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%I #26 Nov 21 2023 18:16:00
%S 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,
%T 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,
%U 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
%N 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.
%C 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.
%C 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.
%H Antti Karttunen, <a href="/A336467/b336467.txt">Table of n, a(n) for n = 1..16384</a>
%H Antti Karttunen, <a href="/A336467/a336467.txt">Data supplement: n, a(n) computed for n = 1..65537</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Cunningham_chain">Cunningham chain</a>
%F For all n >= 1, A331410(a(n)) = A336921(n).
%F From _Antti Karttunen_, Nov 21 2023: (Start)
%F a(n) = A335915(n) / A336466(n).
%F a(1) = 1, and for n > 1, a(n) = A000265(A206787(n)) * a(A336651(n)).
%F (End)
%o (PARI)
%o A000265(n) = (n>>valuation(n,2));
%o A336467(n) = { my(f=factor(n)); prod(k=1,#f~,if(2==f[k,1],1,(A000265(f[k,1]+1))^f[k,2])); };
%Y Cf. A000265, A003959, A206787, A336390, A336392, A336651, A336921, A336922, A351461.
%Y Cf. also A335915, A336466 (similar sequences).
%K nonn,mult
%O 1,5
%A _Antti Karttunen_, Jul 22 2020
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