A382485 - OEIS

%I #64 May 01 2025 01:40:27

%S 1,2,3,1,5,2,7,2,1,3,11,2,13,4,2,1,17,2,19,2,3,6,23,2,1,7,3,2,29,2,31,

%T 2,4,9,2,1,37,10,5,2,41,2,43,3,2,12,47,2,1,2,6,4,53,2,3,2,7,15,59,2,

%U 61,16,2,1,3,2,67,5,8,2,71,2,73,19,3,5,2,3,79,2,1,21,83,2,4,22,10,2,89

%N a(n) = ceiling(n/d^2) where d is the largest divisor of n which is not greater than the square root of n.

%C Conjecture: There exists some constant, k, approximately equal to 1.2, such that a(n) is of average order k*n/log(n). See Tooth Link for evidence.

%H Robert Israel, <a href="/A382485/b382485.txt">Table of n, a(n) for n = 1..10000</a>

%H Clive Tooth, <a href="/A382485/a382485_4.png">[Sum_{i=2..n} a(i)]/[Sum_{i=2..n} i/log(i)] for n=10^7 to 10^9</a>

%F a(n) = ceiling(A033677(n)/A033676(n)).

%F a(n) = 1 iff n is a square.

%F a(n) = n iff n is prime, or 1.

%e a(12)=2 because the largest factor of 12, which is not greater than sqrt(12), is 3; and ceiling(12/3^2)=2.

%p f:= proc(n) local d;

%p      d:= max(select(t -> t^2 <= n, numtheory:-divisors(n)));

%p      ceil(n/d^2)

%p end proc:

%p map(f, [$1..100]); # _Robert Israel_, Apr 30 2025

%t a[n_]:=Ceiling[n/(Select[Divisors[n],#<=Sqrt[n]&][[-1]])^2];Array[a,89] (* _James C. McMahon_, Apr 07 2025 *)

%o (PARI) a(n) = my(d=divisors(n)); ceil(n/d[(length(d)+1)\2]^2); \\ _Michel Marcus_, Apr 07 2025

%Y Cf. A033676, A033677, A056737.

%K nonn,look

%O 1,2

%A _Clive Tooth_, Mar 30 2025