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A156638
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Numbers n such that n^2 + 2 == 0 (mod 9).
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8
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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
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OFFSET
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1,1
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COMMENTS
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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 n such that A010888(n^2) = 7. - V.J. Pohjola, Aug 18 2012.
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REFERENCES
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Henri Cohen, Number Theory Volume I: Tools and Diophantine Equations. Springer Verlag (2007) p. 380. - Artur Jasinski, Apr 30 2010
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LINKS
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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).
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FORMULA
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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
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MAPLE
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A156638:=n->9*n/2 - 9/4 - 7*(-1)^n/4: seq(A156638(n), n=1..80); # Wesley Ivan Hurt, Aug 16 2015
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MATHEMATICA
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LinearRecurrence[{1, 1, -1}, {4, 5, 13}, 50] (* Vincenzo Librandi, Mar 01 2012 *)
intDigSum[n_] := Total[IntegerDigits[n]]; A156638 = {}; Do[If[Nest[intDigSum, n^2, k] == 7, AppendTo[a, n]], {n, 10^5}]; A156638 (* V.J. Pohjola, Aug 18 2012 *)
Flatten[Table[9n - {5, 4}, {n, 30}]] (* Alonso del Arte, Aug 09 2015 *)
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PROG
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(MAGMA) [9*n/2 - 9/4 - 7*(-1)^n/4 : n in [1..80]]; // Wesley Ivan Hurt, Aug 16 2015
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CROSSREFS
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Cf. A060464, A010888, A334521, A334522.
Sequence in context: A283483 A288755 A227897 * A098868 A087882 A338544
Adjacent sequences: A156635 A156636 A156637 * A156639 A156640 A156641
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KEYWORD
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nonn,easy
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AUTHOR
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Vincenzo Librandi, Feb 12 2009
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STATUS
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approved
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