A327503 Number of steps to reach a fixed point starting with n and repeatedly taking the quotient by the maximum divisor that is 1 or not a perfect power (A327501, A327502).

0, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1

Offset: 1

Author

Gus Wiseman, Sep 16 2019

Keywords

• nonn

Comments

First differs from A052409 and A158378 at a(216) = 2, A052409(216) = A158378(216) = 3.

A multiset is aperiodic if its multiplicities are relatively prime. The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k). Heinz numbers of aperiodic multisets are numbers that are not perfect powers (A007916).

a(n) does not depend only on the prime signature of n. A351948 gives the positions where a(A046523(n)) <> a(n). n = 125000 is the first time this happens, see the examples. - Antti Karttunen, Apr 03 2022

Examples

			The transformation A327502 takes 144 -> 2 -> 1, so a(144) = 2.
From _Antti Karttunen_, Apr 03 2022: (Start)
For n = 1728 = 2^6 * 3^3, A327501(1728) = 864 = 2^5 * 3^3, and therefore A327502(1728) = 1728/864 = 2. A327501(2) = 2, thus A327502(2) = 2/2 = 1, so we reached 1 (= A327502(1)) in two steps, and therefore a(1728) = 2.
For n = 125000 = 2^3 * 5^6, A327501(125000) = 31250 = 2^1 * 5^6, and therefore A327502(125000) = 125000/31250 = 4. A327501(4) = 2, thus A327502(4) = 4/2 = 2, from which we reach 1 in one more step, therefore a(125000) = 3.
(End)
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537
• Gus Wiseman, Sequences counting and encoding certain classes of multisets

Crossrefs

See link for additional cross-references.

Cf. A000961, A001597, A007916 (positions of 1's), A052409, A158378, A303386, A327501, A327502, A351948.

Cf. also A327500.

Programs

Mathematica

Table[Length[FixedPointList[#/Max[Select[Divisors[#],GCD@@Last/@FactorInteger[#]==1&]]&,n]]-2,{n,100}]

PARI

A327502(n) = if(n==1, 1, n/vecmax(select(x->((x>1) && !ispower(x)), divisors(n))));
A327503(n) = { my(u=A327502(n)); if(u==n, 0, 1+A327503(u)); }; \\ Antti Karttunen, Apr 02 2022

Formula

a(2^n) = n.

Extensions

Data section extended up to 105 terms by Antti Karttunen, Apr 02 2022