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%I #76 Mar 29 2024 02:13:20
%S 153,370,371,407,165033,221859,336700,336701,340067,341067,407000,
%T 407001,444664,487215,982827,983221,166500333,296584415,333667000,
%U 333667001,334000667,710656413,828538472,142051701000,166650003333,262662141664,333366670000
%N Each number is the sum of the cubes of its 3 sections.
%C 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 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 From _Daniel Forgues_, Jan 30 2015: (Start)
%C The subsequence {153, 165033, 166500333, ...} consists of numbers of the form
%C [(10^n - 4) / 6] * (10^n)^2 + [(10^n) / 2] * (10^n)^1 +
%C [(10^n - 1) / 3] * (10^n)^0 =
%C [(10^n - 4) / 6]^3 + [(10^n) / 2]^3 + [(10^n - 1) / 3]^3, n >= 1,
%C thus equal to the sum of the cube of their "digits" in base 10^n.
%C The subsequence {370, 336700, 333667000, ...} consists of numbers of the form
%C [(10^n - 1) / 3] * (10^n)^2 + {10^n - [(10^n - 1) / 3]} * (10^n)^1 =
%C [(10^n - 1) / 3]^3 + {10^n - [(10^n - 1) / 3]}^3, n >= 1,
%C thus equal to the sum of their "digits" in base 10^n.
%C The subsequence {371, 336701, 333667001, ...} is trivially derived from the subsequence {370, 336700, 333667000, ...}, since 1^3 = 1.
%C The subsequence {407, 340067, 334000667, ...} consists of numbers of the form
%C {10^n - 2 * [(10^n - 1) / 3]} * (10^n)^2 +
%C {10^n - [(10^n - 1) / 3]} * (10^n)^0 =
%C {10^n - 2 * [(10^n - 1) / 3]}^3 + {10^n - [(10^n - 1) / 3]}^3, n >= 1,
%C thus equal to the sum of their "digits" in base 10^n.
%C "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 From _Daniel Forgues_, Feb 04 2015: (Start)
%C 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 The number 407000 is trivially derived from 407, since 40^3 + 70^3 =
%C (4 * 10)^3 + (7 * 10)^3 = (4^3 + 7^3) * 10^3 = 407 * 1000 = 407000.
%C The number 407001 is trivially derived from 407000, since 1^3 = 1. (End)
%C From _Jose M. Arenas_, Mar 08 2017: (Start)
%C The subsequence {340067000000, 334000667000000000, 333400006667000000000000, ...} consists of numbers of the form
%C (4 * 10^(n + 2) + ((10^(n + 1) - 1) / 3) * 10^(n + 3)) * 10^(4 * n + 8) +
%C (7 * 10^(n + 2) + (2 * (10^(n + 1) - 1) / 3) * 10^(n + 3)) * 10^(2 * n + 4) =
%C (4 * 10^(n + 2) + ((10^(n + 1) - 1) / 3) * 10^(n + 3))^3 +
%C (7 * 10^(n + 2) + (2 * (10^(n + 1) - 1) / 3) * 10^(n + 3))^3, n >= 0,
%C thus equal to the sum of their 3 sections, each section of (2 * n + 4) digits.
%C The subsequence {340067000001, 334000667000000001, 333400006667000000000001, ...} is trivially derived from the subsequence {340067000000, 334000667000000000, 333400006667000000000000, ...}, since 1^3 = 1.
%C (End)
%D J. S. Madachy, Madachy's Mathematical Recreations, pp. 166, Dover, NY, 1979.
%D Donald D. Spencer, "Exploring number theory with microcomputers", pp. 65 and 101, Camelot Publishing Co.
%H 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 Jose M. Arenas, <a href="/A056733/a056733.py.txt">Python code</a>.
%e 333667001 = 333^3 + 667^3 + 001^3, so 333667001 is a term.
%t 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 (Python)
%o def a():
%o ..n = 1
%o ..while n < 10**9:
%o ....st = str(n)
%o ....if len(st) % 3 == 0:
%o ......s1 = st[:int(len(st)/3)]
%o ......s2 = st[int(len(st)/3):int(2*len(st)/3)]
%o ......s3 = st[int(2*len(st)/3):int(len(st))]
%o ......if int(s1)**3+int(s2)**3+int(s3)**3 == int(st):
%o ........print(n, end=', ')
%o ........n += 1
%o ......else:
%o ........n += 1
%o ....else:
%o ......n = 10*n
%o a()
%o # Derek Orr, Jul 03 2014
%o (Python)
%o def a():
%o ..for i in range(1,10):
%o ....for j in range(10):
%o ......for k in range(10):
%o ........if i**3 + j**3 + k**3 == i*100 + j*10 + k:
%o ..........print(i*100 + j*10 + k)
%o ..for i in range(10,100):
%o ....for j in range(100):
%o ......for k in range(100):
%o ........if i**3 + j**3 + k**3 == i*10000 + j*100 + k:
%o ..........print(i*10000 + j*100 + k)
%o ..for i in range(100,1000):
%o ....for j in range(1000):
%o ......for k in range(1000):
%o ........if i**3 + j**3 + k**3 == i*1000000 + j*1000 + k:
%o ..........print(i*1000000 + j*1000 + k)
%o a()
%o # _Denys Contant_, Feb 23 2017
%Y Cf. A005188.
%Y See A271730 for a related sequence.
%K nonn,base
%O 1,1
%A _Carlos Rivera_, Aug 13 2000
%E Offset changed to 1 by _N. J. A. Sloane_, Jul 07 2014
%E a(24)-a(27) from _Jose M. Arenas_, Mar 08 2017
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