A330786 Number of steps to reach 1 by iterating the absolute alternating sum-of-divisors function (A206369).

0, 1, 2, 3, 4, 2, 3, 5, 4, 4, 5, 3, 4, 3, 6, 6, 7, 4, 5, 4, 4, 5, 6, 5, 5, 4, 5, 5, 6, 6, 7, 5, 5, 7, 6, 5, 6, 5, 6, 5, 6, 4, 5, 7, 6, 6, 7, 6, 6, 5, 6, 6, 7, 5, 6, 7, 6, 6, 7, 6, 7, 7, 5, 6, 7, 5, 6, 7, 8, 6, 7, 7, 8, 6, 5, 6, 7, 6, 7, 8, 8, 6, 7, 6, 7, 5, 8, 6, 7, 6

Offset: 1

Author

Allan C. Wechsler, Dec 31 2019

Keywords

• nonn

Comments

A preliminary for investigating iterates of A206369.

Examples

			Iterating A206369 with a starting value of 27 gives 20, 12, 6, 2, 1, taking 5 steps to reach 1. So a(27) = 5.
		

References

• See A206369.

Crossrefs

Cf. A206369.

Programs

Mathematica

f[p_,e_] := Sum[(-1)^(e-k)*p^k, {k,0,e}]; s[1] = 1; s[n_] := Times @@ (f @@@ FactorInteger[n]); a[n_] := Length @ FixedPointList[s, n] - 2; Array[a, 90] (* Amiram Eldar, Jan 01 2020 *)

PARI

f(n) = sumdiv(n, d, eulerphi(n/d) * issquare(d)); \\ A206369
a(n) = {if (n==1, return (0)); my(nb = 1); while ((n = f(n)) != 1, nb++); nb;} \\ Michel Marcus, Jan 01 2020

Formula

a(1) = 0; for n > 1, a(n) = 1 + a(A206369(n)).