%I
%S 1,3,5,7,3,11,13
%N Let f denote the map that replaces k with the concatenation of its proper divisors, written in increasing order, each divisor being written in base 10 with its digits in reverse order. 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.
%e n = 7: 2n+1 = 15 = 3*5 > 35 = 5*7 > 57 = 3*19 > 391 = 17*23 > 7132.
%e Then 7132 has proper divisors 2, 4, 1783, 3566, so we get 2438716653.
%e Then 2438716653 has proper divisors 3, 9, 27, 81, 243, 1453, 4359, 6907, 13077, 20721, 39231, 62163, 117693, 186489, 353079, 559467, 1678401, 10035871, 30107613, 90322839, 270968517, 812905551, so we get
%e 397218342354195347096770311270213293361263967119846819703537649551048761178530013167010393822309715869072155509218 = 2*3^4*1217*317539*1211548321*33378971294653*8960783431807*17509226460292689821646170308388500174366980857582533580184934929433.
%Y Cf. A130139, A130141, A130142, A120716.
%K base,more,nonn
%O 0,2
%A _David Applegate_, Jul 30 2007
