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%I #28 Dec 12 2021 20:36:40
%S 3,28,704,768,311296,1507328,3145728,130023424,7516192768,12884901888,
%T 2954937499648,12919261626368,52776558133248,774056185954304,
%U 66428094503714816,31525197391593472,308982963234634989568
%N Numerators in the coefficients that form the even-indexed partial quotients of the continued fraction representation of the inverse tangent of 1/x.
%C Lim_{n->infinity} a(n)/A110260(n) = lim_{n->infinity} A110255(2*n)/A110256(2*n) = Pi.
%C The continued fraction expansion arctan(z) = z/(1 + z^2/(3 + 4*z^2/(5 + 9*z^2/(7 + ...)))) is due to Lambert - see Roegel, Section 1.2.1. - _Peter Bala_, Dec 02 2021
%H Paul D. Hanna, <a href="/A110259/b110259.txt">Table of n, a(n) for n = 1..200</a>
%H Denis Roegel, <a href="https://hal.archives-ouvertes.fr/hal-02984214/document">Lambert’s proof of the irrationality of Pi: Context and translation</a>, [Research Report] LORIA, 2020, hal-02984214.
%F a(n) = A110255(2*n).
%e arctan(1/x) = 1/x - 1/(3*x^3) + 1/(5*x^5) - 1/(7*x^7) +- ...
%e arctan(1/x) = [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, ...]
%e arctan(1/x) = 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: {3, 28/9, 704/225, 768/245, 311296/99225, ...}.
%e The coefficients of x in the odd-indexed partial quotients converge to 4/Pi: {1, 5/4, 81/64, 325/256, 20825/16384, ...}.
%e From _Peter Bala_, Dec 02 2021: (Start)
%e Making use of the expansion 2*arcsin(sqrt(x)/2)^2 = Sum_{n >= 1} x^n/ (n^2*binomial(2*n,n)) we calculate
%e 3 + Pi = Sum_{n >= 1} (2^n)*n/binomial(2*n,n);
%e 28 + 9*Pi = Sum_{n >= 3} (2^n)*n*(n-1)*(n-2)/binomial(2*n,n);
%e 704 + 225*Pi = Sum_{n >= 5} (2^n)*n*(n-1)*...*(n-4)/binomial(2*n,n);
%e 45*(768 + 245*Pi) = Sum_{n >= 7} (2^n)*n*(n-1)*...*(n-6)/binomial(2*n,n);
%e 9*(311296 + 99225*Pi) = Sum_{n >= 9} (2^n)*n*(n-1)*...*(n-8)/ binomial(2*n,n).
%e It appears that Sum_{n >= 2*k+1} (2^n)*n*(n-1)*...*(n-2*k)/binomial(2*n,n) = N(2*k) + D(2*k)*Pi, where the ratios N(2*k)/D(2*k) are equal to the even-indexed partial quotients of Lambert's continued fraction representation of the inverse tangent of 1/x. (End)
%o (PARI) {a(n)=numerator(subst((contfrac( sum(k=0,2*n+2,(-1)^k/x^(2*k+1)/(2*k+1)),2*n+2))[2*n+1],x,1))}
%Y Cf. A110260 (denominators), A110255/A110256 (continued fraction), A110257/A110258.
%K nonn,frac
%O 1,1
%A _Paul D. Hanna_, Jul 18 2005
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