

A060464


Numbers that are not congruent to 4 or 5 mod 9.


13



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
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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
HeathBrown 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 computerhours. 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
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)
JeanLouis ColliotThélène and Olivier Wittenberg, Groupe de Brauer et points entiers de deux familles de surfaces cubiques affines, Amer. J. Math. 134:5 (2012), pp. 13031327.
AndreasStephan 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) 12271230.
D. R. HeathBrown, The density of zeros of forms for which weak approximation fails, Mathematics of Computation 59 (1992), pp. 613623.
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)*(x1)^2 ).  R. J. Mathar, Oct 08 2011
From Wesley Ivan Hurt, Jul 21 2016: (Start)
a(n) = a(n1) + a(n7)  a(n8) for n>8; a(n) = a(n7) + 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) = 9k1, a(7k1) = 9k2, a(7k2) = 9k3, a(7k3) = 9k6, a(7k4) = 9k7, a(7k5) = 9k8, a(7k6) = 9k9. (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 *)


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)*(x1)^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

Cf. A060465, A060466, A060467, A334521, A334522.
A156638 is the complement of this sequence.
Sequence in context: A039189 A039141 A008541 * A039102 A287103 A050023
Adjacent sequences: A060461 A060462 A060463 * A060465 A060466 A060467


KEYWORD

nonn,easy


AUTHOR

N. J. A. Sloane, Apr 10 2001


EXTENSIONS

More terms from James A. Sellers, Apr 11 2001


STATUS

approved



