A333206 a(n) is the least decimal digit of n^3.

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

Author

Robert Israel, Mar 12 2020

Keywords

• nonn
• base

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.

Examples

			The least digit of 6^3=216 is 1, so a(6)=1.
		

References

• R. Guy, Unsolved Problems in Number Theory (Third edition), Springer 2004.

Links

• Robert Israel, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A052044, A054054, A269250, A291639, A291640, A291641, A291642, A291643, A291644.

Programs

Maple

seq(min(convert(n^3,base,10)),n=0..200);

Formula

a(n) = A054054(n^3).