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A258168 Number of ways to write n as floor((p^2+q)/5) with p and q both prime. 2
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

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.

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.

EXAMPLE

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.

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}]

CROSSREFS

Cf. A000040, A258139, A258140, A258141.

Sequence in context: A100867 A220196 A128200 * A061988 A094151 A135800

Adjacent sequences:  A258165 A258166 A258167 * A258169 A258170 A258171

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, May 22 2015

STATUS

approved

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Last modified October 20 07:33 EDT 2019. Contains 328252 sequences. (Running on oeis4.)