

A089592


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.


0



2, 11, 17, 79, 107, 109, 709, 4003, 10009, 11003, 1000039, 1100009, 400000043, 1000000009, 150000000000007
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OFFSET

1,1


COMMENTS

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


LINKS



FORMULA

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.


EXAMPLE

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.


CROSSREFS



KEYWORD

nonn,base,less


AUTHOR



EXTENSIONS



STATUS

approved



