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A356977
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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.
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0
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OFFSET
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0,2
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COMMENTS
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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.
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REFERENCES
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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.
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LINKS
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FORMULA
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a(n) = Sum_{i >= 0} A319394(2^i, k).
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EXAMPLE
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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.]
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CROSSREFS
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KEYWORD
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nonn,hard,more
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AUTHOR
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STATUS
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approved
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