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A089592
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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.
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0
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2, 11, 17, 79, 107, 109, 709, 4003, 10009, 11003, 1000039, 1100009, 400000043, 1000000009, 150000000000007
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OFFSET
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1,1
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COMMENTS
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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*10^45-47 (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
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LINKS
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FORMULA
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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.
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EXAMPLE
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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.
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CROSSREFS
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KEYWORD
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nonn,base,less
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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