A154258 Number of triples <p,s,t> such that p+F_s+(F_t)^2=n, where p is an odd prime, s and t are greater than one and the Fibonacci number F_s or F_t is odd.

0, 0, 0, 0, 1, 1, 2, 2, 3, 3, 2, 4, 3, 5, 4, 5, 6, 3, 5, 6, 5, 7, 4, 5, 7, 4, 6, 7, 6, 6, 6, 5, 11, 6, 8, 6, 6, 7, 6, 9, 9, 4, 9, 5, 9, 10, 6, 8, 8, 7

Offset

1,7

Comments

Zhi-Wei Sun conjectured that a(n)>0 for all n=5,6,... (i.e., any integer n>4 can be written as the sum of an odd prime, a positive Fibonacci number and a square of a positive Fibonacci number, with one of the two Fibonacci numbers odd). He has verified this for n 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 summary concerning my conjecture n=p+F_s+F_t

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

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=10 the a(10)=3 solutions are 3+F_4+(F_3)^2, 5+F_2+(F_3)^2, 7+F_3+(F_2)^2.

Mathematica

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

Crossrefs

Cf. A000040, A000045, A154257

Sequence in context: A276273 A039643 A288887 * A253900 A327487 A105496

Adjacent sequences: A154255 A154256 A154257 * A154259 A154260 A154261

Keyword

nonn

Author

Zhi-Wei Sun, Jan 05 2009

Status

approved