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A382485 a(n) = ceiling(n/d^2) where d is the largest divisor of n which is not greater than the square root of n.

%I A382485 #64 May 01 2025 01:40:27
%S A382485 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 A382485 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 A382485 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 A382485 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 A382485 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 A382485 Robert Israel, <a href="/A382485/b382485.txt">Table of n, a(n) for n = 1..10000</a>
%H A382485 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 A382485 a(n) = ceiling(A033677(n)/A033676(n)).
%F A382485 a(n) = 1 iff n is a square.
%F A382485 a(n) = n iff n is prime, or 1.
%e A382485 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 A382485 f:= proc(n) local d;
%p A382485      d:= max(select(t -> t^2 <= n, numtheory:-divisors(n)));
%p A382485      ceil(n/d^2)
%p A382485 end proc:
%p A382485 map(f, [$1..100]); # _Robert Israel_, Apr 30 2025
%t A382485 a[n_]:=Ceiling[n/(Select[Divisors[n],#<=Sqrt[n]&][[-1]])^2];Array[a,89] (* _James C. McMahon_, Apr 07 2025 *)
%o A382485 (PARI) a(n) = my(d=divisors(n)); ceil(n/d[(length(d)+1)\2]^2); \\ _Michel Marcus_, Apr 07 2025
%Y A382485 Cf. A033676, A033677, A056737.
%K A382485 nonn,look
%O A382485 1,2
%A A382485 _Clive Tooth_, Mar 30 2025