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%I #13 Jun 15 2017 01:55:12
%S 5,8,35,38,50,80,101,102,110,178,196,201,221,227,257,258,279,289,319,
%T 329,331,341,345,350,355,356,371,379,380,384,405,406,408,417,427,455,
%U 457,473,497,500,514,553,564,576,635,638,639,641,644,656,665,668,674,689,709,711,722,725,773,781,792,800,804,836,858,862,894,923,933,943,961,973,978,981,983,992,996
%N Numbers n such that n^3 contains the same digits as some other cube.
%H Robert Israel, <a href="/A188065/b188065.txt">Table of n, a(n) for n = 1..10000</a>
%e 5 is a member since 5^3 = 125, and 512 = 8^3.
%e 178^3 = 5639752 has same digits as 196^3 = 7529536, so 178 and 196 are in the sequence
%p dmax:= 10: # to get all terms of at most dmax digits.
%p N:= 'N':
%p S:= {}:
%p for n from 1 while n^3 < 10^dmax do
%p w:= sort(convert(n^3,base,10));
%p if assigned(N[w]) then
%p if N[w] = 0 then S:= S union {n}
%p else
%p S:= S union {n,N[w]};
%p N[w] = 0
%p fi
%p else N[w]:= n
%p fi
%p od:
%p sort(convert(S,list)); # _Robert Israel_, Jun 15 2017
%t id3Q[n_]:=Module[{idn3=Sort[IntegerDigits[n^3]],perms},perms= FromDigits/@ Permutations[idn3];Length[Select[perms,Sort[IntegerDigits[#]]==idn3 && IntegerQ[#^(1/3)]&]]>1]; Select[Range[1000],id3Q] (* _Harvey P. Dale_, Apr 27 2012 *)
%o (PARI) do(n)=my(v=List()); for(D=1,n, my(m=Map(),low=sqrtnint(10^(D-1)-1,3)+1,high=sqrtnint(10^D-1,3),n3,d); for(n=low,high, d=vecsort(digits(n3=n^3)); if(mapisdefined(m,d), mapput(m,d,1), mapput(m,d,0))); for(n=low,high, if(mapget(m,vecsort(digits(n^3))), listput(v,n)))); Set(v) \\ _Charles R Greathouse IV_, Jun 15 2017
%K nonn,base
%O 1,1
%A _Claudio Meller_, Mar 20 2011
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