A086257 Number of primitive prime factors of 2^n+1.

1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 2, 3, 1, 1, 2, 2, 1, 2, 2, 3, 2, 2, 2, 3, 1, 1, 2, 2, 1, 2, 1, 4, 2, 2, 1, 3, 3, 2, 2, 2, 2, 2, 2, 2, 3, 1, 1, 4, 1, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 4, 1, 3, 3, 4, 1, 2, 3, 4, 5, 2, 1, 4, 1, 3, 3, 3, 3, 1, 2, 3, 2, 1, 4, 3, 2, 4, 1, 4, 2, 1

Offset: 0

Author

T. D. Noe, Jul 14 2003

Keywords

• hard
• nonn

Comments

A prime factor of 2^n+1 is called primitive if it does not divide 2^r+1 for any rA086258 for those n that have a record number of primitive prime factors.

Examples

			a(14) = 2 because 2^14+1 = 5*29*113 and 29 and 113 do not divide 2^r+1 for r < 14.
		

Links

• Max Alekseyev, Table of n, a(n) for n = 0..1128 (terms 0..500 from T. D. Noe; terms 501..1062 from Amiram Eldar)
• Eric Weisstein's World of Mathematics, Zsigmondy Theorem

Crossrefs

Excluding a(0) = 1, forms a bisection of A086251.

Cf. A046799 (number of distinct prime factors of 2^n+1), A054992 (number of prime factors, with repetition, of 2^n+1), A086258.

Programs

Mathematica

nMax=200; pLst={}; Table[f=Transpose[FactorInteger[2^n+1]][[1]]; f=Complement[f, pLst]; cnt=Length[f]; pLst=Union[pLst, f]; cnt, {n, 0, nMax}]

Formula

For n > 0, a(n) = A086251(2*n). - Max Alekseyev, Sep 06 2022