A225974 - OEIS

%I #16 Nov 05 2022 22:24:19

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

%N 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.

%C This sequence is probably finite.

%F a(n) = Min{k >= 0 : A031348(k) = n}. - _Michael S. Branicky_, Oct 13 2022

%e a(1) is not 1, because 1 takes 0 steps to reach 0 or 1. - _N. J. A. Sloane_, Nov 05 2022

%e From _Mohammed Yaseen_, Oct 11 2022: (Start)

%e 5 -> 25 -> 4*25 = 100 -> 1*0*0 = 0. So a(3) = 5.

%e 8 -> 64 -> 36*16 = 576 -> 25*49*36 = 44100 -> 16*16*1*0*0 = 0. So a(4) = 8. (End)

%t 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

%o (Python)

%o from math import prod

%o from itertools import count, islice

%o def f(n): return prod(map(lambda x: x*x, map(int, str(n))))

%o def A031348(n):

%o     c = 0

%o     while n not in {0, 1}: n, c = f(n), c+1

%o     return c

%o def agen():

%o     adict, n = dict(), 0

%o     for k in count(0):

%o         v = A031348(k)

%o         if v not in adict: adict[v] = k

%o         while n in adict: yield adict[n]; n += 1

%o print(list(islice(agen(), 8))) # _Michael S. Branicky_, Oct 13 2022

%Y Cf. A003001, A031348, A031349.

%K nonn,fini,base

%O 0,2

%A _Michel Lagneau_, May 22 2013

%E a(3)-a(6) corrected and a(7) from _Mohammed Yaseen_, Oct 11 2022