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
Examples
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, ...}.
Links
- Paul D. Hanna, Table of n, a(n) for n = 1..200
Programs
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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
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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 *)
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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))}
Formula
a(n) = A110255(2*n-1).
Comments