A188550 Maximal number of divisors d>1 of n-k such that n-d is a multiple of k, when k runs through values 2, 3, ..., floor(sqrt(n)).
1, 1, 2, 1, 2, 2, 3, 2, 2, 1, 4, 3, 2, 3, 4, 2, 3, 4, 4, 3, 2, 3, 6, 4, 4, 3, 4, 3, 4, 4, 5, 4, 4, 3, 6, 6, 3, 3, 6, 3, 4, 4, 4, 5, 4, 4, 8, 5, 6, 3, 4, 4, 4, 6, 6, 4, 4, 1, 8, 6, 4, 6, 6, 3, 5, 4, 4, 3, 4, 3, 9, 8, 6, 5, 6, 3, 4, 4, 8, 5, 6, 5, 8, 6, 4, 3, 6, 6, 6, 8, 6, 6, 4, 3, 10, 6, 8, 5, 6, 4, 6, 6, 6, 7, 4, 3, 8, 9, 4, 4, 8, 5, 6, 6, 4, 5, 4
Offset: 4
Keywords
Links
- Alois P. Heinz, Table of n, a(n) for n = 4..10004
Programs
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Maple
with(numtheory): a:= n-> max(seq(nops(select(x-> irem(x, k)=0, [seq(n-d, d=divisors(n-k) minus{1})])), k=2..floor(sqrt(n)))): seq(a(n), n=4..120); # Alois P. Heinz, Apr 04 2011 -
Mathematica
a[n_] := Max @ Table[ Length @ Select[Table[n-d, {d, Divisors[n-k] // Rest} ], Mod[#, k] == 0&], {k, 2, Floor[Sqrt[n]]}]; Table[a[n], {n, 4, 120}] (* Jean-François Alcover, Feb 06 2016, after Alois P. Heinz *)
Formula
lim sup_{n -> infinity} a(n) = infinity. Indeed, it is easy to show that a(2^(2^n+1)) >= 2^n. Moreover, for n>5, we have a(2^(2^n+1)) > 2^n. - Vladimir Shevelev, Apr 09 2011
Comments