A307417 - OEIS

%I #18 Jul 09 2019 12:44:46

%S 8,9,16,17,27,28,29,35,43,54,55,62,64,65,72,91,92,99,118,125,126,127,

%T 128,133,134,152,153,189,190,216,217,224,243,244,250,251,280,307,341,

%U 342,343,344,351,370,371,407,432,433,468,469,512,513,514,520,539,559

%N 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.

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

%H César Eliud Lozada, <a href="/A307417/b307417.txt">Table of n, a(n) for n = 1..114</a>

%H Allan Wm. Johnson Jr., <a href="https://cms.math.ca/crux/backfile/Crux_v5n01_Jan.pdf">Crux Mathematicorum</a>, Vol. 5, No. 1, January 1979, problem 407, 16.

%H Allan Wm. Johnson Jr., <a href="https://cms.math.ca/crux/backfile/Crux_v5n09_Nov.pdf">Crux Mathematicorum</a>, Vol. 5, No. 9, November 1979, solution to problem 407, 273-277.

%e a(1) = 8 = [2, 0] (base 4) =  2^3 + 0^3

%e a(2) = 9 = [2, 1] (base 4) =  2^3 + 1^3

%e a(3) = 16 = [2, 2] (base 7) =  2^3 + 2^3

%e a(4) = 17 = [1, 2, 2] (base 3) =  1^3 + 2^3 + 2^3

%p sqn:= []; lis:=[];

%p for n to 1000 do

%p   b := 2;

%p   while b < n do #needs to be adjusted

%p     q := convert(n, base, b);

%p     s := convert(map(proc (X) options operator, arrow; X^3 end proc, q), `+`);

%p     if evalb(s = n) then

%p       sqn := [op(sqn), n];

%p       lis := [op(lis), [n, b, ListTools[Reverse](q)]];

%p       break

%p     end if;

%p     b := b+1

%p   end do

%p end do;

%p lis := lis; #list of decompositions [number, base, conversion]

%p sqn := sqn; #sequence

%Y Cf. A005188, A023052, A046197.

%K nonn,base

%O 1,1

%A _César Eliud Lozada_, Apr 07 2019