A335925 a(n) = a(floor((n-1)/a(n-1))) + 1 with a(1) = 1.
1, 2, 2, 2, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 3, 4, 3, 3, 3, 4, 4, 4, 3, 4, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 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, 4, 5, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 4, 5, 5, 5, 5, 5, 4, 5, 4, 5, 4
Offset: 1
Links
- Seiichi Manyama, Table of n, a(n) for n = 1..10000
Programs
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Mathematica
a[1] = 1; a[n_] := a[n] = a[Floor[(n - 1)/a[n - 1]]] + 1 ; Array[a, 100] (* Amiram Eldar, Jun 30 2020 *)
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PARI
a=vector(10^2); a[1]=1; for(n=2, #a, a[n]=a[(n-1)\a[n-1]]+1); a;
Comments