OFFSET
1,1
COMMENTS
From Danny Rorabaugh, Mar 21-24 2015: (Start)
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
LINKS
Danny Rorabaugh, Table of n, a(n) for n = 1..10000
PROG
(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
CROSSREFS
KEYWORD
easy,nonn,base
AUTHOR
Amarnath Murthy and Meenakshi Srikanth (menakan_s(AT)yahoo.com), Apr 29 2003
EXTENSIONS
Corrected and extended by Jason Earls, May 18 2004
STATUS
approved