A254795 Numerators of the convergents of the generalized continued fraction 2 + 1^2/(4 + 3^2/(4 + 5^2/(4 + ... ))).

2, 9, 54, 441, 4410, 53361, 747054, 12006225, 216112050, 4334247225, 95353438950, 2292816782025, 59613236332650, 1671463434096225, 50143903022886750, 1606276360166472225, 54613396245660055650, 1967688541203928475625, 74772164565749282073750

Offset

0,1

Comments

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.

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.

Links

Table of n, a(n) for n=0..18.

T. J. Osler, The missing fractions in Brouncker’s sequence of continued fractions for Pi

Formula

a(2*n-1) = ( A008545(n) )^2 = ( Product {k = 0..n-1} 4*k + 3 )^2.

a(2*n) = (4*n + 2)*( A008545(n) )^2 = (4*n + 2)*( Product {k = 0..n-1} 4*k + 3 )^2.

a(n) = 4*a(n-1) + (2*n - 1)^2*a(n-2) with a(0) = 2, a(1) = 9.

a(2*n) = (4*n + 2)*a(2*n-1); a(2*n+1) = (4*n + 4)*a(2*n) + a(2*n-1).

Maple

a[0] := 2: a[1] := 9:
for n from 2 to 18 do a[n] := 4*a[n-1] + (2*n-1)^2*a[n-2] end do:
seq(a[n], n = 0 .. 18);

Crossrefs

Cf. A008545, A001147, A024199, A142970, A254794, A254796.

Sequence in context: A321974 A127128 A353255 * A064151 A075679 A001757

Adjacent sequences: A254792 A254793 A254794 * A254796 A254797 A254798

Keyword

nonn,frac,easy

Author

Peter Bala, Feb 23 2015

Status

approved