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Find numbers with cubes?

Alexey Podchufarov, own
Posted 7 years ago

Hi, Community! I'm trying to find numbers (calculted with some formula) only with cubes (54=27 * 2, 8 * 7=56 etc). For squares is a function SquareFreeQ in Mathematica, but i found nothing for cubes. I write smthng, but it doesn't works:

For[t = 1, t <= 5, t++,
If[StringFreeQ[ToString[FactorInteger[6 t^2 + 4719 t]], ",3}"],
Print["{", t, ",",FactorInteger[6 t^2 + 4719 t], "},"]]]

Real answer (if construction "StringFreeQ[ToString[FactorInteger..." works OK - answer'll be without first string) : {1,{{3,3},{5,2},{7,1}}}, {2,{{2,1},{3,1},{19,1},{83,1}}}, {3,{{3,2},{1579,1}}}, {4,{{2,2},{3,2},{17,1},{31,1}}}, {5,{{3,1},{5,1},{1583,1}}},

So, what's wrong?
POSTED BY: Alexey Podchufarov
3 Replies
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Thanks a lot - use your idea, it works!!!
POSTED BY: Alexey Podchufarov

Note: It is a mistake to go to strings for this type of coding. Things are simpler and faster as numbers. You probably already know this but for others reading the thread I suggest learning the pattern matching capabilities of Mathematica. They are very powerful. see the documentation pattern matching, also see Select and DeleteCases and other related list handling functions here
POSTED BY: Neil Singer

Alexey,

cubeFind[x_Integer] := Select[FactorInteger[x], #[[2]] >= 3 &]

This will find your cubes. You can return True or False instead by checking for Empty Set:

cubeQ[x_Integer] := 
 Not[Select[FactorInteger[x], #[[2]] >= 3 &] === {}]

You can obviously just return the first part (the factor) or a list of factors by editing this.

Regards
POSTED BY: Neil Singer
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