A288569 Start with n and repeatedly apply the map x -> A087207(x) until we reach 0; a(n) is the number of steps needed, or -1 if 0 is never reached.

1, 2, 3, 2, 3, 4, 3, 2, 3, 4, 3, 4, 3, 4, 5, 2, 3, 4, 3, 4, 5, 4, 3, 4, 3, 6, 3, 4, 3, 4, 3, 2, 5, 6, 5, 4, 3

Offset: 1

Author

N. J. A. Sloane, Jun 25 2017

Keywords

• nonn
• more
• hard

Comments

There is a conjecture that 0 is always reached - see A087207.

The map x -> A087207(x) can also start with A019565(n) until it reaches 1 (which is A019565(0)). - Flávio V. Fernandes, Feb 24 2025

Examples

			10 -> 5 -> 4 -> 1 -> 0 reaches 0 in 4 steps, so a(10)=4.
38 = 2*19 -> 129 = 3*43 -> 8194 = 2*17*241 -> 4503599627370561 = 3^2*37*71*190483425427 -> ..., and a(38) is presently unknown.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000. If a(n) is not known, it is given in the form k + a(x). [A large file]

Crossrefs

Cf. A087207.

Cf. A019565.

Programs

Maple

f:= proc(n) local i; option remember;
  add(2^(numtheory:-pi(t)-1), t = numtheory:-factorset(n)) end proc:
g:= proc(n) local t,count;
   t:= n;
   for count from 0 while t <> 0 do
      t:= f(t)
   od;
   count
end proc:
map(g, [$1..37]); # Robert Israel, Jun 25 2017

Mathematica

f[n_] := Total[2^(PrimePi /@ FactorInteger[n][[All, 1]]-1)]; f[1] = 0;
g[n_] := Module[{t, count}, t = n; For[count = 0, t != 0, count++, t = f[t]]; count];
Table[g[n], {n, 1, 37}] (* Jean-François Alcover, Aug 15 2023, after Robert Israel *)

Python

from sympy import factorint, primepi
def f(n):
    return 0 if n < 2 else sum(1 << int(primepi(i-1)) for i in factorint(n))
def a(n):
    fn, c = n, 0
    while not fn == 0: fn, c = f(fn), c+1
    return c
print([a(n) for n in range(1, 38)]) # Michael S. Branicky, Jul 11 2022