OFFSET
1,3
COMMENTS
Conjecture: n is a sum of three cubes iff n is in this sequence.
As of their 2009 paper, Elsenhans and Jahnel did not know of a sum of three cubes that gives 33 or 42.
The problem with 33 is cracked, see links below: 8866128975287528^3 + (-8778405442862239)^3 + (-2736111468807040)^3 = 33. - Alois P. Heinz, Mar 11 2019
Numbers that are congruent to {0, 1, 2, 3, 6, 7, 8} mod 9. - Wesley Ivan Hurt, Jul 21 2016
Heath-Brown conjectures that n is a sum of three cubes in infinitely many ways iff n is in this sequence (and not at all otherwise). See his paper for a conjectural asymptotic. - Charles R Greathouse IV, Mar 12 2019
The problem with 42 is cracked by Andrew Booker from University of Bristol and Andrew Sutherland from Massachusetts Institute of Technology, see the link below: 42 = (-80538738812075974)^3 + 80435758145817515^3 + 12602123297335631^3. - Jianing Song, Sep 07 2019
A third solution to writing 3 as a sum of three third powers was found by the same team using 4 million computer-hours. 3 = 569936821221962380720^3 + (-569936821113563493509)^3 + (-472715493453327032)^3. - Peter Luschny, Sep 20 2019
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).
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)
EXAMPLE
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
MAPLE
for n from 0 to 100 do if n mod 9 <> 4 and n mod 9 <> 5 then printf(`%d, `, n) fi:od:
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 *)
PROG
(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
(PARI) a(n)=n\7*9+[0, 1, 2, 3, 6, 7, 8][n%7+1] \\ Charles R Greathouse IV, Nov 06 2015
(Magma) [n : n in [0..150] | n mod 9 in [0, 1, 2, 3, 6, 7, 8]]; // Wesley Ivan Hurt, Jul 21 2016
(GAP) A060464:=Filtered([0..100], n->n mod 9 <>4 and n mod 9 <>5); # Muniru A Asiru, Feb 17 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, Apr 10 2001
EXTENSIONS
More terms from James A. Sellers, Apr 11 2001
STATUS
approved