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A225974
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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.
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2
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OFFSET
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0,2
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COMMENTS
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This sequence is probably finite.
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LINKS
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FORMULA
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EXAMPLE
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a(1) is not 1, because 1 takes 0 steps to reach 0 or 1. - N. J. A. Sloane, Nov 05 2022
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)
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MATHEMATICA
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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
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PROG
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(Python)
from math import prod
from itertools import count, islice
def f(n): return prod(map(lambda x: x*x, map(int, str(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):
if v not in adict: adict[v] = k
while n in adict: yield adict[n]; n += 1
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CROSSREFS
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KEYWORD
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nonn,fini,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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