

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.


3



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,7


COMMENTS

ZhiWei 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.
ZhiWei 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), 103107.
Z. W. Sun and M. H. Le, Integers not of the form c(2^a+2^b)+p^{alpha}, Acta Arith. 99(2001), 183190.


LINKS

ZhiWei SUN, Table of n, a(n), n=1..50000.
D. S. McNeil, Sun's strong conjecture
ZhiWei Sun, A promising conjecture: n=p+F_s+F_t
ZhiWei 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^m2^n and x^2F_{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]==0Mod[x, 3]>0)&&PQ[n(Fibonacci[x])^3Fibonacci[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: A304081 A101312 A241273 * A293435 A294901 A305818
Adjacent sequences: A154260 A154261 A154262 * A154264 A154265 A154266


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Jan 06 2009


STATUS

approved



