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Right leading diagonal of the triangle described in comments.
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%I #10 Oct 28 2021 10:00:25

%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 n-th 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, n-1, 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