A167234 Smallest number such that no two divisors of n are congruent modulo a(n).

1, 2, 3, 4, 3, 6, 4, 5, 5, 6, 3, 7, 5, 8, 8, 9, 3, 10, 4, 7, 8, 6, 3, 13, 7, 7, 5, 11, 3, 11, 4, 9, 7, 6, 8, 13, 5, 5, 7, 11, 3, 16, 4, 12, 13, 6, 3, 17, 5, 11, 9, 7, 3, 10, 7, 15, 5, 5, 3, 21, 7, 7, 11, 11, 7, 14, 4, 7, 7, 16, 3, 13, 5, 10, 13, 7, 8, 14, 4, 17, 7, 6, 3, 23, 9, 8, 5, 13, 3, 19, 8, 12

Offset: 1

Author

Franklin T. Adams-Watters, Oct 31 2009

Keywords

• nonn

Comments

What can we say about the asymptotic behavior of this sequence? Does it contain every integer > 2 infinitely often?

For n > 6, a(n) <= floor(n/2) + 1; but this seems to be a very crude estimate.

Links

• Paul Tek, Table of n, a(n) for n = 1..10000

Programs

Mathematica

allDiffQ[l_List] := (Length[l] == Length[DeleteDuplicates[l]]);
a[n_Integer] := Module[{ds = Divisors[n]},
   Catch[Do[If[allDiffQ[Mod[#, m] & /@ ds], Throw[m]], {m, n}]]];
a /@ Range[92] (* Peter Illig, Jul 11 2018 *)

PARI

alldiff(v)=v=vecsort(v);for(k=1,#v-1,if(v[k]==v[k+1],return(0)));1
a(n)=local(ds);ds=divisors(n);for(k=#ds,n,if(alldiff(vector(#ds,i,ds[i]%k)),return(k)))

Python

from sympy import divisors
from itertools import count
def a(n):
    d = divisors(n)
    return next(k for k in count(1) if len(set(di%k for di in d))==len(d))
print([a(n) for n in range(1, 93)]) # Michael S. Branicky, Jan 30 2023