A079645 Numbers j such that the integer part of the cube root of j divides j.
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, 51, 54, 57, 60, 63, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 104, 108, 112, 116, 120, 124, 125, 130, 135, 140, 145, 150, 155, 160, 165, 170, 175, 180, 185, 190, 195, 200, 205, 210
Offset: 1
Keywords
Examples
252^(1/3) = 6.316359597656... and 252/6 = 42 hence 252 is in the sequence.
References
- R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. 2nd Edition. Addison-Wesley, Reading, MA, 1994. Section 3.2, pp. 74-76.
Links
- G. C. Greubel, Table of n, a(n) for n = 1..10000
- B. Cloitre, Some divisibility sequences.
Programs
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Magma
[n: n in [1..250] | n mod Floor(n^(1/3)) eq 0 ]; // G. C. Greubel, Jul 20 2023
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Maple
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
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Mathematica
Select[Range[1000], Mod[#, Floor[Power[#, 1/3]]] == 0 &] Select[Range[1000],Divisible[#,Floor[CubeRoot[#]]]&] (* Harvey P. Dale, Jun 19 2023 *)
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SageMath
[n for n in (1..250) if n%(floor(n^(1/3)))==0 ] # G. C. Greubel, Jul 20 2023
Formula
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
Comments