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A047420
Numbers that are congruent to {0, 1, 2, 3, 4, 6} mod 8.
5
0, 1, 2, 3, 4, 6, 8, 9, 10, 11, 12, 14, 16, 17, 18, 19, 20, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 40, 41, 42, 43, 44, 46, 48, 49, 50, 51, 52, 54, 56, 57, 58, 59, 60, 62, 64, 65, 66, 67, 68, 70, 72, 73, 74, 75, 76, 78, 80, 81, 82, 83, 84, 86, 88
OFFSET
1,3
COMMENTS
The product of any two terms belongs to the sequence and therefore also a(n)^2, a(n)^3, a(n)^4, etc. - Bruno Berselli, Nov 28 2012
FORMULA
a(n+1) = 6*floor(n/3)+(n mod 3). - Gary Detlefs, Mar 09 2010
G.f.: x^2*(1+x^2+2*x^4) / ( (1+x+x^2)*(x^2-x+1)*(x-1)^2 ). - R. J. Mathar, Dec 05 2011
From Wesley Ivan Hurt, Jun 15 2016: (Start)
a(n) = 2*a(n-1)-2*a(n-2)+2*a(n-3)-2*a(n-4)+2*a(n-5)-a(n-6) for n>6.
a(n) = (12*n-18+sqrt(3)*(3*sin(n*Pi/3)+sin(2*n*Pi/3)))/9.
a(6k) = 8k-2, a(6k-1) = 8k-4, a(6k-2) = 8k-5, a(6k-3) = 8k-6, a(6k-4) = 8k-7, a(6k-5) = 8k-8. (End)
Sum_{n>=2} (-1)^n/a(n) = (sqrt(2)-1)*Pi/8 + 3*log(2)/4. - Amiram Eldar, Dec 26 2021
MAPLE
A047420:=n->(12*n-18+sqrt(3)*(3*sin(n*Pi/3)+sin(2*n*Pi/3)))/9: seq(A047420(n), n=1..100); # Wesley Ivan Hurt, Jun 15 2016
MATHEMATICA
Select[Range[0, 100], MemberQ[{0, 1, 2, 3, 4, 6}, Mod[#, 8]] &] (* Wesley Ivan Hurt, Jun 15 2016 *)
LinearRecurrence[{2, -2, 2, -2, 2, -1}, {0, 1, 2, 3, 4, 6}, 70] (* Harvey P. Dale, Aug 11 2021 *)
PROG
(Magma) [n : n in [0..100] | n mod 8 in [0..4] cat [6]]; // Wesley Ivan Hurt, Jun 15 2016
CROSSREFS
Sequence in context: A039147 A039107 A317544 * A039075 A050078 A075449
KEYWORD
nonn,easy
STATUS
approved