A102566 a(n) = {minimal k such that f^k(prime(n)) = 1} where f(m) = (m+1)/2^r, 2^r is the highest power of two dividing m+1.

2, 1, 2, 1, 2, 2, 4, 3, 2, 2, 1, 4, 4, 3, 2, 3, 2, 2, 5, 4, 5, 3, 4, 4, 5, 4, 3, 3, 3, 4, 1, 6, 6, 5, 5, 4, 4, 5, 4, 4, 4, 4, 2, 6, 5, 4, 4, 2, 4, 4, 4, 2, 4, 2, 8, 6, 6, 5, 6, 6, 5, 6, 5, 4, 5, 4, 5, 6, 4, 4, 6, 4, 3, 4, 3, 2, 6, 5, 6, 5, 5, 5, 3, 5, 3, 3, 6, 5, 4, 3, 4, 2, 3, 3, 3, 2, 2, 8, 7, 6, 7, 6, 6, 6, 5

Offset: 1

Author

Yasutoshi Kohmoto, Feb 25 2005

Keywords

• nonn
• base

Comments

A066195(n+1) is the prime corresponding to the first n in this sequence. - David Wasserman, Apr 08 2008

Examples

			f(f(f(f(17)))) = 1, prime(7) = 17, so a(7) = 4.
prime(16) = 53 = (2*27-1) = (2*(2^2*7-1)-1) = (2*(2^2*(2^3*1-1)-1)-1), has 3 levels, so a(16) = 3.
		

Crossrefs

Cf. A023416, A035103, A066195.

Programs

PARI

f(n) = (n+1)/2^(valuation(n+1, 2));
a(n) = {my(k = 1, p = prime(n)); while((q=f(p)) != 1, k++; p = q); k;} \\ Michel Marcus, Nov 20 2016

PARI

a(n) = my(p=prime(n)); 2 + logint(p, 2) - hammingweight(p); \\ Kevin Ryde, Nov 06 2023

Formula

a(n) = A023416(prime(n)) + 1. - David Wasserman, Apr 08 2008

a(n) = A035103(n) + 1. - Filip Zaludek, Nov 19 2016

Extensions

More terms from David Wasserman, Apr 08 2008