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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; a = 0; a[n_] := a[n] = 1 + Min[a/@divs[n]]; Array[a, 100] (* Amiram Eldar, Nov 29 2018 *) PROG (C) #include int main () {     const int N = 100;     int steps[N + 1];     steps = 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

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