OFFSET
1,1
COMMENTS
Brouncker gave the generalized continued fraction expansion 4/Pi = 1 + 1^2/(2 + 3^2/(2 + 5^2/(2 + ... ))). More generally, Osler shows that the continued fraction n + 1^2/(2*n + 3^2/(2*n + 5^2/(2*n + ... ))) equals a rational multiple of 4/Pi or its reciprocal when n is a positive odd integer, and equals a rational multiple of L^2/Pi or its reciprocal when n is a positive even integer.
REFERENCES
O. Perron, Die Lehre von den Kettenbrüchen, Band II, Teubner, Stuttgart, 1957
LINKS
Muniru A Asiru, Table of n, a(n) for n = 1..2000
Peter Bala, Notes on the constants A096427 and A224268
B. C. Berndt, R. L. Lamphere and B. M. Wilson, Chapter 12 of Ramanujan's second notebook: Continued fractions, Rocky Mountain Journal of Mathematics, Volume 15, Number 2 (1985), 235-310.
T. J. Osler, The missing fractions in Brouncker's sequence of continued fractions for Pi, The Mathematical Gazette, 96(2012), pp. 221-225.
FORMULA
L^2/Pi = 2*( (1/4)!/(1/2)! )^4 = 9/4*( (1/4)!/(3/4)! )^2.
L^2/Pi = lim_{n -> oo} (4*n + 2) * Product {k = 0..n} ( (4*k - 1)/(4*k + 1) )^2
Generalized continued fraction: L^2/Pi = 2 + 1^2/(4 + 3^2/(4 + 5^2/(4 + ... ))). This is the particular case n = 0, x = 2 of a result of Ramanujan - see Berndt et al., Entry 25. See also Perron, p. 35.
The sequence of convergents to Ramanujan's continued fraction begins [2/1, 9/4, 54/25, 441/200, 4410/2025, ...]. See A254795 for the numerators and A254796 for the denominators.
Another continued fraction is L^2/Pi = 1 + 2/(1 + 1*3/(2 + 3*5/(2 + 5*7/(2 + 7*9/(2 + ... ))))), which can be transformed into the slowly converging series: L^2/Pi = 1 + 4 * Sum {n >= 0} P(n)^2/(4*n + 5), where P(n) = Product {k = 1..n} (4*k - 1)/(4*k + 1).
(L^2/Pi)^2 = 3 + 2*( 1^2/(1 + 1^2/(3 + 3^2/(1 + 3^2/(3 + 5^2/(1 + 5^2/(3 + ... )))))) ) follows by setting n = 0, x = 2 in Entry 26 of Berndt et al.
From Peter Bala, Feb 28 2019: (Start)
For m = 0,1,2,..., C = 4*(m + 1)*P(m)/Q(m), where P(m) = Product_{n >= 1} ( 1 - (4*m + 3)^2/(4*n + 1)^2 ) and Q(m) = Product_{n >= 0} ( 1 - (4*m + 1)^2/(4*n + 3)^2 ).
For m = 0,1,2,..., C = - Product_{k = 1..m} (1 - 4*k)/(1 + 4*k) * Product_{n >= 0} ( 1 - (4*m + 2)^2/(4*n + 1)^2 ) and
1/C = Product_{k = 0..m} (1 + 4*k)/(1 - 4*k) * Product_{n >= 0} ( 1 - (4*m + 2)^2/(4*n + 3)^2 ).
C = (Pi/2) * ( Sum_{n = -oo..oo} exp(-Pi*n^2) )^4. (End)
Equals A133748/Pi. - Hugo Pfoertner, Apr 13 2024
EXAMPLE
2.18843961522647663883676994070446454325937272282556672211928621....
MAPLE
MATHEMATICA
RealDigits[2*(Gamma[5/4]/Gamma[3/2])^4, 10, 110][[1]] (* G. C. Greubel, Mar 06 2019 *)
PROG
(PARI) default(realprecision, 110); 2*(gamma(5/4)/gamma(3/2))^4 \\ G. C. Greubel, Mar 06 2019
(Magma) SetDefaultRealField(RealField(110)); 2*(Gamma(5/4)/Gamma(3/2))^4; // G. C. Greubel, Mar 06 2019
(Sage) numerical_approx(2*(gamma(5/4)/gamma(3/2))^4, digits=110) # G. C. Greubel, Mar 06 2019
CROSSREFS
KEYWORD
AUTHOR
Peter Bala, Feb 22 2015
STATUS
approved