A047234 Numbers that are congruent to {0, 1, 4} mod 6.
0, 1, 4, 6, 7, 10, 12, 13, 16, 18, 19, 22, 24, 25, 28, 30, 31, 34, 36, 37, 40, 42, 43, 46, 48, 49, 52, 54, 55, 58, 60, 61, 64, 66, 67, 70, 72, 73, 76, 78, 79, 82, 84, 85, 88, 90, 91, 94, 96, 97, 100, 102, 103, 106, 108, 109, 112, 114, 115, 118, 120, 121, 124
Offset: 1
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).
Programs
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Magma
[n : n in [0..150] | n mod 6 in [0, 1, 4]]; // Wesley Ivan Hurt, Jun 14 2016
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Maple
A047234:=n->(6*n-7+cos(2*n*Pi/3)+sqrt(3)*sin(2*n*Pi/3))/3: seq(A047234(n), n=1..100); # Wesley Ivan Hurt, Jun 14 2016
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Mathematica
Select[Range[0, 200], Mod[#, 6] == 0 || Mod[#, 6] == 1 || Mod[#, 6] == 4 &] (* Vladimir Joseph Stephan Orlovsky, Jul 07 2011 *) LinearRecurrence[{1, 0, 1, -1}, {0, 1, 4, 6}, 100] (* Vincenzo Librandi, Jun 15 2016 *) #+{0,1,4}&/@(6*Range[0,20])//Flatten (* Harvey P. Dale, Jul 25 2019 *)
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PARI
a(n)=(n-1)\3*6+[4,0,1][n%3+1] \\ Charles R Greathouse IV, Jun 11 2015
Formula
Equals partial sums of (0, 1, 3, 2, 1, 3, 2, 1, 3, 2, ...). - Gary W. Adamson, Jun 19 2008
G.f.: x^2*(1+x)*(2*x+1)/((1+x+x^2)*(x-1)^2). - R. J. Mathar, Oct 08 2011
From Wesley Ivan Hurt, Jun 14 2016: (Start)
a(n) = a(n-1) + a(n-3) - a(n-4) for n>4.
a(n) = (6*n-7+cos(2*n*Pi/3)+sqrt(3)*sin(2*n*Pi/3))/3.
a(3k) = 6k-2, a(3k-1) = 6k-5, a(3k-2) = 6k-6. (End)
Sum_{n>=2} (-1)^n/a(n) = (3-sqrt(3))*Pi/18 + log(2)/3 + log(2+sqrt(3))/(2*sqrt(3)). - Amiram Eldar, Dec 14 2021
E.g.f.: (6 + exp(x)*(6*x - 7) + exp(-x/2)*(cos(sqrt(3)*x/2) + sqrt(3)*sin(sqrt(3)*x/2)))/3. - Stefano Spezia, Jul 26 2024