|
%I #4 Aug 28 2017 20:15:49
%S 1,5,6,8,11,12,13,17,21,23,24,25,26,27,28,31,39,41,44,45,49,51,53,54,
%T 55,56,57,58,61,64,68,71,75,81,82,83,85,88,91,97,98,104,105,108,111,
%U 113,114,119,121,122,125,129,136,137,139,146,147,148,151,153,156
%N Numbers k such that 1 is the smallest decimal digit of k^3.
%e 11 is in the sequence because 11^3 = 1331, the smallest decimal digit of which is 1.
%o (PARI) select(k->vecmin(digits(k^3))==1, vector(500, k, k))
%Y Cf. A291639, A291641, A291642, A291643, A291644.
%K nonn,base
%O 1,2
%A _Colin Barker_, Aug 28 2017
|
