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%I #8 Feb 24 2015 05:31:28
%S 2,9,54,441,4410,53361,747054,12006225,216112050,4334247225,
%T 95353438950,2292816782025,59613236332650,1671463434096225,
%U 50143903022886750,1606276360166472225,54613396245660055650,1967688541203928475625,74772164565749282073750
%N Numerators of the convergents of the generalized continued fraction 2 + 1^2/(4 + 3^2/(4 + 5^2/(4 + ... ))).
%C Brouncker gave the generalized continued fraction expansion 4/Pi = 1 + 1^2/(2 + 3^2/(2 + 5^2/(2 + ... ))). The sequence of convergents begins [1/1, 3/2, 15/13, 105/76, ... ]. The numerators of the convergents are in A001147, the denominators in A024199.
%C In extending Brouckner's result, Osler showed that 2 + 1^2/(4 + 3^2/(4 + 5^2/(4 + ... ))) = L^2/Pi, where L is the lemniscate constant A062539. The sequence of convergents to Osler's continued fraction begins [2/1, 9/4, 54/25, 441/200, 4410/2025, ...]. Here we list the (unreduced) numerators of these convergents. See A254796 for the sequence of denominators. See A254794 for the decimal expansion of L^2/Pi.
%H T. J. Osler, <a href="http://www.rowan.edu/colleges/csm/departments/math/facultystaff/osler/130%20Missing%20fractions%20in%20Brouncker.pdf">The missing fractions in Brouncker’s sequence of continued fractions for Pi</a>
%F a(2*n-1) = ( A008545(n) )^2 = ( Product {k = 0..n-1} 4*k + 3 )^2.
%F a(2*n) = (4*n + 2)*( A008545(n) )^2 = (4*n + 2)*( Product {k = 0..n-1} 4*k + 3 )^2.
%F a(n) = 4*a(n-1) + (2*n - 1)^2*a(n-2) with a(0) = 2, a(1) = 9.
%F a(2*n) = (4*n + 2)*a(2*n-1); a(2*n+1) = (4*n + 4)*a(2*n) + a(2*n-1).
%p a[0] := 2: a[1] := 9:
%p for n from 2 to 18 do a[n] := 4*a[n-1] + (2*n-1)^2*a[n-2] end do:
%p seq(a[n], n = 0 .. 18);
%Y Cf. A008545, A001147, A024199, A142970, A254794, A254796.
%K nonn,frac,easy
%O 0,1
%A _Peter Bala_, Feb 23 2015
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