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%I #11 Dec 05 2013 19:57:02
%S 1,0,0,0,0,10,3,7,0,3,0,10,3,0,22,0,51,7,9,10,0,412,0,16,18,0,3,0,3,3,
%T 0,9,0,3,0,3,4,0,3,0,0
%N Number of times repeated reverse concatenation of n followed by n gives a prime, where n == 1,3,7 or 9 (mod 10), or 0 if no such prime exists.
%C Except for the first term every nonzero term is >1.
%C The larger numbers are probable primes. - _Joshua Zucker_, May 10 2006
%C The sequence probably continues 0 0 10 0 0 0 0 0 0 4 0 0 0 130 6 0 0 0 4 0 0 0 0 0 6 6 0 0 4 0 4 0 0 0 0 0 10 10 34 0 0 0 0 but the 0's in that list that correspond to 103, 107, 113, 119, 133, 143, 157, 169, 187, 203, 209, are not proved (but if there is a term there, it is more than 500). - _Joshua Zucker_, May 10 2006
%C Not only must each nonzero term be >1 (to avoid divisibility by 11), it also cannot equal 2 (mod 3) to avoid divisibility by 3. - _Joshua Zucker_, May 10 2006
%e The term corresponding to 19 is 7, as 7 concatenation of 91 followed by 19 is the least such prime. (9191919191919119 is a prime).
%Y Cf. A110408.
%K base,more,nonn
%O 0,6
%A _Amarnath Murthy_, Jul 30 2005
%E More terms from _Joshua Zucker_, May 10 2006
%E Edited by _T. D. Noe_, Oct 30 2008