|
|
A070471
|
|
a(n) = n^3 mod 5.
|
|
3
|
|
|
0, 1, 3, 2, 4, 0, 1, 3, 2, 4, 0, 1, 3, 2, 4, 0, 1, 3, 2, 4, 0, 1, 3, 2, 4, 0, 1, 3, 2, 4, 0, 1, 3, 2, 4, 0, 1, 3, 2, 4, 0, 1, 3, 2, 4, 0, 1, 3, 2, 4, 0, 1, 3, 2, 4, 0, 1, 3, 2, 4, 0, 1, 3, 2, 4, 0, 1, 3, 2, 4, 0, 1, 3, 2, 4, 0, 1, 3, 2, 4, 0, 1, 3, 2, 4, 0, 1, 3, 2, 4, 0, 1, 3, 2, 4, 0, 1, 3, 2, 4, 0
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,3
|
|
COMMENTS
|
Also, a(n) = n^7 mod 5 since 7 == 3 (mod 5-1).
|
|
LINKS
|
|
|
FORMULA
|
G.f.: -x*(1+3*x+2*x^2+4*x^3) / ( (x-1)*(1+x+x^2+x^3+x^4) ). - R. J. Mathar, Dec 10 2010
|
|
MATHEMATICA
|
CoefficientList[Series[-x (1 + 3 x + 2 x^2 + 4 x^3)/((x - 1) (1 + x + x^2 + x^3 + x^4)), {x, 0, 100}], x] (* Vincenzo Librandi, May 21 2014 *)
|
|
PROG
|
(Sage) [power_mod(n, 3, 5) for n in (0..101)] # Zerinvary Lajos, Oct 29 2009
(PARI) x='x+O('x^99); concat(0, Vec(-x*(1+3*x+2*x^2+4*x^3)/((x-1)*(1+x+x^2+x^3+x^4)))) \\ Altug Alkan, Mar 27 2016
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|