcover, Jun 13 2019, from Maple *)

Prog

(PARI) {a(n)=numerator(subst((contfrac( sum(k=0, 2*n+1, (-1)^k/x^(2*k+1)/(2*k+1)), 2*n+2))[2*n], x, 1))}

Crossrefs

Cf. A110258 (denominators), A110255/A110256 (continued fraction), A110259/A110260.

Sequence in context: A189443 A275091 A275347 * A370197 A335177 A135918

Adjacent sequences: A110254 A110255 A110256 * A110258 A110259 A110260

Keyword

frac,nonn

Author

Paul D. Hanna, Jul 18 2005

Status

approved