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A060464
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Numbers that are not congruent to 4 or 5 mod 9.
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14
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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
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1,3
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COMMENTS
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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
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REFERENCES
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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
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LINKS
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Tim Browning and Brady Haran, 74 is cracked, Numberphile video (2016)
Andrew Sutherland, Sums of three cubes, Slides of a talk given May 07 2020 on the Number Theory Web.
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FORMULA
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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
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)
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EXAMPLE
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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
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MAPLE
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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
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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 *)
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PROG
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(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
(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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CROSSREFS
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A156638 is the complement of this sequence.
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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