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

Author

Vladimir Shevelev, Apr 04 2011

Keywords

• nonn

Comments

Conjecture: if the definition is changed so that k runs through values 2, 3, ..., floor((n-2)/2) then, beginning with n=6, the sequence remains without changes. - Vladimir Shevelev, Apr 10 2011

From Vladimir Shevelev, Jan 21 2013: (Start)

Other conjectures:

1) Primes 5, 7, 13 are only primes p for which a(p) = 1;

2) Primes 11 and 19 are only primes p for which a(p) = 2;

3) Let n = m^2 and m be the least value of k for which the number of divisors d > 1 of n-k, such that k|(n-d), equals a(n). Then m is prime or even power of a prime. (End)

Links

• Alois P. Heinz, Table of n, a(n) for n = 4..10004

Crossrefs

Cf. A000005, A188579, A188586, A188591, A188592.

Programs

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