A154263 Number of triples <p, s,t> such that p+F_s+(F_t)^3=n, where p is an odd prime, s and t are greater than one and F_s or F_t is odd.

0, 0, 0, 0, 1, 1, 2, 1, 3, 1, 2, 2, 2, 4, 2, 5, 3, 2, 3, 4, 3, 4, 2, 3, 4, 5, 3, 4, 2, 2, 3, 7, 6, 5, 6, 3, 4, 5, 4, 9, 4, 6, 6, 3, 7, 7, 5, 5, 4, 5

Offset

1,7

Comments

Zhi-Wei Sun conjectured that a(n)>0 for every n=5,6,...; in other words, any integer n>4 can be written as the sum of an odd prime, a positive Fibonacci number and a cube of a positive Fibonacci number, with one of two Fibonacci numbers odd. He has verified this up to 3*10^7.

Zhi-Wei Sun has offered a monetary reward for settling this conjecture.

References

R. Crocker, On a sum of a prime and two powers of two, Pacific J. Math. 36(1971), 103-107.

Z. W. Sun and M. H. Le, Integers not of the form c(2^a+2^b)+p^{alpha}, Acta Arith. 99(2001), 183-190.

Links

Zhi-Wei SUN, Table of n, a(n), n=1..50000.

D. S. McNeil, Sun's strong conjecture

Zhi-Wei Sun, A promising conjecture: n=p+F_s+F_t

Zhi-Wei Sun, A summary concerning my conjecture n=p+F_s+F_t

K. J. Wu and Z. W. Sun, Covers of the integers with odd moduli and their applications to the forms x^m-2^n and x^2-F_{3n}/2, Math. Comp., in press. arXiv:math.NT/0702382

Example

For n=14 the a(14)=4 solutions are 3+F_4+(F_3)^3, 5+F_2+(F_3)^3, 5+F_6+(F_2)^3, 11+F_3+(F_2)^3

Mathematica

PQ[m_]:=m>2&&PrimeQ[m] RN[n_]:=Sum[If[(Mod[n, 2]==0||Mod[x, 3]>0)&&PQ[n-(Fibonacci[x])^3-Fibonacci[y]], 1, 0], {x, 2, 2*Log[2, n^(1/3)+1]}, {y, 2, 2*Log[2, Max[2, n-(Fibonacci[x])^3]]}] Do[Print[n, " ", RN[n]]; Continue, {n, 1, 50000}]

Crossrefs

Cf. A154257, A154258, A000040, A000045

Sequence in context: A101312 A241273 A353646 * A293435 A294901 A347042

Adjacent sequences: A154260 A154261 A154262 * A154264 A154265 A154266

Keyword

nonn

Author

Zhi-Wei Sun, Jan 06 2009

Status

approved