A335898 a(n) = a(floor((n-1)/a(n-1))) + a(floor((n-2)/a(n-2))) with a(1) = a(2) = 1.

1, 1, 2, 2, 2, 2, 3, 3, 2, 3, 4, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 4, 5, 5, 4, 5, 6, 5, 5, 6, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 5, 6, 6, 4, 6, 6, 5, 6, 6, 6, 6, 6, 7, 6, 6, 8, 7, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 6, 7, 7, 6, 7

Offset: 1

Author

Altug Alkan, following a suggestion from Andrew R. Booker, Jun 29 2020

Keywords

• nonn
• look
• easy

Comments

This sequence is a_1(n) where a_i(n) = Sum_{k=1..i+1} a_i(floor((n-k)/a_i(n-k))) with a_i(n) = 1 for n <= i+1.

Conjecture: This sequence hits every positive integer.

Links

• Altug Alkan, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A130535, A283207.

Programs

Mathematica

a[1] = a[2] = 1; a[n_] := a[n] = a[Floor[(n-1)/a[n-1]]] + a[Floor[(n-2)/a[n-2]]]; Array[a, 100] (* Amiram Eldar, Jun 29 2020 *)

PARI

a=vector(10^2); a[1]=a[2]=1; for(n=3, #a, a[n]=a[(n-1)\a[n-1]]+a[(n-2)\a[n-2]]); a