A202279 Numbers k such that the sum of digits^3 of k equals Sum_{d|k, 1

142, 160, 1375, 6127, 12643, 51703, 86833, 103039, 104647, 112093, 137317, 218269, 261883, 266923, 449881, 505891, 617569, 907873

Offset: 1

Author

Michel Lagneau, Dec 15 2011

Keywords

• nonn
• base
• fini
• full

Comments

The sequence is finite because the restricted sum of divisors of n, for n composite, is at least sqrt(n), while the sum of the cubes of the digits of n is at most 9^3*log_10(n+1). - Giovanni Resta, Oct 05 2018

Examples

			160 is in the sequence because 1^3 + 6^3 + 0^3 = 217, and the sum of the divisors 1< d<160 is 2 + 4 + 5 + 8 + 10 + 16 + 20 + 32 + 40 + 80 = 217.
		

Crossrefs

Cf. A070308, A202279, A202147, A202285, A202240.

Programs

Maple

A055012 := proc(n)
        add(d^3,d=convert(n,base,10)) ;
end proc:
A048050 := proc(n)
        if n > 1 then
        numtheory[sigma](n)-1-n ;
        else
                0;
        end if;
end proc:
isA202279 := proc(n)
        A055012(n) = A048050(n) ;
end proc:
for n from 1 do
        if isA202279(n) then
                printf("%d,\n",n);
        end if;
end do; # R. J. Mathar, Dec 15 2011

Mathematica

Q[n_]:=Module[{a=Total[Rest[Most[Divisors[n]]]]}, a == Total[IntegerDigits[n]^3]]; Select[Range[2, 5*10^7], Q]
Select[Range[1000000],DivisorSigma[1,#]-#-1==Total[IntegerDigits[#]^3]&] (* Harvey P. Dale, Jul 19 2014 *)

Formula

{n: A055012(n) = A048050(n)}. - R. J. Mathar, Dec 15 2011