A130549 Numerators of partial sums for a series for 2*Zeta(2)/3 = (Pi^2)/9.
1, 13, 197, 1105, 9211, 130277, 82987349, 331950131, 16929464521, 29241805241, 3538258509761, 6259995854281, 1057939300471201, 1057939300716589, 51133732870640471, 372975463296151087, 107789908892879155343
Offset: 1
Examples
Rationals r(n): [1, 13/12, 197/180, 1105/1008, 9211/8400, 130277/118800, ...]. r(3) = 1/(1 - 1/(13 - 48/(34))) = 197/180. - _Peter Bala_, Feb 17 2024
References
- L. Berggren, T. Borwein and P. Borwein, Pi: A Source Book, Springer, New York, 1997, p. 687.
- A. van der Poorten, A proof that Euler missed..., reprinted in Pi: A Source Book, pp. 439-447, eq. 2', with a hint for the proof in footnote 4.
Links
- C. Elsner, On recurrence formulas for sums involving binomial coefficients, Fib. Q., 43,1 (2005), 31-45.
- Wolfdieter Lang, Rationals and limit.
- A. J. van der Poorten, A proof that Euler missed ... Apery's proof of the irrationality of zeta(3), Math. Intelligencer 1 (1978/1979), 195-203.
- Herbert S. Wilf, Accelerated series for universal constants, by the WZ method, Discrete Mathematics & Theoretical Computer Science, Vol 3, No 4 (1999).
Programs
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Maple
seq(numer(add(2/(k^2*binomial(2*k, k)), k = 1 .. n)), n = 1 .. 17); # Peter Bala, Mar 03 2015
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Mathematica
Table[2*Sum[1/(i^2*Binomial[2*i, i]), {i, 1, n}], {n, 1, 20}] // Numerator Accumulate[Table[1/(n^2 Binomial[2n,n]),{n,20}]]//Numerator (* Harvey P. Dale, Jan 27 2019 *) -
PARI
a(n) = numerator(2*sum(i=1, n, 1/(i^2*binomial(2*i, i)))); \\ Michel Marcus, Mar 10 2016
Formula
a(n) = numerator(r(n)), n>=1, with the rationals r(n) defined above.
Numerator of 2*Sum_{i=1..n} 1/(i^2*C(2*i,i)). - Wolfdieter Lang, Oct 07 2008; edited by Michel Marcus, Mar 10 2016
a(n) = A112093(n) for n >= 2. - Georg Fischer, Nov 03 2018
From Peter Bala, Feb 17 2024: (Start)
The sequences {(2*n)! : n >= 1} and {(2*n)!*r(n) : n >= 1} satisfy the same second-order recurrence u(n) = (5*n^2 - 4*n + 1)*u(n-1) - 2*(n - 1)^3*(2*n - 3)*u(n-2) leading to the continued fraction representations r(n) = 1/(1 - 1/(13 - 48/(34 - 270/(65 - ... - 2*(2*n - 3)*(n - 1)^3/(5*n^2 - 4*n + 1 ))))) and Pi^2/9 = 1/(1 - 1/(13 - 48/(34 - 270/(65 - ... - 2*(2*n - 3)*(n - 1)^3/((5*n^2 - 4*n + 1) - ... ))))). (End)
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