%I
%S 2,5,7,13,13,31,31,59,59,59,59,179,179,179,179,179,179,179,179,179,
%T 179,179,179,179,179,179,1381,1381,1381,1381,1381,1381,1381,1381,1381,
%U 1381,1381,1381,1381,1381,1381,1381,1381,1381,1381,1381,1381,1381,24671
%N Right leading diagonal of the triangle described in comments.
%C Below is the triangle in which the left half of the nth row contains numbers from 2 to n and the mirror image about the leading column contains numbers in increasing order with the same prime signature.
%C Initial triangle starts:
%C 2;
%C 2 3;
%C 2 3 4;
%C 2 3 4 5 9;
%C Final triangle starts:
%C 2;
%C 2 3 5;
%C 2 3 4 5 7;
%C 2 3 4 5 9 11 13;
%C 2 3 4 5 6 7 9 11 13;
%C 2 3 4 5 6 7 10 11 25 29 31;
%C ...
%C All terms of the sequence are primes since they are numbers with same prime signature as the numbers of the first column, that are all equal to 2, a prime.
%e In the row for n = 5 the terms corresponding to 4,3,2 are 9,11,13 respectively.
%o (PARI) findsps(last, ps) = {new = last+1; while(factor(new)[,2] != ps, new++); new;}
%o a(n) = {last = n; orig = last  1; for (i = 1, n1, ps = factor(orig)[,2]; last = findsps(last, ps); orig ;); return (last);} \\ _Michel Marcus_, Feb 05 2014
%Y Cf. A093496.
%K nonn
%O 2,1
%A _Amarnath Murthy_, Apr 27 2004
%E Edited and extended by _Michel Marcus_, Feb 05 2014
