%I #16 Nov 22 2014 00:21:32
%S 2,11,17,79,107,109,709,4003,10009,11003,1000039,1100009,400000043,
%T 1000000009,150000000000007
%N For any prime p, define a sequence S_p: S_p(1) = p and S_p(n+1) is the least prime > S_p(n) that begins with the last digit of S_p(n). Let f(p) be the first member of S_p that is the digit reversal of the previous member. Sequence contains primes p that such that f(p) does not equal f(q) for any q < p.
%C The corresponding values f(a(n)) are 9001, 31, 71, 97, 701, 70001, 907, 7*10^8+1, 90001, 30011, 7*10^9+1, 9000011, 9*10^22+1, 9*10^9+1, 7*10^14+51, 7*10^19+13, 9*10^46+7, 7*10^50+43, 9*10^60+227. p = 4*10^4547 (between a(18) and a(19)) appears to be the first prime such that f(p) doesn't exist: the digit reversal doesn't occur < 10^300 and is unlikely to occur later.  _David Wasserman_, Oct 03 2005
%F Begin with any prime, continue with the next prime having same beginning digit as that of the last digit of the prime preceding until the first prime reversal is found.
%e In the sequence beginning with the prime 2, continue 23 31 101 103 307 701 1009 9001 . . . . [A061448]. The first occurrence of a prime reversal beginning with the prime 2 is 1009 and 9001. This is a different first occurrence prime reversal than that found in the sequence beginning with 11 which continues to 13 31.
%Y Cf. A061448.
%K nonn,base,less
%O 1,1
%A _Enoch Haga_, Dec 29 2003
%E More terms from _David Wasserman_, Oct 03 2005
%E Incorrect terms removed by _Charles R Greathouse IV_, Nov 21 2014
