A356928 - OEIS

%I #69 Nov 01 2022 13:48:29

%S 0,4,9,15,60,106

%N 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.

%C a(6) >= 298. We do not have information about whether 298 has been proved to be a(6). - _Peter Munn_, Sep 08 2022

%C a(7) >= 772. - _Jon E. Schoenfield_, Sep 05 2022

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

%D 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.

%H E. F. Bravo and J. J. Bravo, <a href="https://www.researchgate.net/publication/282537969_Powers_of_two_as_sums_of_three_Fibonacci_numbers">Powers of two as sums of three Fibonacci numbers</a>, Lithuanian Mathematical Journal, 55, pp. 301-311 (2015).

%H Pagdame Tiebekabe and Ismaïla Diouf, <a href="https://arxiv.org/abs/2209.07871">On the Diophantine equation Sum_{k=1..5} F(n_k) = 2^a</a>, arXiv:2209.07871 [math.NT], 2022.

%e 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.]

%Y Cf. A007000.

%K nonn,hard,more

%O 0,2

%A _Pagdame Tiebekabe_, Sep 05 2022

%E a(0)=0 added by _Peter Munn_, Sep 05 2022

%E Name and example edited by _Peter Munn_, Sep 06 2022