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a(n) = A356886(2^n+1)/A356886(2^n-1).
1

%I #64 Oct 02 2022 00:10:41

%S 3,3,3,5,5,7,11,11,13,17,19,19,23,29,31,37,37,41,43,47,53,59,59,61,67,

%T 71,73,79,83,83,89

%N a(n) = A356886(2^n+1)/A356886(2^n-1).

%C All terms are odd primes; some of them are repeated.

%C Conjecture: This sequence has the pattern: a prime repeated, a run of m times primes standing alone, the next prime repeated, a run of m+1 times primes standing alone, ... . All primes will be repeated once or follow in sequence. We know that if A356886(2^n-1) = p1 then A356886(2^n+1) will be p1*p2. p2 will be the smallest possible prime such that p1*p2 is not yet in the sequence A356886, thus p2 = a(n). Let p1*pn be already in A356886 and pn < p2, then we know that p1*pn will be on a position A356886(2^n-(2^(k+1)-2)) with some k > 0. This should explain this pattern. - _Thomas Scheuerle_, Sep 14 2022

%Y Cf. A000051, A000225, A356886.

%Y Cf. also A065091 (odd primes).

%K nonn,more

%O 1,1

%A _Paul Curtz_, Sep 09 2022

%E a(14)-a(24) from _Michel Marcus_, Sep 13 2022

%E a(25)-a(31) from _Chai Wah Wu_, Oct 01 2022