A188550 - OEIS

%I #39 Feb 06 2016 12:10:49

%S 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,

%T 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,

%U 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

%N 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)).

%C 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

%C From _Vladimir Shevelev_, Jan 21 2013: (Start)

%C Other conjectures:

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

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

%C 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)

%H Alois P. Heinz, <a href="/A188550/b188550.txt">Table of n, a(n) for n = 4..10004</a>

%F 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

%p with(numtheory):

%p a:= n-> max(seq(nops(select(x-> irem(x, k)=0,

%p     [seq(n-d, d=divisors(n-k) minus{1})])), k=2..floor(sqrt(n)))):

%p seq(a(n), n=4..120);  # _Alois P. Heinz_, Apr 04 2011

%t 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_ *)

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

%K nonn

%O 4,3

%A _Vladimir Shevelev_, Apr 04 2011