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.

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, 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, 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

Offset: 1

Author

Clive Tooth, Mar 30 2025

Keywords

• nonn
• look

Comments

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.

Examples

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

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Clive Tooth, [Sum_{i=2..n} a(i)]/[Sum_{i=2..n} i/log(i)] for n=10^7 to 10^9

Crossrefs

Cf. A033676, A033677, A056737.

Programs

Maple

f:= proc(n) local d;
     d:= max(select(t -> t^2 <= n, numtheory:-divisors(n)));
     ceil(n/d^2)
end proc:
map(f, [$1..100]); # Robert Israel, Apr 30 2025

Mathematica

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

PARI

a(n) = my(d=divisors(n)); ceil(n/d[(length(d)+1)\2]^2); \\ Michel Marcus, Apr 07 2025

Formula

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

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

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