A299959 Least prime factor of (4^(2n+1)+1)/5, a(0) = 1.

1, 13, 5, 29, 13, 397, 53, 5, 137, 229, 13, 277, 5, 13, 107367629, 5581, 13, 5, 149, 13, 10169, 173, 5, 3761, 29, 13, 15358129, 5, 13, 1181, 733, 13, 5, 269, 13, 569, 293, 5, 29, 317, 13, 997, 5, 13, 1069, 29, 13, 5, 389, 13, 809, 41201, 5, 857, 5669, 13, 58309, 5, 13, 29, 397, 13, 5, 509, 13

Offset: 0

Author

M. F. Hasler, Feb 22 2018

Keywords

• nonn

Comments

The range of this sequence with a(0) = 1 omitted, {5, 13, 29, 53, 137, ...}, appears to be a subset of A261580 (and of the Pythagorean primes A002144). Is there a smaller superset sequence in OEIS? - M. F. Hasler, Jan 07 2025

Examples

			For n = 0, A299960(0) = (4^1+1)/5 = 5/5 = 1, therefore we let a(0) = 1.
For n = 1, A299960(1) = (4^3+1)/5 = 65/5 = 13 is prime, therefore a(1) = 13.
For n = 2, A299960(2) = (4^5+1)/5 = 1025/5 = 205 = 5*41, therefore a(2) = 5.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 0..715

Crossrefs

Cf. A299960, A052539.

Programs

PARI

a(n)=A020639(4^(2*n+1)\5+1) \\ Using factor(...)[1,1] requires complete factorization and is much less efficient for large n.

Formula

a(n) = A020639(A299960(n)) = A020639(A052539(2n+1)/5).

a(n) = 5 iff n = 2 (mod 5); otherwise, a(n) = 13 if n = 1 (mod 3).

Otherwise, a(n) = 29 if n = 3 (mod 7), else a(n) = 53 if n = 6 (mod 13), else a(n) = 137 if n = 8 (mod 17), else a(n) = 149 if n = 18 (mod 27), else a(n) = 173 if n = 21 (mod 43), etc... - M. F. Hasler, Jan 07 2025