A334521 -id:A334521 - cp's OEIS frontend

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

N. J. A. Sloane, Apr 10 2001

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

			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

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

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).

Cf. A060465, A060466, A060467, A334521, A334522.

A156638 is the complement of this sequence.

A060464:=Filtered([0..100],n->n mod 9 <>4 and n mod 9 <>5); # Muniru A Asiru, Feb 17 2018

[n : n in [0..150] | n mod 9 in [0, 1, 2, 3, 6, 7, 8]]; // Wesley Ivan Hurt, Jul 21 2016

for n from 0 to 100 do if n mod 9 <> 4 and n mod 9 <> 5 then printf(`%d,`, n) fi:od:

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 *)

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

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

a(n)=n\7*9+[0, 1, 2, 3, 6, 7, 8][n%7+1] \\ Charles R Greathouse IV, Nov 06 2015

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)

More terms from James Sellers, Apr 11 2001

A156638 Numbers k such that k^2 + 2 == 0 (mod 9).

Original entry on oeis.org

4, 5, 13, 14, 22, 23, 31, 32, 40, 41, 49, 50, 58, 59, 67, 68, 76, 77, 85, 86, 94, 95, 103, 104, 112, 113, 121, 122, 130, 131, 139, 140, 148, 149, 157, 158, 166, 167, 175, 176, 184, 185, 193, 194, 202, 203, 211, 212, 220, 221, 229, 230, 238, 239, 247, 248, 256

Offset: 1

Vincenzo Librandi, Feb 12 2009

From Artur Jasinski, Apr 30 2010: (Start)
Numbers congruent to 4 or 5 mod 9.
Numbers which are not the sum of 3 cubes.
Complement to A060464. (End)
Numbers k such that A010888(k^2) = 7. - V.J. Pohjola, Aug 18 2012

Henri Cohen, Number Theory Volume I: Tools and Diophantine Equations. Springer Verlag (2007) p. 380. - Artur Jasinski, Apr 30 2010

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Andrew Sutherland, Sums of three cubes, Slides of a talk given May 07 2020 on the Number Theory Web.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A060464, A010888, A332437, A334521, A334522.

[9*n/2 - 9/4 - 7*(-1)^n/4 : n in [1..80]]; // Wesley Ivan Hurt, Aug 16 2015

A156638:=n->9*n/2 - 9/4 - 7*(-1)^n/4: seq(A156638(n), n=1..80); # Wesley Ivan Hurt, Aug 16 2015

LinearRecurrence[{1, 1, -1}, {4, 5, 13}, 50] (* Vincenzo Librandi, Mar 01 2012 *)
Flatten[Table[9n - {5, 4}, {n, 30}]] (* Alonso del Arte, Aug 09 2015 *)
Select[Range[300],PowerMod[#,2,9]==7&] (* Harvey P. Dale, Jan 31 2023 *)

a(n) = (18*n - 9 - 7*(-1)^n)/4 \\ David Lovler, Aug 21 2022

For n > 2, a(n) = a(n-2) + 9.
G.f.: x*(4*x^2 + x + 4)/(x^3 - x^2 - x + 1). - Alexander R. Povolotsky, Feb 15 2009
From R. J. Mathar, Feb 19 2009: (Start)
a(n) = a(n-1) + a(n-2) - a(n-3), n>3.
a(n) = 9*n/2 - 9/4 - 7*(-1)^n/4.
G.f.: x*(4 + x + 4*x^2)/((1 + x)*(1 - x)^2). (End)
a(n) = -a(-n+1). - Bruno Berselli, Jan 08 2012
E.g.f.: 4 + ((18*x - 9)*exp(x) - 7*exp(-x))/4. - David Lovler, Aug 21 2022
Sum_{n>=1} (-1)^(n+1)/a(n) = tan(Pi/18)*Pi/9. - Amiram Eldar, Sep 26 2022
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = 1.
Product_{n>=1} (1 + (-1)^n/a(n)) = 2*cos(Pi/9) - 1 (= A332437 - 1). (End)

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A060464 Numbers that are not congruent to 4 or 5 mod 9.

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A156638 Numbers k such that k^2 + 2 == 0 (mod 9).

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A334522 Numbers that are either congruent to +-4 mod 9, or else are cubes or twice cubes.

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