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

4,3

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

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

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 *)

Crossrefs

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

Sequence in context: A355661 A109129 A304486 * A064122 A323424 A334098

Adjacent sequences: A188547 A188548 A188549 * A188551 A188552 A188553

Keyword

nonn

Author

Vladimir Shevelev, Apr 04 2011

Status

approved