A130141 Let f denote the map that replaces k with the concatenation of its nontrivial divisors, written in decreasing order, each divisor being written in base 10 in the normal way. Then a(n) = prime reached when starting at 2n+1 and iterating f.

1, 3, 5, 7, 3, 11, 13, 53, 17, 19, 73, 23, 5, 313, 29, 31, 113

Offset: 0

Author

Carsten Lund (lund(AT)research.att.com), Jul 30 2007, Aug 01 2007

Keywords

• base
• more
• nonn

Comments

If 2n+1 is 1 or a prime, set a(n) = 2n+1. If no prime is ever reached, set a(n) = -1.

The value of a(17) is currently unknown.

From Klaus Brockhaus, Aug 01 2007: (Start)

"The sequence, starting from the beginning but writing 0 when a number with more than 90 digits is reached, is:

1,3,5,7,3,11,13,53,17,19,73,23,5,313,29,31,113,0,37,197,41,43,0,47,7,173,53,0,193,

59,61,0,0,67,233,71,73,0,391393,79,9313991471335749211973,83,0,293,89,137,313,

0,97,0,101,103,0,107,109,373,113,0,391393,593,11,1993,0,127,433,131,197,0,137,139,

0,0,0,0,149,151,511793,0,157,1141931201,6113,163,0,167,13,0,173,0,593,179,181,

613,12575251553,0,0,191,193,0,197,199,673,..." (End)

From Andrew Carter (acarter09(AT)newarka.edu), Dec 16 2008: (Start)

If "when starting at 2n+1" in the definition were replaced by "when starting at n", all even-indexed terms except for a(2) and a(4) would be -1.

A simple proof is that an even number's smallest nontrivial divisor is 2, and the result of iterating f for any even number except for 2 and 4 would have factors in front, so the resulting number would be of the form 10m+2 where m is an integer greater than 1, and therefore an even number ending in 2 other than two itself, so no prime would ever be reached. (End)

Examples

			n = 13: 2n+1 = 27 has nontrivial divisors 3 and 9, so we get 93, which has proper divisors 3 and 31, so we get 313, prime. So a(13) = 313.
		

Crossrefs

Cf. A130139, A130140, A130142, A120716.

Extensions

Edited by Michel Marcus, Mar 09 2023