%I
%S 1,3,5,7,3,11,13,53,17,19,73,23,5,313,29,31,113
%N Let f denote the map that replaces k with the concatenation of its proper 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.
%C If 2n+1 is 1 or a prime, set a(n) = 2n+1. If no prime is ever reached, set a(n) = 1.
%C Comment from _Klaus Brockhaus_, Aug 01 2007:
%C "The sequence, starting from the beginning but writing 0 when a number with more than 90 digits is reached, is:
%C 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,
%C 59,61,0,0,67,233,71,73,0,391393,79,9313991471335749211973,83,0,293,89,137,313,
%C 0,97,0,101,103,0,107,109,373,113,0,391393,593,11,1993,0,127,433,131,197,0,137,139,
%C 0,0,0,0,149,151,511793,0,157,1141931201,6113,163,0,167,13,0,173,0,593,179,181,
%C 613,12575251553,0,0,191,193,0,197,199,673,..."
%C Contribution from Andrew Carter (acarter09(AT)newarka.edu), Dec 16 2008: (Start)
%C If this sequence were to do all terms, all even terms except 2 and 4 would be 1:
%C a simple proof is that an even number's smallest proper divisor is 2,
%C all even numbers except 2 and 4 have factors in front so the resulting number would be 10n+2 and n is an integer greater than 1,
%C and therefore an even number endng in 2 other than two  rinse and repeat (End)
%e n = 13: 2n+1 = 27 has proper divisors 3 and 9, so we get 93, which has proper divisors 3 and 31, so we get 313, prime. So a(13) = 313.
%Y Cf. A130139, A130140, A130142, A120716.
%K base,more,nonn
%O 0,2
%A Carsten Lund (lund(AT)research.att.com), Jul 30 2007, Aug 01 2007
%E The value of a(17) is currently unknown.
