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A047240
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Numbers that are congruent to {0, 1, 2} mod 6.
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13
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0, 1, 2, 6, 7, 8, 12, 13, 14, 18, 19, 20, 24, 25, 26, 30, 31, 32, 36, 37, 38, 42, 43, 44, 48, 49, 50, 54, 55, 56, 60, 61, 62, 66, 67, 68, 72, 73, 74, 78, 79, 80, 84, 85, 86, 90, 91, 92, 96, 97, 98, 102, 103, 104, 108, 109, 110, 114, 115, 116, 120, 121, 122
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OFFSET
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1,3
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COMMENTS
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Partial sums of 0,1,1,4,1,1,4,... - Paul Barry, Feb 19 2007
Numbers k such that floor(k/3) = 2*floor(k/6). - Bruno Berselli, Oct 05 2017
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LINKS
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FORMULA
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G.f.: x*(1 + x + 4*x^2)/((1 - x)*(1 - x^3)).
a(n) = 2*n - 3 - cos(2*n*Pi/3) + sin(2*n*Pi/3)/sqrt(3). (End)
a(n) = 6*floor(n/3) + (n mod 3). - Gary Detlefs, Mar 09 2010
a(n) = a(n-1) + a(n-3) - a(n-4) for n>4.
a(3*k) = 6*k-4, a(3*k-1) = 6*k-5, a(3*k-2) = 6*k-6. (End)
Sum_{n>=2} (-1)^n/a(n) = (3-sqrt(3))*Pi/18 + log(2+sqrt(3))/(2*sqrt(3)). - Amiram Eldar, Dec 14 2021
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MAPLE
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select(k -> modp(iquo(k, 3), 2) = 0, [$0..122]); # Peter Luschny, Oct 05 2017
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MATHEMATICA
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a047240[n_] := Flatten[Map[6 # + {0, 1, 2} &, Range[0, n]]]; a047240[20] (* data *) (* Hartmut F. W. Hoft, Mar 06 2017 *)
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PROG
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(Magma) [0], [6*Floor(n/3) + (n mod 3): n in [1..65]]; // Vincenzo Librandi, Oct 23 2011
(Python 3)
[k for k in range(123) if (k//3) % 2 == 0] # Peter Luschny, Oct 05 2017
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CROSSREFS
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Cf. similar sequences with formula n+i*floor(n/3) listed in A281899.
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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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