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A082615
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Palindromes such that the GCD of any pair of successive terms is a palindromic prime.
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2
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2, 4, 6, 8, 22, 33, 44, 55, 66, 77, 88, 99, 111, 141, 171, 222, 232, 242, 252, 262, 272, 282, 292, 414, 424, 434, 444, 454, 464, 474, 484, 494, 606, 616, 626, 636, 646, 656, 666, 676, 686, 696, 717, 747, 777, 828, 838, 848, 858, 868, 878, 888, 898, 2002, 3113
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OFFSET
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1,1
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COMMENTS
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This sequence is generated by setting a(1)=2 and letting a(n+1) be the least palindromic integer strictly greater than a(n) such that GCD(a(n),a(n+1)) is a palindromic prime.
By initiating with a different palindrome, a different sequence is generated: for example, [3, 6, 8, ...] or [151, 1661, 1771, 1881, 1991, 2002, 3113, ...]. 323 is the smallest palindrome greater than 1 whose sequence does not eventually coincide with the present sequence. Indeed, 323 is the smallest palindrome greater than 1 with no palindromic prime factor (thus, as with 1, the sequence beginning with 323 has only one term).
I conjecture that every sequence generated in this fashion, beginning with a palindrome that has a palindromic prime factor, is infinite and eventually coincides with this sequence.
(End)
The sequence, started with any palindrome having a palindromic prime factor, is certainly infinite: if a(n) is a palindrome divisible by palindromic prime p, then M = p*(1+10^m) for large enough m is another palindrome; if m is divisible by the order of 10 mod q for every prime q other than 2 and 5 dividing a(n), then gcd(M,a(n)) = p. - Robert Israel, Mar 24 2015
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LINKS
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PROG
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(PARI) ispal(n) = d = digits(n); Vecrev(d) == d;
lista(nn) = {prev = 2; print1(prev, ", "); for (n=2, nn, k = prev+1; while(!(ispal(k) && (p=gcd(prev, k)) && isprime(p) && ispal(p)), k++); print1(k, ", "); prev = k; ); } \\ Michel Marcus, Mar 24 2015
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CROSSREFS
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KEYWORD
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easy,nonn,base
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AUTHOR
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Amarnath Murthy and Meenakshi Srikanth (menakan_s(AT)yahoo.com), Apr 29 2003
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EXTENSIONS
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STATUS
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approved
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