A047399 Numbers that are congruent to {0, 3, 6} mod 8.
0, 3, 6, 8, 11, 14, 16, 19, 22, 24, 27, 30, 32, 35, 38, 40, 43, 46, 48, 51, 54, 56, 59, 62, 64, 67, 70, 72, 75, 78, 80, 83, 86, 88, 91, 94, 96, 99, 102, 104, 107, 110, 112, 115, 118, 120, 123, 126, 128, 131, 134, 136, 139, 142, 144, 147, 150, 152, 155, 158
Offset: 1
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..10000
- Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).
Programs
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Magma
[Floor((8*n-6)/3): n in [1..60]]; // Vincenzo Librandi, Jul 11 2011
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Maple
seq(floor((8*n-6)/3), n=1..52); # Gary Detlefs, Mar 07 2010
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Mathematica
f[n_] := 3 n - Floor[n/3]; Array[f, 52, 0] (* Or *) Cases[ Range[0, 136], n_ /; MatchQ[ Mod[n, 8], 0 | 3 | 6]] (* Robert G. Wilson v, Jul 10 2011 *)
Formula
a(n) = a(n-1) + a(n-3) - a(n-4) for n>4.
a(n) = floor((8*n-6)/3). [Gary Detlefs, Mar 07 2010]
a(n) = 3*n-floor(n/3). [Gary Detlefs, Jul 09 2011]
G.f. x^2*(3+3*x+2*x^2) / ((1+x+x^2)*(x-1)^2). - R. J. Mathar, Oct 08 2011
From Wesley Ivan Hurt, Jun 13 2016: (Start)
a(n) = (24*n-21+3*cos(2*n*Pi/3)-sqrt(3)*sin(2*n*Pi/3))/9.
a(3k) = 8k-2, a(3k-1) = 8k-5, a(3k-2) = 8k-8. (End)