A259942 Numbers that are larger than or equal to the sum of the cubes of their prime factors (with multiplicity).

1, 64, 96, 108, 128, 144, 162, 192, 216, 240, 243, 256, 270, 288, 300, 320, 324, 360, 384, 400, 405, 432, 448, 450, 480, 486, 500, 504, 512, 540, 560, 567, 576, 600, 625, 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

Offset

1,2

Comments

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

Links

Robert Israel, Table of n, a(n) for n = 1..10000

Eric Weisstein's World of Mathematics, Waring's_problem

Wikipedia, Smooth number

Example

64 = 2*2*2*2*2*2 >= 6 * 2^3, so 64 is in the sequence.
96 = 3*2*2*2*2*2 >= 3^3 + 5 * 2^3, so 96 is in the sequence.
256 = 4*4*4*4 >= 4*4^3, so 256 is in the sequence.

Maple

isA259942 := proc(n)
    local ifa;
    ifa := ifactors(n)[2] ;
    return (n >= add( op(2, p)*op(1, p)^3, p=ifa)) ;
end proc:
for n from 0 to 1000 do
    if isA259942(n) then
        printf("%d, ", n);
    end if;
end do: # R. J. Mathar, Nov 27 2015

Mathematica

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

Prog

(Python)
from sympy import factorint
for j in range(1, 1001):
    k =  factorint(j)
    it = k.keys()
    va = k.values()
    alfa = 0
    for l in range(0, len(k)):
        alfa = alfa + va[l]*(it[l]**3)
    if alfa <=j:
        print j
(PARI) isok(n) = {my(f = factor(n)); n >= sum(k=1, #f~, f[k, 2]*f[k, 1]^3); } \\ Michel Marcus, Nov 28 2015

Crossrefs

Cf. A002804, A166319.

Sequence in context: A262743 A176819 A114407 * A114417 A046305 A114828

Adjacent sequences: A259939 A259940 A259941 * A259943 A259944 A259945

Keyword

nonn

Author

Francesco Di Matteo, Nov 08 2015

Status

approved