A336467 - OEIS

A336467

Fully multiplicative with a(2) = 1 and a(p) = A000265(p+1) for odd primes p, with A000265(k) giving the odd part of k.

1, 1, 1, 1, 3, 1, 1, 1, 1, 3, 3, 1, 7, 1, 3, 1, 9, 1, 5, 3, 1, 3, 3, 1, 9, 7, 1, 1, 15, 3, 1, 1, 3, 9, 3, 1, 19, 5, 7, 3, 21, 1, 11, 3, 3, 3, 3, 1, 1, 9, 9, 7, 27, 1, 9, 1, 5, 15, 15, 3, 31, 1, 1, 1, 21, 3, 17, 9, 3, 3, 9, 1, 37, 19, 9, 5, 3, 7, 5, 3, 1, 21, 21, 1, 27, 11, 15, 3, 45, 3, 7, 3, 1, 3, 15, 1, 49, 1, 3, 9, 51, 9, 13, 7, 3

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OFFSET

1,5

COMMENTS

For the comment here, we extend the definition of the first kind of Cunningham chain (see Wikipedia-article) so that also isolated primes for which neither (p-1)/2 nor 2p+1 is a prime are considered to be in singular chains, that is, in chains of the length one. If we replace one or more instances of any particular odd prime factor p in n with any odd prime q of the same Cunningham chain, so that m = (q^k)*n / p^(e-k), where e is the exponent of p of n, and k <= e is the number of instances of p replaced with q, then it holds that a(m) = a(n), and by induction, the value stays invariant for any number of such replacements. Note also that A001222, but not necessarily A001221 will stay invariant in such changes.

For example, if some of the odd prime divisors p of n are Sophie Germain primes (in A005384), then replacing any of them with 2p+1 ("safe primes", i.e., the corresponding terms of A005385), gives a new number m, for which a(m) = a(n). And vice versa, the same is true for any safe prime factors > 5 of n (that are in A005385), then replacing any one of them with (p-1)/2 will not affect the result. For example, a(5*11*23*47) = a(11*11*23*23) = a(5^4) = a(11^4) = a(23^4) = 81, as 5, 11, 23 and 47 are in the same Cunningham chain of the first kind.

LINKS

Antti Karttunen, Table of n, a(n) for n = 1..16384

Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537

Wikipedia, Cunningham chain

FORMULA

For all n >= 1, A331410(a(n)) = A336921(n).

From Antti Karttunen, Nov 21 2023: (Start)

a(n) = A335915(n) / A336466(n).

a(1) = 1, and for n > 1, a(n) = A000265(A206787(n)) * a(A336651(n)).

(End)

PROG

(PARI)