A306261 Least k > 0 such that 2n - p is prime where p is some prime divisor of 4n^2 - (2k-1)^2 for n >= 4.

1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 3, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 2, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 3, 1, 2, 3, 1, 1, 1, 1, 3, 1, 1, 1, 1, 3, 4, 1, 1, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 1, 2, 3, 1, 1, 1, 1, 2, 2, 1, 1, 2, 1

Offset: 4

Author

Juri-Stepan Gerasimov, Feb 01 2019

Keywords

• nonn

Comments

Conjecture: a(n) exists for n >= 4.

The conjecture holds up to 10^6. Records: a(4) = 1, a(6) = 2, a(34) = 3, a(75) = 4, a(154) = 9, a(1027) = 10, a(1097) = 11, a(1477) = 14, a(1552) = 17, a(5179) = 18, a(10684) = 29, a(70201) = 32, a(79861) = 43, a(519632) = 45, a(1018804) = 46, a(1713031) = 47, .... - Charles R Greathouse IV, Feb 17 2019

Examples

			a(4) = 1 because 4*4^2 - (2*1-1)^2 = 63 = 3^2*7 and 2*4 - 3 = 5 is prime;
a(5) = 1 because 4*5^2 - (2*1-1)^2 = 99 = 3^2*11 and 2*5 - 3 = 7 is prime;
a(6) = 2 because 4*6^2 - (2*1-1)^2 = 143 = 11*13 and 2*6 - 11 = 1 is not a prime, 2*6 - 13 = -1 is not a prime, but 4*6^2 -(2*2-1)^2 = 135 = 3^3*5 and 2*6 - 5 = 7 is prime.
		

Crossrefs

Cf. A020481, A306247.

Programs

PARI

a(n)=for(k=1,2*n,my(f=factor(4*n^2-(2*k-1)^2)[,1]);for(i=1,#f,if(isprime(2*n-f[i]),return(k)))); "does not exist" \\ Charles R Greathouse IV, Feb 17 2019