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 A060464 Numbers that are not congruent to 4 or 5 mod 9. 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 (list; graph; refs; listen; history; text; internal format)
 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. As of 2015, the problem is still open, see the Numberphile video. - M. F. Hasler, Nov 10 2015 Numbers that are congruent to {0, 1, 2, 3, 6, 7, 8} mod 9. - Wesley Ivan Hurt, Jul 21 2016 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 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. Sander G. Huisman, Newer sums of three cubes, arXiv:1604.07746 [math.NT], 2016. H. Mishima, About n=x^3+y^3+z^3 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 *) 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 Cf. A060465, A060466, A060467. 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

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Last modified February 20 00:28 EST 2019. Contains 320329 sequences. (Running on oeis4.)