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A333206
a(n) is the least decimal digit of n^3.
1
0, 1, 8, 2, 4, 1, 1, 3, 1, 2, 0, 1, 1, 1, 2, 3, 0, 1, 2, 5, 0, 1, 0, 1, 1, 1, 1, 1, 1, 2, 0, 1, 2, 3, 0, 2, 4, 0, 2, 1, 0, 1, 0, 0, 1, 1, 3, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 2, 0, 1, 2, 2, 0, 1, 0, 0, 1, 2, 0, 0, 1, 3, 3, 2, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 6, 0, 0, 3, 3, 1, 1
OFFSET
0,3
COMMENTS
Dean Hickerson found an infinite sequence of n such that a(n) > 0 (see Guy, sec F24). Are there infinitely many such that a(n) > 1? If not, what is the greatest n with a(n)=k for each k > 1?
Heuristically, we should expect on the order of ((10-m)^3/100)^d terms n with d digits and a(n) >= m. Since 5^3/100 > 1 > 4^3/100 we should expect infinitely many terms with a(n) >= 5 but only finitely many terms with a(n) >= 6. See A291644 for a(n) = 5. There are only two n <= 10^6 with a(n) >= 6, namely a(2) = 8 and a(92) = 6.
REFERENCES
R. Guy, Unsolved Problems in Number Theory (Third edition), Springer 2004.
LINKS
FORMULA
a(n) = A054054(n^3).
EXAMPLE
The least digit of 6^3=216 is 1, so a(6)=1.
MAPLE
seq(min(convert(n^3, base, 10)), n=0..200);
KEYWORD
nonn,base
AUTHOR
Robert Israel, Mar 12 2020
STATUS
approved