A006221 From Apery continued fraction for zeta(3): zeta(3)=6/(5-1^6/(117-2^6/(535-3^6/(1463...)))).
5, 117, 535, 1463, 3105, 5665, 9347, 14355, 20893, 29165, 39375, 51727, 66425, 83673, 103675, 126635, 152757, 182245, 215303, 252135, 292945, 337937, 387315, 441283, 500045, 563805, 632767, 707135, 787113, 872905, 964715, 1062747, 1167205, 1278293, 1396215, 1521175
Offset: 0
Examples
zeta(3) = 1.20205690315959428539973816151..., while eight terms of the sequence gives 6/(5-1^6/(117-2^6/(535-3^6/(1463-4^6/(3105-5^6/(9347-6^6/(14355))))))) = 1.20205690315959366144848279245...
References
- G. V. Chudnovsky, Transcendental numbers, pp. 45-69 of Number Theory Carbondale 1979, Lect. Notes Math. 751 (1982).
- S. R. Finch, Mathematical Constants, Cambridge, 2003, p. 46.
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
- P. Flajolet, B. Vallee, and I. Vardi, Continued fractions from Euclid to the present day, preprint, 2000.
- Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
- Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
- Xavier Viennot, Introduction to Chapter 3 on continued fractions, 2013
- Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
- Index entries for two-way infinite sequences
- Index entries for zeta function.
Crossrefs
Apéry's number or Apéry's constant zeta(3) is A002117. - N. J. A. Sloane, Jul 11 2023
Cf. A005259.
Programs
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Haskell
a006221 n = (17 * n * (n + 1) + 5) * (2 * n + 1) -- Reinhard Zumkeller, Mar 13 2014
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Maple
A006221:=z*(z+1)*(5*z**2+92*z+5)/(z-1)**4; [Conjectured by Simon Plouffe in his 1992 dissertation.]
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Mathematica
a[n_] := (2n + 1)(17n^2 + 17n + 5); a /@ Range[0, 31] (* Jean-François Alcover, Sep 03 2019 *)
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PARI
a(n)=34*n^3+51*n^2+27*n+5
Formula
G.f.: (5 + 97*x + 97*x^2 + 5*x^3)/(1-x)^4.
a(n) = 34*n^3 + 51*n^2 + 27*n + 5 = (2*n + 1)*(17*n*(n+1) + 5) [Viennot, p.2].
Can be extended to negative indices by: a(n) = -a(-1-n).
E.g.f.: exp(x)*(5 + 112*x + 153*x^2 + 34*x^3). - Stefano Spezia, Nov 03 2024
Extensions
Typo in description corrected Apr 09 2006 (1436 should have been 1463). Thanks to Simon Plouffe for this correction.