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%I #14 Nov 12 2012 23:39:38
%S 1,1,4,9,64,225,256,245,16384,99225,65536,480249,1048576,1002001,
%T 4194304,41409225,1073741824,2393453205,4294967296,4102737925,
%U 68719476736,940839860961,274877906944,4113258565689,17592186044416
%N Denominators in the fractional coefficients that form the partial quotients of the continued fraction representation of the inverse tangent of 1/x.
%C Limit A110255(2*n-1)/a(2*n-1) = limit A110257(n)/A110258(n) = 4/Pi.
%C Limit A110255(2*n)/a(2*n) = limit A110259(n)/A110260(n) = Pi.
%H Paul D. Hanna, <a href="/A110256/b110256.txt">Table of n, a(n) for n = 1..400</a>
%e arctan(1/x) = 1/x - 1/(3*x^3) + 1/(5*x^5) - 1/(7*x^7) +-...
%e = [0; x, 3*x, 5/4*x, 28/9*x, 81/64*x, 704/225*x, 325/256*x,
%e 768/245*x, 20825/16384*x, 311296/99225*x, 83349/65536*x,
%e 1507328/480249*x, 1334025/1048576*x, 3145728/1002001*x,...]
%e = 1/(x + 1/(3*x + 1/(5/4*x + 1/(28/9*x + 1/(81/64*x +...))))).
%e The coefficients of x in the even-indexed partial quotients converge to Pi:
%e {3, 28/9, 704/225, 768/245, 311296/99225, ...}.
%e The coefficients of x in the odd-indexed partial quotients converge to 4/Pi:
%e {1, 5/4, 81/64, 325/256, 20825/16384, ...}.
%o (PARI) {a(n)=denominator(subst((contfrac( sum(k=0,n,(-1)^k/x^(2*k+1)/(2*k+1)),n+1))[n+1],x,1))}
%Y Cf. A110255 (numerators), A110257/A110258 (odd-indexed), A110259/A110260 (even-indexed).
%Y Cf. A095175. [From _R. J. Mathar_, Aug 18 2008]
%K cofr,frac,nonn
%O 1,3
%A _Paul D. Hanna_, Jul 18 2005
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