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

Author

Peter Munn and Jon E. Schoenfield, Sep 07 2022

Keywords

• nonn
• hard
• more

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.

Examples

			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.]
		

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

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

Crossrefs

Cf. A020908, A319394, A356928.

Formula

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