%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 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.
%A Carsten Lund (lund(AT)research.att.com), Jul 30 2007, Aug 01 2007
%E The value of a(17) is currently unknown.