|
| |
|
|
A165983
|
|
Period 16: repeat 1,1,1,2,1,1,1,2,1,1,1,4,1,1,1,4.
|
|
0
| |
|
|
1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 2, 1
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 0,4
|
|
|
COMMENTS
| The numerator of the reduced fraction A061037(n+3)/A061041(2n+6).
|
|
|
FORMULA
| a(n)= +a(n-4) -a(n-8) +a(n-12). [R. J. Mathar, Dec 17 2010]
G.f.: ( -1-x-x^2-2*x^3-x^8-x^9-x^10-4*x^11 ) / ( (x-1)*(1+x)*(1+x^2)*(x^8+1) ). [R. J. Mathar, Dec 17 2010]
a(n)=(1/80)*{16*(n mod 16)-14*[(n+1) mod 16]+[(n+2) mod 16]+[(n+3) mod 16]+16*[(n+4) mod 16]-14*[(n+5) mod 16]+[(n+6) mod 16]+[(n+7) mod 16]+6*[(n+8) mod 16]-4*[(n+9) mod 16]+[(n+10) mod 16]+[(n+11) mod 16]+6*[(n+12) mod 16]-4*[(n+13) mod 16]+[(n+14) mod 16]+[(n+15) mod 16]}, with n>=0 [From Paolo P. Lava (paoloplava(AT)gmail.com), Oct 19 2009]
a(4n) = a(4n+1) = a(4n+2) = 1. a(4n+3) = A165207(n).
|
|
|
CROSSREFS
| Cf. A064038.
Sequence in context: A143654 A170981 A161096 * A083894 A128257 A198254
Adjacent sequences: A165980 A165981 A165982 * A165984 A165985 A165986
|
|
|
KEYWORD
| nonn,easy,less
|
|
|
AUTHOR
| Paul Curtz (bpcrtz(AT)free.fr), Oct 03 2009
|
| |
|
|
