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A047227 Numbers that are congruent to {1, 2, 3, 4} mod 6. 5
1, 2, 3, 4, 7, 8, 9, 10, 13, 14, 15, 16, 19, 20, 21, 22, 25, 26, 27, 28, 31, 32, 33, 34, 37, 38, 39, 40, 43, 44, 45, 46, 49, 50, 51, 52, 55, 56, 57, 58, 61, 62, 63, 64, 67, 68, 69, 70, 73, 74, 75, 76, 79, 80, 81, 82, 85, 86, 87, 88, 91, 92, 93, 94, 97, 98 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
a(k)^m is a term for k and m in N. - Jerzy R Borysowicz, Apr 18 2023
LINKS
FORMULA
From Johannes W. Meijer, Jul 07 2011: (Start)
a(n) = floor((n+2)/4) + floor((n+1)/4) + floor(n/4) + 2*floor((n-1)/4) + floor((n+3)/4).
G.f.: x*(1 + x + x^2 + x^3 + 2*x^4)/(x^5 - x^4 - x + 1). (End)
From Wesley Ivan Hurt, May 20 2016: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(n) = (6n - 5 - i^(2n) + (1+i)*i^(1-n) + (1-i)*i^(n-1))/4 where i=sqrt(-1).
a(2n) = A047235(n), a(2n-1) = A047241(n). (End)
E.g.f.: (4 + sin(x) - cos(x) + (3*x - 2)*sinh(x) + 3*(x - 1)*cosh(x))/2. - Ilya Gutkovskiy, May 21 2016
From Wesley Ivan Hurt, May 21 2016: (Start)
a(n) = A047246(n) + 1.
a(n+2) - a(n+1) = A093148(n) for n>0.
a(1-n) = - A047247(n). (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(3)*Pi/12 + 2*log(2)/3 - log(3)/4. - Amiram Eldar, Dec 17 2021
MAPLE
A047227:=n->(6*n-5-I^(2*n)+(1+I)*I^(1-n)+(1-I)*I^(n-1))/4: seq(A047227(n), n=1..100); # Wesley Ivan Hurt, May 20 2016
MATHEMATICA
Complement[Range[100], Flatten[Table[{6n - 1, 6n}, {n, 0, 15}]]] (* Alonso del Arte, Jul 07 2011 *)
Select[Range[100], MemberQ[{1, 2, 3, 4}, Mod[#, 6]]&] (* Vincenzo Librandi, Jan 06 2013 *)
PROG
(Magma) [n: n in [0..100] | n mod 6 in [1..4]]; // Vincenzo Librandi, Jan 06 2013
(PARI) a(n)=([0, 1, 0, 0, 0; 0, 0, 1, 0, 0; 0, 0, 0, 1, 0; 0, 0, 0, 0, 1; -1, 1, 0, 0, 1]^(n-1)*[1; 2; 3; 4; 7])[1, 1] \\ Charles R Greathouse IV, May 03 2023
CROSSREFS
Complement of A047264. Equals A203016 divided by 3.
Sequence in context: A104576 A332485 A031477 * A317047 A228102 A261412
KEYWORD
nonn,easy
AUTHOR
STATUS
approved

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Last modified April 19 19:02 EDT 2024. Contains 371798 sequences. (Running on oeis4.)