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

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

0,1

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).

References

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

Links

Table of n, a(n) for n=0..86.

Gert Almkvist, A solution to a tantalizing problem, The Fibonacci Quarterly, Vol. 24, No. 4 (1986), pp. 316-322.

Richard Andre-Jeannin, Summation of certain reciprocal series related to Fibonacci and Lucas numbers, The Fibonacci Quarterly, Vol. 29, No. 3 (1991), pp. 200-204.

Robert P. Backstrom, On reciprocal series related to Fibonacci numbers with subscripts in arithmetic progression, The Fibonacci Quarterly, Vol. 19, No. 1 (1981), pp. 14-21. See section 6, pp. 19-20.

Daniel Duverney and Iekata Shiokawa, On series involving Fibonacci and Lucas numbers I, AIP Conference Proceedings, Vol. 976, No. 1 (2008), pp. 62-76, alternative link.

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).

Example

0.64452178306727444209927311903801690292890812387799...

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]]]

Crossrefs

Cf. A000032, A001622, A002390, A094214, A160509, A240926.

Sequence in context: A201285 A317127 A195359 * A316162 A332431 A198840

Adjacent sequences: A338300 A338301 A338302 * A338304 A338305 A338306

Keyword

nonn,cons

Author

Amiram Eldar, Oct 21 2020

Status

approved