A225974 Multiplicative persistence with squares of decimal digits: smallest number such that the number of iterations of "multiply digits squared" needed to reach 0 or 1 equals n.

0, 10, 25, 5, 8, 6, 3, 2

Offset: 0

Author

Michel Lagneau, May 22 2013

Keywords

• nonn
• fini
• base

Comments

This sequence is probably finite.

Examples

			a(1) is not 1, because 1 takes 0 steps to reach 0 or 1. - _N. J. A. Sloane_, Nov 05 2022
From _Mohammed Yaseen_, Oct 11 2022: (Start)
5 -> 25 -> 4*25 = 100 -> 1*0*0 = 0. So a(3) = 5.
8 -> 64 -> 36*16 = 576 -> 25*49*36 = 44100 -> 16*16*1*0*0 = 0. So a(4) = 8. (End)
		

Crossrefs

Cf. A003001, A031348, A031349.

Programs

Mathematica

lst = {}; n = 0; Do[While[True, k = n; c = 0; While[k > 9, k = Times @@ IntegerDigits[k]^2; c++]; If[c == l, Break[]]; n++]; AppendTo[lst, n], {l, 0, 7}]; lst

Python

from math import prod
from itertools import count, islice
def f(n): return prod(map(lambda x: x*x, map(int, str(n))))
def A031348(n):
    c = 0
    while n not in {0, 1}: n, c = f(n), c+1
    return c
def agen():
    adict, n = dict(), 0
    for k in count(0):
        v = A031348(k)
        if v not in adict: adict[v] = k
        while n in adict: yield adict[n]; n += 1
print(list(islice(agen(), 8))) # Michael S. Branicky, Oct 13 2022

Formula

a(n) = Min{k >= 0 : A031348(k) = n}. - Michael S. Branicky, Oct 13 2022

Extensions

a(3)-a(6) corrected and a(7) from Mohammed Yaseen, Oct 11 2022