A356928 a(n) is the number of solutions, j >= 0 and 1 <= 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, 4, 9, 15, 60, 106
Offset: 0
Examples
For n=2, the a(2) = 9 solutions are 1 with (1,1), 1 with (1,2), 2 with (1,4), 1 with (2,2), 2 with (2,4), 2 with (3,3), 3 with (4,5), 4 with (4,7), and 4 with (6,6) according to the paper of Bravo and Luca. [That is, 2=1+1, 2=1+1 (again), 4=1+3, 2=1+1 (again), 4=1+3 (again), 4=2+2, 8=3+5, 16=3+13, and 16=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).
- Pagdame Tiebekabe and Ismaïla Diouf, On the Diophantine equation Sum_{k=1..5} F(n_k) = 2^a, arXiv:2209.07871 [math.NT], 2022.
Crossrefs
Cf. A007000.
Extensions
a(0)=0 added by Peter Munn, Sep 05 2022
Name and example edited by Peter Munn, Sep 06 2022
Comments