A307417 Numbers that can be expressed in a base in such a way that the sum of cubes of their digits in this base equals the original number.

8, 9, 16, 17, 27, 28, 29, 35, 43, 54, 55, 62, 64, 65, 72, 91, 92, 99, 118, 125, 126, 127, 128, 133, 134, 152, 153, 189, 190, 216, 217, 224, 243, 244, 250, 251, 280, 307, 341, 342, 343, 344, 351, 370, 371, 407, 432, 433, 468, 469, 512, 513, 514, 520, 539, 559

Offset: 1

Author

César Eliud Lozada, Apr 07 2019

Keywords

• nonn
• base

Comments

There are infinitely many such numbers (proof in the second Johnson link).

Examples

			a(1) = 8 = [2, 0] (base 4) =  2^3 + 0^3
a(2) = 9 = [2, 1] (base 4) =  2^3 + 1^3
a(3) = 16 = [2, 2] (base 7) =  2^3 + 2^3
a(4) = 17 = [1, 2, 2] (base 3) =  1^3 + 2^3 + 2^3
		

Links

• César Eliud Lozada, Table of n, a(n) for n = 1..114
• Allan Wm. Johnson Jr., Crux Mathematicorum, Vol. 5, No. 1, January 1979, problem 407, 16.
• Allan Wm. Johnson Jr., Crux Mathematicorum, Vol. 5, No. 9, November 1979, solution to problem 407, 273-277.

Crossrefs

Cf. A005188, A023052, A046197.

Programs

Maple

sqn:= []; lis:=[];
for n to 1000 do
  b := 2;
  while b < n do #needs to be adjusted
    q := convert(n, base, b);
    s := convert(map(proc (X) options operator, arrow; X^3 end proc, q), `+`);
    if evalb(s = n) then
      sqn := [op(sqn), n];
      lis := [op(lis), [n, b, ListTools[Reverse](q)]];
      break
    end if;
    b := b+1
  end do
end do;
lis := lis; #list of decompositions [number, base, conversion]
sqn := sqn; #sequence