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Numbers that are larger than or equal to the sum of the cubes of their prime factors (with multiplicity).
1

%I #61 Feb 18 2021 01:35:49

%S 1,64,96,108,128,144,162,192,216,240,243,256,270,288,300,320,324,360,

%T 384,400,405,432,448,450,480,486,500,504,512,540,560,567,576,600,625,

%U 630,640,648,672,675,700,720,729,750,756,768,784,800,810,840,864,875,882,896,900,945,960,972,980,1000

%N Numbers that are larger than or equal to the sum of the cubes of their prime factors (with multiplicity).

%C This sequence is analogous to A166319 but with cubes instead of squares.

%H Robert Israel, <a href="/A259942/b259942.txt">Table of n, a(n) for n = 1..10000</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/WaringsProblem.html">Waring's_problem</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Smooth_number">Smooth number</a>

%e 64 = 2*2*2*2*2*2 >= 6 * 2^3, so 64 is in the sequence.

%e 96 = 3*2*2*2*2*2 >= 3^3 + 5 * 2^3, so 96 is in the sequence.

%e 256 = 4*4*4*4 >= 4*4^3, so 256 is in the sequence.

%p isA259942 := proc(n)

%p local ifa;

%p ifa := ifactors(n)[2] ;

%p return (n >= add( op(2,p)*op(1,p)^3,p=ifa)) ;

%p end proc:

%p for n from 0 to 1000 do

%p if isA259942(n) then

%p printf("%d,",n);

%p end if;

%p end do: # _R. J. Mathar_, Nov 27 2015

%t scpfQ[n_]:=n>=Total[Flatten[Table[#[[1]],{#[[2]]}]&/@ FactorInteger[ n]]^3]; Select[Range[1000],scpfQ] (* _Harvey P. Dale_, Dec 25 2015 *)

%o (Python)

%o from sympy import factorint

%o for j in range(1, 1001):

%o k = factorint(j)

%o it = k.keys()

%o va = k.values()

%o alfa = 0

%o for l in range(0,len(k)):

%o alfa = alfa + va[l]*(it[l]**3)

%o if alfa <=j:

%o print j

%o (PARI) isok(n) = {my(f = factor(n)); n >= sum(k=1, #f~, f[k,2]*f[k,1]^3);} \\ _Michel Marcus_, Nov 28 2015

%Y Cf. A002804, A166319.

%K nonn

%O 1,2

%A _Francesco Di Matteo_, Nov 08 2015