

A338303


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


1



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
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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.
AndreJeannin (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. 316322.
Richard AndreJeannin, Summation of certain reciprocal series related to Fibonacci and Lucas numbers, The Fibonacci Quarterly, Vol. 29, No. 3 (1991), pp. 200204.
Robert P. Backstrom, On reciprocal series related to Fibonacci numbers with subscripts in arithmetic progression, The Fibonacci Quarterly, Vol. 19, No. 1 (1981), pp. 1421. See section 6, pp. 1920.
Daniel Duverney and Iekata Shiokawa, On series involving Fibonacci and Lucas numbers I, AIP Conference Proceedings, Vol. 976, No. 1 (2008), pp. 6276, 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



