%I #43 Mar 17 2021 17:59:13
%S 1,1,1,1,1,1,3,1,1,1,5,1,3,3,1,1,1,1,9,1,3,5,11,1,1,3,1,3,7,1,15,1,5,
%T 1,3,1,9,9,3,1,5,3,21,5,1,11,23,1,9,1,1,3,13,1,5,3,9,7,29,1,15,15,3,1,
%U 3,5,33,1,11,3,35,1,9,9,1,9,15,3,39,1,1,5,41,3,1,21,7,5,11,1,9,11,15,23,9,1,3,9,5,1,25,1,51,3,3
%N Fully multiplicative with a(p) = A000265(p-1) for any prime p, where A000265(k) gives the odd part of k.
%C For the comment here, we extend the definition of the second 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 in such a 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 in A005382, then replacing it with 2p-1 (i.e., the corresponding terms of A005383), gives a new number m, for which a(m) = a(n). And vice versa, the same is true for any of the prime divisors > 3 of n that are in A005383, then replacing any one of them with (p+1)/2 will not affect the result. For example, a(37*37*37) = a(19*37*73) = 729 as 37 is both in A005382 and in A005383.
%C a(n) = A053575(n) for squarefree n (A005117). - _Antti Karttunen_, Mar 16 2021
%H Antti Karttunen, <a href="/A336466/b336466.txt">Table of n, a(n) for n = 1..16384</a>
%H Antti Karttunen, <a href="/A336466/a336466.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 a(n) = A000265(A003958(n)) = A000265(A333787(n)).
%F a(A000010(n)) = A336468(n) = a(A053575(n)).
%F A329697(a(n)) = A336396(n) = A329697(n) - A087436(n).
%F a(n) = A335915(n) / A336467(n). - _Antti Karttunen_, Mar 16 2021
%t Array[Times @@ Map[If[# <= 2, 1, (# - 1)/2^IntegerExponent[# - 1, 2]] &, Flatten[ConstantArray[#1, #2] & @@@ FactorInteger[#]]] &, 105] (* _Michael De Vlieger_, Jul 24 2020 *)
%o (PARI)
%o A000265(n) = (n>>valuation(n,2));
%o A336466(n) = { my(f=factor(n)); prod(k=1,#f~,A000265(f[k,1]-1)^f[k,2]); };
%Y Cf. A000265, A003958, A005117, A005382, A005383, A053575, A329697, A333787, A335915, A336396, A336467, A336468, A339870, A339876 [= a(A122111(n))], A339903 [= a(A003961(n))], A340084 [= gcd(n-1, a(n))], A340085, A340086.
%Y Cf. also A336460, A336470, A339974.
%K nonn,mult,look
%O 1,7
%A _Antti Karttunen_, Jul 22 2020