A099468 Numbers n such that there are no primes < 2n in the sequence m(0)=n, m(k+1)=m(k)+4k.

1, 21, 45, 51, 81, 213, 249, 315, 477, 525, 681, 891, 1143, 1221, 1851, 1965, 2415, 5133

Offset: 1

Author

T. D. Noe, Oct 17 2004

Keywords

• nonn

Comments

No others < 10^8. Note that 3 divides all these n > 1. This sequence is conjectured to be complete. Related to a question posed in A036468 by Zhang Ming-Zhi. Let r=2s+1 be an odd number. If n = (s+1)^2+s^2, then the sequence m(0)=n, m(k+1)=m(k)+4k for k=0,1,...s calculates the s+1 distinct sums of two squares (r-i)^2+i^2.

Examples

			45 is here because 45, 49, 57, 69 and 85 are all composite.
		

Crossrefs

Cf. A036468 (number of ways to represent 2n+1 as a+b with a^2+b^2 prime).

Programs

Mathematica

lst={}; Do[n=m; found=False; k=0; While[n=n+4k; !found && n<2m, found=PrimeQ[n]; k++ ]; If[ !found, AppendTo[lst, m]], {m, 1, 10000, 2}]; lst