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A079645 Numbers j such that the integer part of the cube root of j divides j.

%I A079645 #30 Oct 07 2023 21:36:50
%S A079645 1,2,3,4,5,6,7,8,10,12,14,16,18,20,22,24,26,27,30,33,36,39,42,45,48,
%T A079645 51,54,57,60,63,64,68,72,76,80,84,88,92,96,100,104,108,112,116,120,
%U A079645 124,125,130,135,140,145,150,155,160,165,170,175,180,185,190,195,200,205,210
%N A079645 Numbers j such that the integer part of the cube root of j divides j.
%C A079645 Concrete Mathematics Casino Problem - Winners.
%D A079645 R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. 2nd Edition. Addison-Wesley, Reading, MA, 1994. Section 3.2, pp. 74-76.
%H A079645 G. C. Greubel, <a href="/A079645/b079645.txt">Table of n, a(n) for n = 1..10000</a>
%H A079645 B. Cloitre, <a href="http://dl.dropbox.com/u/46675017/divisibility_sequences.pdf">Some divisibility sequences</a>.
%F A079645 For n = (k/2)*(3*k+11) - m for some fixed m >= 0 with n > ((k-1)/2)*(3*(k-1) + 11) we have a(n) = k^3 + 3*k^2 + (3-m)*k. - _Benoit Cloitre_, Jan 22 2012
%e A079645 252^(1/3) = 6.316359597656... and 252/6 = 42 hence 252 is in the sequence.
%p A079645 t1:=[]; for n from 1 to 500 do t2:=floor(n^(1/3)); if n mod t2 = 0 then t1:=[op(t1),n]; fi; od: t1; # _N. J. A. Sloane_, Oct 29 2006
%t A079645 Select[Range[1000], Mod[#, Floor[Power[#, 1/3]]] == 0 &]
%t A079645 Select[Range[1000],Divisible[#,Floor[CubeRoot[#]]]&] (* _Harvey P. Dale_, Jun 19 2023 *)
%o A079645 (Magma) [n: n in [1..250] | n mod Floor(n^(1/3)) eq 0 ]; // _G. C. Greubel_, Jul 20 2023
%o A079645 (SageMath) [n for n in (1..250) if n%(floor(n^(1/3)))==0 ] # _G. C. Greubel_, Jul 20 2023
%Y A079645 Cf. A006446, A032378, A079631, A112873, A120721.
%K A079645 nonn
%O A079645 1,2
%A A079645 _Benoit Cloitre_, Jan 31 2003