A258168 Number of ways to write n as floor((p^2+q)/5) with p and q both prime.

3, 4, 3, 4, 5, 4, 4, 4, 4, 7, 4, 5, 6, 4, 5, 5, 4, 5, 4, 3, 5, 6, 4, 6, 5, 6, 5, 5, 3, 6, 6, 7, 3, 7, 5, 8, 8, 5, 5, 9, 5, 4, 6, 7, 4, 7, 5, 6, 7, 5, 4, 5, 4, 7, 8, 6, 6, 8, 4, 8, 7, 5, 8, 7, 4, 7, 5, 7, 4, 6, 6, 13, 7, 7, 6, 8, 4, 10, 10, 9

Offset: 1

Author

Zhi-Wei Sun, May 22 2015

Keywords

• nonn

Comments

Conjecture: Let n be any positive integer. Then a(n) > 0. Moreover, one of the four consecutive numbers 5*n, 5*n+1, 5*n+2, 5*n+3 can be written as p^2+q with p and q both prime.

It seems that there are infinitely many positive integers n such that none of n, n+1, n+2, n+3, n+4 has the form p^2 + q with p and q both prime.

See also A258141 for a similar conjecture.

Note that neither 3763 nor 5443 can be written as floor((p^2+q)/4) with p and q both prime.

Examples

			a(1) = 3 since 1 = floor((2^2+2)/5) = floor((2^2+3)/5) = floor((2^2+5)/5) with 2, 3, 5 all prime.
a(2) = 4 since 2 = floor((2^2+7)/5) = floor((3^2+2)/5) = floor((3^2+3)/5) = floor((3^2+5)/5) with 2, 3, 5, 7 all prime.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Zhi-Wei Sun, Natural numbers represented by floor(x^2/a) + floor(y^2/b) + floor(z^2/c), arXiv:1504.01608 [math.NT], 2015.

Crossrefs

Cf. A000040, A258139, A258140, A258141.

Programs

Mathematica

Do[m=0;Do[If[PrimeQ[5n+r-Prime[k]^2],m=m+1],{r,0,4},{k,1,PrimePi[Sqrt[5n+r]]}];Print[n," ",m];Continue,{n,1,80}]