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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. 1
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,3

LINKS

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

EXAMPLE

For 1 there is nothing to do. Hence a(1)=0.

For 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.

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

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Last modified March 25 01:17 EDT 2019. Contains 321450 sequences. (Running on oeis4.)