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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.
4
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.
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
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jan 05 2009
STATUS
approved