A349410 Length of cycle reached when iterating the mapping x-> n*A000005(x) on 1.

1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 3, 2, 1, 3, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 1, 4, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 5, 1, 2, 4, 2, 1, 1, 1, 2, 1, 2, 1, 2, 2, 4, 1, 2, 1, 2, 1, 2, 3, 3, 2, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 1, 4, 2, 2, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 1, 1, 3, 2

Offset: 1

Author

Tejo Vrush, Nov 16 2021

Keywords

• nonn

Examples

			For n = 9, 1 --> 9 --> 27 --> 36 --> 81 --> 45 --> 54 --> 72 --> 108 --> 108. The cycle reached has just one term: 108. Therefore, a(9) = 1.
		

Crossrefs

Cf. A000005, A000040, A062011.

Programs

Mathematica

a[n_] := Module[{s = NestWhileList[n * DivisorSigma[0, #] &, 1, UnsameQ, All]}, Differences[Position[s, s[[-1]]]][[1, 1]]]; Array[a, 100] (* Amiram Eldar, Nov 17 2021 *)

PARI

f(n, x) = n*numdiv(x);
find(nm, v) = {forstep (n=#v-1, 1, -1, if (v[#v] == v[n], return(#v-n);););}
a(n) = {my(list = List(), found=0, m=n); listput(list, m); while (! found, my(nm = f(n, m)); listput(list, nm); found = find(nm, list); m = nm;); found;} \\ Michel Marcus, Nov 17 2021

Python

from sympy import divisor_count
terms = []
for n in range(1, 101):
    s, t = [1], True
    while t:
        for i in range(2, len(s)):
            if s[-i] == s[-1]:
                t = False
                terms.append(i - 1)
                break
        s.append(n*divisor_count(s[-1]))
print(terms) # Gleb Ivanov, Nov 17 2021

Formula

a(A000040(n)) = 1.