A356977 a(n) is the number of solutions, j >= 0 and 2 <= m_1 <= ... <= m_n, of the equation Sum_{k=1..n} F(m_k) = 2^j where F(i) is the i-th Fibonacci number.

0, 3, 6, 10, 36, 66

Offset

0,2

Comments

The difficulty of this sequence comes in determining which is the largest j in a solution for a(n), equivalently the last nonzero term in each sum from the A319394-based formula in the formula section.

a(2) derives from Bravo and Luca, a(3) from Bravo and Bravo, a(4) from Pagdame Tiebekabe and Diouf. Pagdame has indicated A356928(5), from which a(5) is derived, has been determined.

a(6) >= 178, a(7) >= 478.

References

J. J. Bravo, and F. Luca, On the Diophantine equation F_n+F_m=2^a, Quaest. Math. 39 (2016) 391-400.

P. Tiebekabe and I. Diouf, On solutions of Diophantine equation F_{n_1}+F_{n_2}+F_{n_3}+F_{n_4}=2^a, Journal of Algebra and Related Topics, Volume 9, Issue 2 (2021), 131-148.

Links

Table of n, a(n) for n=0..5.

E. F. Bravo and J. J. Bravo, Powers of two as sums of three Fibonacci numbers, Lithuanian Mathematical Journal, 55, pp. 301-311 (2015).

Formula

a(n) = Sum_{i >= 0} A319394(2^i, k).

Example

For n = 2, the a(2) = 6 solutions are j = 1 with (2,2), j = 2 with (2,4) and (3,3), j = 3 with (4,5), j = 4 with (4,7) and (6,6) according to the paper of Bravo and Luca. [That is, 2 = 1+1, 4 = 1+3 = 2+2, 8 = 3+5, 16 = 3+13 = 8+8.]

Crossrefs

Cf. A020908, A319394, A356928.

Sequence in context: A094276 A151376 A066245 * A068821 A062100 A262242

Adjacent sequences: A356974 A356975 A356976 * A356978 A356979 A356980

Keyword

nonn,hard,more

Author

Peter Munn and Jon E. Schoenfield, Sep 07 2022

Status

approved