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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. 2
0, 10, 25, 5, 8, 6, 3, 2 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
This sequence is probably finite.
LINKS
FORMULA
a(n) = Min{k >= 0 : A031348(k) = n}. - Michael S. Branicky, Oct 13 2022
EXAMPLE
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)
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
PROG
(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
CROSSREFS
Sequence in context: A265150 A057462 A180043 * A251317 A251201 A048195
KEYWORD
nonn,fini,base
AUTHOR
Michel Lagneau, May 22 2013
EXTENSIONS
a(3)-a(6) corrected and a(7) from Mohammed Yaseen, Oct 11 2022
STATUS
approved

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Last modified March 28 16:58 EDT 2024. Contains 371254 sequences. (Running on oeis4.)