|
%I #32 Mar 13 2020 16:34:59
%S 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,
%T 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,
%U 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
%N a(n) is the least decimal digit of n^3.
%C 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?
%C 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.
%D R. Guy, Unsolved Problems in Number Theory (Third edition), Springer 2004.
%H Robert Israel, <a href="/A333206/b333206.txt">Table of n, a(n) for n = 0..10000</a>
%F a(n) = A054054(n^3).
%e The least digit of 6^3=216 is 1, so a(6)=1.
%p seq(min(convert(n^3,base,10)),n=0..200);
%Y Cf. A052044, A054054, A269250, A291639, A291640, A291641, A291642, A291643, A291644.
%K nonn,base
%O 0,3
%A _Robert Israel_, Mar 12 2020
|
