A060464 Numbers that are not congruent to 4 or 5 mod 9.
0, 1, 2, 3, 6, 7, 8, 9, 10, 11, 12, 15, 16, 17, 18, 19, 20, 21, 24, 25, 26, 27, 28, 29, 30, 33, 34, 35, 36, 37, 38, 39, 42, 43, 44, 45, 46, 47, 48, 51, 52, 53, 54, 55, 56, 57, 60, 61, 62, 63, 64, 65, 66, 69, 70, 71, 72, 73, 74, 75, 78, 79, 80, 81, 82, 83, 84, 87, 88, 89, 90, 91
Offset: 1
Examples
30 belongs to this sequence because it has the partition as sum of 3 cubes 30 = (-283059965)^3 + (-2218888517)^3 + (2220422932)^3. - _Artur Jasinski_, Apr 30 2010, edited by _M. F. Hasler_, Nov 10 2015
References
- R. K. Guy, Unsolved Problems in Number Theory, Section D5.
- Cohen H. 2007. Number Theory Volume I: Tools and Diophantine Equations. Springer Verlag p. 380. - Artur Jasinski, Apr 30 2010
Links
- Harry J. Smith, Table of n, a(n) for n = 1..2000
- Nikos Bagis, On the numbers that are sums of three cubes, arXiv:2009.11972 [math.GM], 2020.
- Andrew R. Booker, Cracking the problem with 33, March 2019
- Andrew R. Booker and Brady Haran, 42 is the new 33, Numberphile video (2019)
- Andrew R. Booker and Brady Haran, NEWS: The Mystery of 42 is Solved, Numberphile video (2019)
- Tim Browning and Brady Haran, The Uncracked Problem with 33, Numberphile video (2015)
- Tim Browning and Brady Haran, 74 is cracked, Numberphile video (2016)
- Jean-Louis Colliot-Thélène and Olivier Wittenberg, Groupe de Brauer et points entiers de deux familles de surfaces cubiques affines, Amer. J. Math. 134:5 (2012), pp. 1303-1327.
- Andreas-Stephan Elsenhans and Jörg Jahnel, List of solutions of x^3 + y^3 + z^3 = n for n < 1000 neither a cube nor twice a cube
- A.-S. Elsenhans, J. Jahnel, New sums of three cubes, Math. Comp. 78 (2009) 1227-1230.
- Brady Haran, 569936821221962380720, Numberphile video (2020)
- D. R. Heath-Brown, The density of zeros of forms for which weak approximation fails, Mathematics of Computation 59 (1992), pp. 613-623.
- Sander G. Huisman, Newer sums of three cubes, arXiv:1604.07746 [math.NT], 2016.
- H. Mishima, About n=x^3+y^3+z^3
- Andrew Sutherland, Sums of three cubes, Slides of a talk given May 07 2020 on the Number Theory Web.
- University of Bristol, Sum of three cubes for 42 finally solved - using real life planetary computer
- Wikipedia, Manin obstruction
- Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,1,-1).
Crossrefs
Programs
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GAP
A060464:=Filtered([0..100],n->n mod 9 <>4 and n mod 9 <>5); # Muniru A Asiru, Feb 17 2018
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Magma
[n : n in [0..150] | n mod 9 in [0, 1, 2, 3, 6, 7, 8]]; // Wesley Ivan Hurt, Jul 21 2016
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Maple
for n from 0 to 100 do if n mod 9 <> 4 and n mod 9 <> 5 then printf(`%d,`, n) fi:od:
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Mathematica
a = {}; Do[If[(Mod[n, 9] == 4) || (Mod[n, 9] == 5), , AppendTo[a, n]], {n, 1, 300}]; a (* Artur Jasinski, Apr 30 2010 *) Which[Mod[#,9]==4,Nothing,Mod[#,9]==5,Nothing,True,#]&/@Range[0,100] (* Harvey P. Dale, Jul 31 2023 *) -
PARI
n=-1; for (m=0, 4000, if (m%9!=4 && m%9!=5, write("b060464.txt", n++, " ", m)); if (n==2000, break)) \\ Harry J. Smith, Jul 05 2009 -
PARI
concat(0, Vec(x^2*(x^3+x^2+1)*(x^3+x+1)/((1+x+x^2+x^3+x^4+x^5+x^6)*(x-1)^2) + O(x^100))) \\ Altug Alkan, Nov 06 2015
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PARI
a(n)=n\7*9+[0, 1, 2, 3, 6, 7, 8][n%7+1] \\ Charles R Greathouse IV, Nov 06 2015
Formula
G.f.: x^2*(x^3+x^2+1)*(x^3+x+1) / ( (1+x+x^2+x^3+x^4+x^5+x^6)*(x-1)^2 ). - R. J. Mathar, Oct 08 2011
From Wesley Ivan Hurt, Jul 21 2016: (Start)
a(n) = a(n-1) + a(n-7) - a(n-8) for n>8; a(n) = a(n-7) + 9 for n>7.
a(n) = (63*n - 63 + 2*(n mod 7) + 2*((n+1) mod 7) - 12*((n+2) mod 7) + 2*((n+3) mod 7) + 2*((n+4) mod 7) + 2*((n+5) mod 7) + 2*((n+6) mod 7))/49.
a(7k) = 9k-1, a(7k-1) = 9k-2, a(7k-2) = 9k-3, a(7k-3) = 9k-6, a(7k-4) = 9k-7, a(7k-5) = 9k-8, a(7k-6) = 9k-9. (End)
Extensions
More terms from James Sellers, Apr 11 2001
Comments