A261205
Numbers k such that floor(k^(1/m)) divides k for all integers m >= 1.
Original entry on oeis.org
1, 2, 3, 4, 6, 8, 12, 16, 20, 24, 30, 36, 42, 48, 64, 72, 80, 120, 210, 240, 288, 324, 420, 528, 552, 576, 600, 624, 900, 1260, 1764, 1848, 1980, 3024, 6480, 8100, 8280, 11880, 14160, 14280, 14400, 14520, 14640, 28560, 43680, 44520, 46872, 50400, 175560, 331200, 346920, 491400, 809100, 3418800, 4772040, 38937600, 203918400, 2000862360
Offset: 1
From _Michel Marcus_, Aug 13 2015: (Start)
For k=1 to 9, we have the following floored roots:
k=1: 1, 1, ...
k=2: 2, 1, 1, ...
k=3: 3, 1, 1, ...
k=4: 4, 2, 1, 1, ...
k=5: 5, 2, 1, 1, ...
k=6: 6, 2, 1, 1, ...
k=7: 7, 2, 1, 1, ...
k=8: 8, 2, 2, 1, 1, ...
k=9: 9, 3, 2, 1, 1, ...
where one can see that 5, 7 and 9 are not terms. (End)
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fQ[n_] := Block[{d, k = 2, lst = {}}, While[d = Floor[n^(1/k)]; d > 1, AppendTo[lst, d]; k++]; Union[ IntegerQ@# & /@ (n/Union[lst])] == {True}]; k = 4; lst = {1, 2, 3}; While[k < 10^6, If[fQ@ k, AppendTo[lst, k]; Print@ k]; k++]; lst (* Robert G. Wilson v, Aug 15 2015 *)
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is(n) = my(k,t); k=2; while( (t=sqrtnint(n, k)) > 1, if(n%t, return(0)); k++); 1
n=1; while(n<10^5,if(is(n),print1(n,", "));n++) /* Able to generate terms < 10^5 */ \\ Derek Orr, Aug 12 2015
A261342
Numbers n such that either floor(n^(1/k)) or ceiling(n^(1/k)) divides n for all integers k >= 1.
Original entry on oeis.org
1, 2, 3, 4, 6, 8, 9, 12, 15, 16, 20, 24, 30, 36, 42, 48, 56, 63, 64, 72, 80, 90, 100, 120, 132, 144, 156, 168, 195, 210, 224, 240, 288, 324, 360, 400, 420, 440, 528, 552, 576, 600, 624, 675, 702, 756, 840, 870, 900, 930, 960, 1056, 1155, 1260, 1332, 1368, 1560, 1680, 1764, 1848, 1980, 2352, 2600, 2704
Offset: 1
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{ isA261342(n) = my(k,t1,t2); k=2; until(t2<=2, t1=floor(sqrtn(n+.5,k)); t2=ceil(sqrtn(n-.5,k)); if(n%t1 && n%t2, return(0)); k++); 1; }
A261341
Numbers n such that round(n^(1/k)) divides n for all integers k>=1.
Original entry on oeis.org
1, 2, 4, 6, 12, 30, 36, 42, 72, 240, 420, 600, 900, 1560, 1764, 3600, 6084, 8100, 46440, 1742400, 4062240, 35814240
Offset: 1
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isA[n_] :=
Block[{t},
For[k = 2, (t = Floor[1/2 + n^(1/k)]) >= 2, k++,
If[Mod[n, t] != 0, Return[False]]]; Return[True]]
Select[Range[1, 100000], isA[#] &] (* Julien Kluge, Apr 04 2016 *)
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{ isA261341(n) = my(k,t); k=2; until(t<=2, t=round(sqrtn(n,k)); if(n%t,return(0)); k++); 1; }
A261417
Numbers n such that both ceiling(sqrt(n)) and ceiling(n^(1/3)) divide n.
Original entry on oeis.org
1, 2, 4, 6, 9, 12, 36, 56, 64, 90, 100, 110, 132, 144, 156, 210, 400, 576, 702, 729, 870, 900, 930, 1056, 1089, 1122, 1332, 1560, 2352, 2450, 2970, 3600, 4032, 4096, 4556, 4624, 4692, 5112, 5184, 5256, 5852, 7140, 8190, 9702, 9900, 12432, 14400, 15500, 15625, 16770, 16900, 17030, 18090, 18225, 18360, 19740
Offset: 1
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[n: n in [1..2000] | n mod Ceiling((n^(1/2))) eq 0 and n mod Ceiling((n^(1/3))) eq 0 ];
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Select[Range[200000], Mod[#, Ceiling[#^(1/2)]] == Mod[#, Ceiling[#^(1/3)]] == 0 &] (* Vincenzo Librandi, Aug 21 2016 *)
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is(n) = Mod(n, ceil(sqrt(n)))==0 && Mod(n, ceil(n^(1/3)))==0 \\ Felix Fröhlich, Aug 21 2016
Showing 1-4 of 4 results.
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