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

%I #16 Nov 14 2024 02:06:20

%S 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,

%T 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,

%U 4,6,6,13,7,7,6,8,4,10,10,9

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

%C 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.

%C 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.

%C See also A258141 for a similar conjecture.

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

%H Zhi-Wei Sun, <a href="/A258168/b258168.txt">Table of n, a(n) for n = 1..10000</a>

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1504.01608">Natural numbers represented by floor(x^2/a) + floor(y^2/b) + floor(z^2/c)</a>, arXiv:1504.01608 [math.NT], 2015.

%e 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.

%e 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.

%t 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}]

%Y Cf. A000040, A258139, A258140, A258141.

%K nonn,changed

%O 1,1

%A _Zhi-Wei Sun_, May 22 2015