A373203 a(n) = minimum k>1 such that n^k contains all distinct decimal digits of n.
2, 2, 5, 5, 3, 2, 2, 5, 5, 3, 2, 2, 3, 5, 4, 6, 5, 5, 5, 7, 5, 3, 4, 7, 3, 2, 8, 2, 5, 3, 5, 4, 3, 3, 3, 6, 6, 5, 4, 3, 3, 6, 7, 4, 3, 4, 4, 4, 4, 3, 2, 3, 7, 5, 3, 2, 3, 5, 5, 3, 2, 3, 5, 2, 2, 3, 2, 3, 4, 5, 5, 3, 3, 3, 2, 3, 2, 5, 5, 5, 5
Offset: 0
Examples
For n=12, a(12)=3 because 12^3=1728 contains all decimal digits of n. Compare to A253600(12)=2 because 12^2=144 contains any digit of n.
Links
- Michel Marcus, Table of n, a(n) for n = 0..10000
Programs
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Mathematica
seq={}; Do[k=1;Until[ContainsAll[IntegerDigits[n^k],IntegerDigits[n] ],k++];AppendTo[seq,k] ,{n,0,80}];seq -
PARI
a(n) = my(k=2, d=Set(digits(n))); while(setintersect(Set(digits(n^k)), d) != d, k++); k; \\ Michel Marcus, Jun 01 2024
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Python
from itertools import count def a(n): s = set(str(n)) return next(k for k in count(2) if s <= set(str(n**k))) print([a(n) for n in range(81)]) # Michael S. Branicky, May 27 2024
Formula
A111442(n) = n^a(n).
Comments