%I
%S 1,3,5,7,3,11,13,53,17,19,73,23,5,9343,29,31,113,1,37,313,41,43
%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 with its digits in reverse order. Then a(n) = first 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.
%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 133.
%e Then 133 has proper divisors 7 and 19, so we get 917.
%e Then 917 has proper divisors 7 and 131, so we get 1317.
%e Then 1317 has proper divisors 3 and 439, so we get 9343, a prime and a(13) = 9343.
%e From _Sean A. Irvine_, Sep 11 2009: (Start)
%e Proof chain for a(17). The following gives the argument to f at each step, followed by its factorization.
%e 35 factors as 5 * 7.
%e 75 has factors 3 * 5 * 5.
%e 525153 has factors 3 * 193 * 907.
%e 15057112727099753913 has factors 3 * 4463 * 17215189 * 65325353.
%e 179719996575730910515106159846737337176838928854211713151146478934050745192125561494032705883138679506795913535235676554615981512719833136443 has factors 29 * 29 * 5546454298803948416569 * 8370112457804191610629 * 13338101723922940394396774098231 * 345111672681489292530961043464303237918570147336150469919363833
%e 765...4892 (3249 digits) is divisible by 2, and hence all subsequent steps will be divisible by 2, therefore no prime is ever reached, therefore a(17)=1. (End)
%Y Cf. A130139, A130140, A130141, A120716.
%K base,more,sign
%O 0,2
%A Adam L. Buchsbaum (alb(AT)research.att.com), Jul 30 2007, Aug 01 2007
%E The value of a(17) is currently unknown.
%E 5 more terms (details for a(17) in example). Next term requires factoring a 1478digit number.  _Sean A. Irvine_, Sep 11 2009
