A254794 Decimal expansion of L^2/Pi where L is the lemniscate constant A062539.
2, 1, 8, 8, 4, 3, 9, 6, 1, 5, 2, 2, 6, 4, 7, 6, 6, 3, 8, 8, 3, 6, 7, 6, 9, 9, 4, 0, 7, 0, 4, 4, 6, 4, 5, 4, 3, 2, 5, 9, 3, 7, 2, 7, 2, 2, 8, 2, 5, 5, 6, 6, 7, 2, 2, 1, 1, 9, 2, 8, 6, 2, 1, 0, 5, 7, 9, 4, 5, 1, 9, 3, 8, 4, 4, 5, 9, 3, 2, 9, 4, 7, 7, 7, 1, 0, 3, 3, 1, 4, 9, 6, 7, 7, 5, 6, 0, 8, 6, 3, 1, 8, 0, 2
Offset: 1
Examples
2.18843961522647663883676994070446454325937272282556672211928621....
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.
Programs
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Magma
SetDefaultRealField(RealField(110)); 2*(Gamma(5/4)/Gamma(3/2))^4; // G. C. Greubel, Mar 06 2019
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Maple
#A254794 digits:=105: 2*( GAMMA(5/4)/GAMMA(3/2) )^4: evalf(%);
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Mathematica
RealDigits[2*(Gamma[5/4]/Gamma[3/2])^4, 10, 110][[1]] (* G. C. Greubel, Mar 06 2019 *)
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PARI
default(realprecision, 110); 2*(gamma(5/4)/gamma(3/2))^4 \\ G. C. Greubel, Mar 06 2019
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Sage
numerical_approx(2*(gamma(5/4)/gamma(3/2))^4, digits=110) # G. C. Greubel, Mar 06 2019
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
Comments