cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A338303 Decimal expansion of Sum_{k>=0} 1/(L(2*k) + 2), where L(k) is the k-th Lucas number (A000032).

Original entry on oeis.org

6, 4, 4, 5, 2, 1, 7, 8, 3, 0, 6, 7, 2, 7, 4, 4, 4, 2, 0, 9, 9, 2, 7, 3, 1, 1, 9, 0, 3, 8, 0, 1, 6, 9, 0, 2, 9, 2, 8, 9, 0, 8, 1, 2, 3, 8, 7, 7, 9, 9, 1, 8, 5, 7, 6, 5, 1, 4, 2, 5, 5, 2, 7, 5, 7, 7, 6, 8, 6, 8, 6, 1, 6, 8, 3, 6, 7, 8, 7, 4, 3, 3, 4, 1, 4, 0, 8
Offset: 0

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Author

Amiram Eldar, Oct 21 2020

Keywords

Comments

Backstrom (1981) found that the sum is approximately equal to 1/8 + 1/(4*log(phi)), where phi is the golden ratio (A001622). The difference is less than 1/10^7. He called this difference "a tantalizing problem".
Almkvist (1986) added a term to Backstrom's formula to get an even better approximation which differs from the exact value by less than 1/10^33: 1/8 + 1/(4*log(phi)) + Pi^2/(log(phi)^2 * (exp(Pi^2/log(phi)) - 2)). He found an exact formula from a quotient of two Jacobi theta functions (see the FORMULA section), and showed that both approximations are just the first terms in a rapidly converging series.
Since exp(-Pi^2/log(phi)) = 1.23...*10^(-9) is small, the convergence is rapid: the number of terms needed in each of the two series in the formula to get 10^2, 10^3 and 10^4 decimal digits are merely 3, 10 and 32, respectively.
Andre-Jeannin (1991) noted that if the summand 2 that is added to the Lucas numbers is replaced with sqrt(5), then the sum is Sum_{k>=0} 1/(L(2*k) + sqrt(5)) = 1/phi (A094214).

Examples

			0.64452178306727444209927311903801690292890812387799...
		

References

  • Jonathan M. Borwein and Peter B. Borwein, Pi and the AGM, Wiley, 1987, p. 99.

Crossrefs

Programs

  • Mathematica
    With[{lg = Log[GoldenRatio], kmax = 3}, sum[m_, k_] := (-1)^k*k^m*Exp[-Pi^2*k^2/lg]; RealDigits[1/8 + ( 1/(4*lg))*(1 - (4*Pi^2/lg)*(Sum[sum[2, k], {k, 1, kmax}]/(1 + 2*Sum[sum[0, k], {k, 1, kmax}]))), 10, 100][[1]]]

Formula

Equals Sum_{k>=0} 1/A240926(k).
Equals 1/4 + Sum_{k>=1} q^(2*k)/(1 + q^(2*k))^2, where q = 1/phi.
Equals 1/8 + (1/(4*log(phi))) * (1 - (4*Pi^2/log(phi)) * (S(2)/(1 + 2*S(0)))), where S(m) = Sum_{k>=1} (-1)^k * k^m * exp(-Pi^2*k^2/log(phi)) (Almkvist, 1986).