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

1,2

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.

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

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.

Example

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

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 *)

Prog

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

Crossrefs

Cf. A033676, A033677, A056737.

Sequence in context: A086112 A138798 A277708 * A360496 A356473 A328661

Adjacent sequences: A382482 A382483 A382484 * A382486 A382487 A382488

Keyword

nonn,look

Author

Clive Tooth, Mar 30 2025

Status

approved