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Restricted growth sequence transform of the ordered pair [A336466(n), A365426(n)].

For all i, j: A003602(i) = A003602(j) => a(i) = a(j).

Multiplicative because A336466 is.

Restricted growth sequence transform of the triplet [A336158(n), A336466(n), A336467(n)].

For all i, j >= 1:

A003602(i) = A003602(j) => a(i) = a(j),

a(i) = a(j) => A366381(i) = A366381(j),

a(i) = a(j) => A335880(i) = A335880(j),

a(i) = a(j) => A336390(i) = A336390(j),

a(i) = a(j) => A336470(i) = A336470(j).

Restricted growth sequence transform of the ordered pair [A336466(n), A336467(n)].

For all i, j: A003602(i) = A003602(j) => A366380(i) = A366380(j) => a(i) = a(j).

From Antti Karttunen, Apr 07 2020: (Start)

Also the least number of iterations of nondeterministic map k -> k-(k/p) needed to reach a power of 2, when any prime factor p of k can be used. The minimal length path to the nearest power of 2 (= 2^A064415(n)) is realized whenever one uses any of the A005087(k) distinct odd prime factors of the current k, at any step of the process. For example, this could be done by iterating with the map k -> k-(k/A078701(k)), i.e., by using the least odd prime factor of k (instead of the largest prime).

Proof: Viewing the prime factorization of changing k as a multiset ("bag") of primes, we see that liquefying any odd prime p with step p -> (p-1) brings at least one more 2 to the bag, while applying p -> (p-1) to any 2 just removes it from the bag, but gives nothing back. Thus the largest (and thus also the nearest) power of 2 is reached by eliminating - step by step - all odd primes from the bag, but none of 2's, and it doesn't matter in which order this is done.

The above implies also that the sequence is totally additive, which also follows because both A064097 and A064415 are. That A064097(n) = A329697(n) + A054725(n) for all n > 1 can be also seen by comparing the initial conditions and the recursion formulas of these three sequences.

For any n, A333787(n) is either the nearest power of 2 reached (= 2^A064415(n)), or occurs on some of the paths from n to there.

(End)

A003401 gives the numbers k where a(k) = A005087(k). See also A336477. - Antti Karttunen, Mar 16 2021

This is not necessarily the squarefree kernel. E.g., for n=19, phi(19)=18 is divisible by 9, an odd square. Values at which this kernel is 1 correspond to A003401 (polygons constructible with ruler and compass).

Multiplicative with a(2^e) = 1, a(p^e) = p^(e-1)*A000265(p-1). - Christian G. Bower, May 16 2005

For all i, j: A324400(i) = A324400(j) => a(i) = a(j) => A336118(i) = A336118(j).

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.