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%I #16 Mar 30 2024 12:33:02
%S 1,1,2,2,2,3,3,4,5,5,5,6,6,6,7,7,7,8,8,8,9,9,9,10,10,10,11,11,11,12,
%T 12,12,13,13,13,14,14,14,15,15,15,16,16,16,17,17,17,18,18,18,19,19,19,
%U 20,20,20,21,21,21,22,22,22,23,24,24,25,25,25,26,26,26,27,27,27,28,28
%N a(n) is the number of integers m from 1 to n inclusive such that m^m is a cube.
%C Dick Hess gave a puzzle at a "Gathering for Gardner" meeting asking for a(40).
%C a(n) is the number of integers not exceeding n that are divisible by 3 plus the number of cubes in the same range that are not divisible by 3.
%F a(n) = floor(n/3) + floor(n^(1/3)) - floor(n^(1/3)/3).
%e Suppose n = 40. There are 13 numbers in the range that are divisible by 3 and should be counted. In addition, there are two cubes 1 and 8 that are not divisible by 3. Thus, a(40) = 15.
%t Table[Floor[n/3] + Floor[n^(1/3)] - Floor[n^(1/3)/3], {n, 100}]
%Y Cf. A000578, A329547.
%K nonn
%O 1,3
%A _Tanya Khovanova_, Mar 28 2024
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