A229747 Largest prime factor of 4^(2*n+1)+1.

5, 13, 41, 113, 109, 2113, 1613, 1321, 26317, 525313, 14449, 30269, 268501, 279073, 536903681, 384773, 4327489, 47392381, 231769777, 21841, 43249589, 1759217765581, 29247661, 140737471578113, 4981857697937, 1326700741, 1801439824104653, 3630105520141

Offset: 0

Author

Colin Barker, Sep 28 2013

Keywords

• nonn

Comments

4^(2*n+1)+1 = 2^(2*(2*n+1))+1 = (2^(2*n+1)-2^(n+1)+1) * (2^(2*n+1)+2^(n+1)+1).

For all n, the smallest prime factor of 4^(2*n+1)+1 is 5.

Therefore, the present sequence also gives the largest prime factor of (4^(2*n+1)+1)/5 = A299960(n), for all n > 0. See A299959 for the smallest prime factor of this. - M. F. Hasler, Feb 27 2018

Examples

			For n=7, 4^(2*n+1)+1 = 1073741825 = 5*5*13*41*61*1321. So a(7)=1321.
		

Links

• Daniel Suteu, Table of n, a(n) for n = 0..547

Crossrefs

Cf. A207262. Bisection of A274903.

Cf. A299960, A299959, A052539.

Programs

Mathematica

Table[FactorInteger[4^(2n+1)+1][[-1,1]],{n,0,30}] (* Harvey P. Dale, Mar 10 2018 *)

PARI

a(n) = {
  f=factor(2^(2*n+1)-2^(n+1)+1);
  g=factor(2^(2*n+1)+2^(n+1)+1);
  max(f[matsize(f)[1],1], g[matsize(g)[1],1])
}

Formula

a(n) = A006530(A052539(2n+1)) = A006530(A207262(n+1)), and for n > 1, a(n) = A006530(A299960(n)) = A006530(A052539(2n+1)/5). \\ M. F. Hasler, Feb 27 2018

a(n) = max(A229767(n), A229768(n)), for n >= 1. - Daniel Suteu, Jun 08 2022