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

1,2

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

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 *)

Prog

(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

Crossrefs

Sequence in context: A079065 A097272 A126630 * A088043 A332931 A248376

Adjacent sequences: A167231 A167232 A167233 * A167235 A167236 A167237

Keyword

nonn

Author

Franklin T. Adams-Watters, Oct 31 2009

Status

approved