The first four terms are also called Narcissistic or Armstrong numbers. The first 16 terms are found in the Spencer's book, pages 65 and 101.
The sequence contains several infinite subsequences such as 153, 165033, 166500333, 166650003333, ...; 370, 336700, 333667000, 333366670000, ... or 371, 336701, 333667001, 333366670001, ...  Ulrich Schimke (ulrschimke(AT)aol.com), Jun 08 2001
From Daniel Forgues, Jan 30 2015: (Start)
The subsequence {153, 165033, 166500333, ...} are numbers of the form
[(10^n  4) / 6] * (10^n)^2 + [(10^n) / 2] * (10^n)^1 +
[(10^n  1) / 3] * (10^n)^0 =
[(10^n  4) / 6]^3 + [(10^n) / 2]^3 + [(10^n  1) / 3]^3, n >= 1,
thus equal to the sum of the cube of their "digits" in base 10^n.
The subsequence {370, 336700, 333667000, ...} are numbers of the form
[(10^n  1) / 3] * (10^n)^2 + {10^n  [(10^n  1) / 3]} * (10^n)^1 =
[(10^n  1) / 3]^3 + {10^n  [(10^n  1) / 3]}^3, n >= 1,
thus equal to the sum of their "digits" in base 10^n.
The subsequence {371, 336701, 333667001, ...} is trivially derived from the subsequence {370, 336700, 333667000, ...}, since 1^3 = 1.
The subsequence {407, 340067, 334000667, ... } are numbers of the form
{10^n  2 * [(10^n  1) / 3]} * (10^n)^2 +
{10^n  [(10^n  1) / 3]} * (10^n)^0 =
{10^n  2 * [(10^n  1) / 3]}^3 + {10^n  [(10^n  1) / 3]}^3, n >= 1,
thus equal to the sum of their "digits" in base 10^n.
"There are just four numbers (after 1) which are the sums of the cubes of their digits, viz. 153 = 1^3 + 5^3 + 3^3, 370 = 3^3 + 7^3 + 0^3, 371 = 3^3 + 7^3 + 1^3, and 407 = 4^3 + 0^3 + 7^3. This is an odd fact, very suitable for puzzle columns and likely to amuse amateurs, but there is nothing in it which appeals much to a mathematician. The proof is neither difficult nor interestingmerely a little tiresome. The theorem is not serious; and it is plain that one reason (though perhaps not the most important) is the extreme speciality of both the enunciation and the proof, which is not capable of any significant generalization."  G. H. Hardy, "A Mathematicianâ€™s Apology" (End)
From Daniel Forgues, Feb 04 2015: (Start)
The subsequence {341067, 333401006667, 333334001000666667, ...} is trivially derived from the even indexed terms 2n, n >= 1, of the subsequence {407, 340067, 334000667, 333400006667, ...}, since (10^n)^3 = 10^n * 10^(2n). These numbers are equal to the sum of the cube of their "digits" in base 10^(2n), n >= 1.
The number 407000 is trivially derived from 407, since 40^3 + 70^3 =
(4 * 10)^3 + (7 * 10)^3 = (4^3 + 7^3) * 10^3 = 407 * 1000 = 407000.
The number 407001 is trivially derived from 407000, since 1^3 = 1. (End)
From Jose M. Arenas, Mar 08 2017: (Start)
The subsequence {340067000000, 334000667000000000, 333400006667000000000000, ...} are numbers of the form
(4 * 10^(n + 2) + ((10^(n + 1)  1) / 3) * 10^(n + 3)) * 10^(4 * n + 8) +
(7 * 10^(n + 2) + (2 * (10^(n + 1)  1) / 3) * 10^(n + 3)) * 10^(2 * n + 4) =
(4 * 10^(n + 2) + ((10^(n + 1)  1) / 3) * 10^(n + 3))^3 +
(7 * 10^(n + 2) + (2 * (10^(n + 1)  1) / 3) * 10^(n + 3))^3, n >= 0,
thus equal to the sum of their 3 sections, each section of (2 * n + 4) digits.
The subsequence {340067000001, 334000667000000001, 333400006667000000000001, ...} is trivially derived from the subsequence {340067000000, 334000667000000000, 333400006667000000000000, ...}, since 1^3 = 1.
(End)
