A288914 a(1) = 2; a(n) = a(floor(n/a(n-1))) + 1 for n > 1.

2, 3, 3, 3, 3, 4, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 4, 5, 4, 4, 4, 4, 4, 5, 5, 5, 4, 5, 4, 5, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5

Offset: 1

Author

Altug Alkan, Jun 19 2017

Keywords

• nonn
• easy

Comments

Least values of k such that a(k) = n are 1, 2, 6, 24, 120, 720, 5040, ... (n > 1).

These appear to be (n-1)!. Verified for 2 <= n <= 11. - Robert Israel, Jun 22 2017

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000142, A002024, A130147.

Programs

Maple

f:= proc(n) option remember;
  procname(floor(n/procname(n-1)))+1
end proc:
f(1):= 2:
map(f, [$1..200]);# Robert Israel, Jun 22 2017

Mathematica

a = {2}; Do[AppendTo[a, a[[Floor[n/a[[n - 1]] ] ]] + 1], {n, 2, 105}]; a (* Michael De Vlieger, Jun 21 2017 *)

PARI

q=vector(10000); q[1]=2; for(n=2, #q, q[n] = q[n\q[n-1]]+1); q