|
%I #20 Feb 14 2021 15:25:22
%S 142,160,1375,6127,12643,51703,86833,103039,104647,112093,137317,
%T 218269,261883,266923,449881,505891,617569,907873
%N Numbers k such that the sum of digits^3 of k equals Sum_{d|k, 1<d<k} d.
%C 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
%F {n: A055012(n) = A048050(n)}. - _R. J. Mathar_, Dec 15 2011
%e 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.
%p A055012 := proc(n)
%p add(d^3,d=convert(n,base,10)) ;
%p end proc:
%p A048050 := proc(n)
%p if n > 1 then
%p numtheory[sigma](n)-1-n ;
%p else
%p 0;
%p end if;
%p end proc:
%p isA202279 := proc(n)
%p A055012(n) = A048050(n) ;
%p end proc:
%p for n from 1 do
%p if isA202279(n) then
%p printf("%d,\n",n);
%p end if;
%p end do; # _R. J. Mathar_, Dec 15 2011
%t Q[n_]:=Module[{a=Total[Rest[Most[Divisors[n]]]]}, a == Total[IntegerDigits[n]^3]]; Select[Range[2, 5*10^7], Q]
%t Select[Range[1000000],DivisorSigma[1,#]-#-1==Total[IntegerDigits[#]^3]&] (* _Harvey P. Dale_, Jul 19 2014 *)
%Y Cf. A070308, A202279, A202147, A202285, A202240.
%K nonn,base,fini,full
%O 1,1
%A _Michel Lagneau_, Dec 15 2011
|
