A327502 a(n) = n/A327501(n), where A327501(n) is the maximum divisor of n that is 1 or not a perfect power.
1, 1, 1, 2, 1, 1, 1, 4, 3, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 9, 1, 1, 1, 1, 16, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 32, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 27, 1, 1, 1, 1, 1
Offset: 1
Examples
The divisors of 36 that are 1 or not a perfect power are {1, 2, 3, 6, 12, 18}, so a(36) = 36/18 = 2.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..16383
- Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537
- Gus Wiseman, Sequences counting and encoding certain classes of multisets
Crossrefs
Programs
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Mathematica
Table[n/Max[Select[Divisors[n],GCD@@Last/@FactorInteger[#]==1&]],{n,100}]
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PARI
A327502(n) = if(n==1, 1, n/vecmax(select(x->((x>1) && !ispower(x)), divisors(n)))); \\ Antti Karttunen, Sep 19 2019 (after program given by Michel Marcus for A327501)
Formula
a(n) = n/A327501(n).
Comments