A322163 Minimal number of steps needed to get from n to 1, where for n > 1 the next step is to either n-1 or max(a,b) for any a > 1 and b > 1 such that ab=n.

0, 1, 2, 2, 3, 3, 4, 3, 3, 4, 5, 3, 4, 5, 4, 3, 4, 4, 5, 4, 5, 6, 7, 4, 4, 5, 4, 5, 6, 4, 5, 4, 5, 5, 5, 4, 5, 6, 5, 4, 5, 5, 6, 6, 4, 5, 6, 4, 5, 5, 5, 5, 6, 4, 5, 4, 5, 6, 7, 4, 5, 6, 4, 4, 5, 6, 7, 5, 6, 5, 6, 4, 5, 6, 5, 6, 6, 5, 6, 4, 4, 5, 6, 4, 5, 6, 7

Offset

1,3

Links

Antoine Mathys, Table of n, a(n) for n = 1..20000

Example

For n=1, there is nothing to do. Hence a(1)=0.
For n=4, the possible sequences of steps are 4->3->2->1 and 4->2->1. Thus the minimal number of steps needed to reach 1 is a(4)=2.
For n=6, the possible sequences of steps are 6->5->4->3->2->1, 6->5->4->2->1 and 6->3->2->1. Thus the minimal number of steps needed to reach 1 is a(6)=3.

Mathematica

divs[n_] := Append[Select[Most[Divisors[n]], #>= Sqrt[n] &], n-1]; a[0] = 0; a[1] = 0; a[n_] := a[n] = 1 + Min[a/@divs[n]]; Array[a, 100] (* Amiram Eldar, Nov 29 2018 *)

Prog

(C)
#include <stdio.h>
int main ()
{
    const int N = 100;
    int steps[N + 1];
    steps[1] = 0;
    for (int n = 2; n <= N; n++) {
        int next = n - 1;
        for (int i = n - 1; i * i >= n; i--) {
            if (n % i == 0) {
                if (steps[i] < steps[next]) {
                    next = i;
                }
            }
        }
        steps[n] = 1 + steps[next];
    }
    for (int n = 1; n <= N; n++) {
        printf ("%d %d\n", n, steps[n]);
    }
}
(PARI) seq(n)={my(v=vector(n)); for(n=2, n, my(m=v[n-1]); fordiv(n, d, if(d>=n/d && d<n, m=min(m, v[d]))); v[n]=m+1); v} \\ Andrew Howroyd, Nov 29 2018

Crossrefs

Sequence in context: A173419 A099053 A230697 * A075167 A253555 A252464

Adjacent sequences: A322160 A322161 A322162 * A322164 A322165 A322166

Keyword

nonn

Author

Antoine Mathys, Nov 29 2018

Status

approved