This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A056733 #81 Sep 04 2025 21:25:43
%S A056733 153,370,371,407,165033,221859,336700,336701,340067,341067,407000,
%T A056733 407001,444664,487215,982827,983221,166500333,296584415,333667000,
%U A056733 333667001,334000667,710656413,828538472,142051701000,166650003333,262662141664,333366670000
%N A056733 Each number is the sum of the cubes of its 3 sections.
%C A056733 The first four terms are also called Narcissistic or Armstrong numbers. The first 16 terms are found in Spencer's book, pages 65 and 101.
%C A056733 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
%C A056733 From _Daniel Forgues_, Jan 30 2015: (Start)
%C A056733 The subsequence {153, 165033, 166500333, ...} consists of numbers of the form
%C A056733 [(10^n - 4) / 6] * (10^n)^2 + [(10^n) / 2] * (10^n)^1 +
%C A056733 [(10^n - 1) / 3] * (10^n)^0 =
%C A056733 [(10^n - 4) / 6]^3 + [(10^n) / 2]^3 + [(10^n - 1) / 3]^3, n >= 1,
%C A056733 thus equal to the sum of the cube of their "digits" in base 10^n.
%C A056733 The subsequence {370, 336700, 333667000, ...} consists of numbers of the form
%C A056733 [(10^n - 1) / 3] * (10^n)^2 + {10^n - [(10^n - 1) / 3]} * (10^n)^1 =
%C A056733 [(10^n - 1) / 3]^3 + {10^n - [(10^n - 1) / 3]}^3, n >= 1,
%C A056733 thus equal to the sum of their "digits" in base 10^n.
%C A056733 The subsequence {371, 336701, 333667001, ...} is trivially derived from the subsequence {370, 336700, 333667000, ...}, since 1^3 = 1.
%C A056733 The subsequence {407, 340067, 334000667, ...} consists of numbers of the form
%C A056733 {10^n - 2 * [(10^n - 1) / 3]} * (10^n)^2 +
%C A056733 {10^n - [(10^n - 1) / 3]} * (10^n)^0 =
%C A056733 {10^n - 2 * [(10^n - 1) / 3]}^3 + {10^n - [(10^n - 1) / 3]}^3, n >= 1,
%C A056733 thus equal to the sum of their "digits" in base 10^n.
%C A056733 "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 interesting--merely 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 Mathematician’s Apology" (End)
%C A056733 From _Daniel Forgues_, Feb 04 2015: (Start)
%C A056733 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.
%C A056733 The number 407000 is trivially derived from 407, since 40^3 + 70^3 =
%C A056733 (4 * 10)^3 + (7 * 10)^3 = (4^3 + 7^3) * 10^3 = 407 * 1000 = 407000.
%C A056733 The number 407001 is trivially derived from 407000, since 1^3 = 1. (End)
%C A056733 From _Jose M. Arenas_, Mar 08 2017: (Start)
%C A056733 The subsequence {340067000000, 334000667000000000, 333400006667000000000000, ...} consists of numbers of the form
%C A056733 (4 * 10^(n + 2) + ((10^(n + 1) - 1) / 3) * 10^(n + 3)) * 10^(4 * n + 8) +
%C A056733 (7 * 10^(n + 2) + (2 * (10^(n + 1) - 1) / 3) * 10^(n + 3)) * 10^(2 * n + 4) =
%C A056733 (4 * 10^(n + 2) + ((10^(n + 1) - 1) / 3) * 10^(n + 3))^3 +
%C A056733 (7 * 10^(n + 2) + (2 * (10^(n + 1) - 1) / 3) * 10^(n + 3))^3, n >= 0,
%C A056733 thus equal to the sum of their 3 sections, each section of (2 * n + 4) digits.
%C A056733 The subsequence {340067000001, 334000667000000001, 333400006667000000000001, ...} is trivially derived from the subsequence {340067000000, 334000667000000000, 333400006667000000000000, ...}, since 1^3 = 1. (End)
%D A056733 J. S. Madachy, Madachy's Mathematical Recreations, pp. 166, Dover, NY, 1979.
%D A056733 Donald D. Spencer, "Exploring number theory with microcomputers", pp. 65 and 101, Camelot Publishing Co.
%H A056733 Jose M. Arenas and Giovanni Resta, <a href="/A056733/b056733.txt">Table of n, a(n) for n = 1..69</a> (terms < 10^18, first 49 terms from Jose M. Arenas)
%H A056733 Jose M. Arenas, <a href="/A056733/a056733.py.txt">Python code</a>.
%e A056733 333667001 = 333^3 + 667^3 + 001^3, so 333667001 is a term.
%t A056733 f[n_] := Block[{len = IntegerLength@ n}, If[IntegerQ[len/3], n == Plus @@ Flatten[(FromDigits /@ Partition[IntegerDigits@ n, len/3])^3], False]]; Select[Range[10^6], f] (* _Michael De Vlieger_, Jan 31 2015 *)
%o A056733 (Python)
%o A056733 def a():
%o A056733 n = 1
%o A056733 while n < 10**9:
%o A056733 st = str(n)
%o A056733 if len(st) % 3 == 0:
%o A056733 s1 = st[:int(len(st)/3)]
%o A056733 s2 = st[int(len(st)/3):int(2*len(st)/3)]
%o A056733 s3 = st[int(2*len(st)/3):int(len(st))]
%o A056733 if int(s1)**3+int(s2)**3+int(s3)**3 == int(st):
%o A056733 print(n, end=', ')
%o A056733 n += 1
%o A056733 else:
%o A056733 n += 1
%o A056733 else:
%o A056733 n = 10*n
%o A056733 a()
%o A056733 # _Derek Orr_, Jul 03 2014
%o A056733 (Python)
%o A056733 def a():
%o A056733 for i in range(1,10):
%o A056733 for j in range(10):
%o A056733 for k in range(10):
%o A056733 if i**3 + j**3 + k**3 == i*100 + j*10 + k:
%o A056733 print(i*100 + j*10 + k)
%o A056733 for i in range(10,100):
%o A056733 for j in range(100):
%o A056733 for k in range(100):
%o A056733 if i**3 + j**3 + k**3 == i*10000 + j*100 + k:
%o A056733 print(i*10000 + j*100 + k)
%o A056733 for i in range(100,1000):
%o A056733 for j in range(1000):
%o A056733 for k in range(1000):
%o A056733 if i**3 + j**3 + k**3 == i*1000000 + j*1000 + k:
%o A056733 print(i*1000000 + j*1000 + k)
%o A056733 a()
%o A056733 # _Denys Contant_, Feb 23 2017
%Y A056733 Cf. A005188.
%Y A056733 See A271730 for a related sequence.
%K A056733 nonn,base,changed
%O A056733 1,1
%A A056733 _Carlos Rivera_, Aug 13 2000
%E A056733 Offset changed to 1 by _N. J. A. Sloane_, Jul 07 2014
%E A056733 a(24)-a(27) from _Jose M. Arenas_, Mar 08 2017