|
|
A130782
|
|
Period 5, repeat [1, 1, 2, 1, 1].
|
|
6
|
|
|
1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,3
|
|
COMMENTS
|
Continued fraction expansion of (4 + sqrt(65))/7. - Klaus Brockhaus, Apr 30 2010
|
|
LINKS
|
|
|
FORMULA
|
G.f.: (x^4 + x^3 + 2*x^2 + x + 1)/(1-x^5). - Ralf Stephan, Mar 28 2014
Euler transform of length 5 sequence [1, 1, -1, -1, 1]. - Michael Somos, Jun 17 2015
Moebius transform is length 10 sequence [1, -1, 0, 0, 1, 0, 0, 0, 0, -1]. - Michael Somos, Jun 17 2015
G.f.: Sum_{k>0} a(2*k-1) * q^n = f(q) + f(q^5) where f(q) := q / (1 - q^2). - Michael Somos, Jun 17 2015
a(n) = b(2*n + 1) where b() is multiplicative with b(2^e) = 0^3, b(5^e) = 2 if e>0, b(p^e) = 1 otherwise. - Michael Somos, Jun 17 2015
a(n) = a(-n) = a(n+5) for all n in Z. - Michael Somos, Jun 17 2015
|
|
EXAMPLE
|
a(5) = floor((5 + 3)/5) - ceiling((5 - 7)/5) = floor(8/5) - ceil(-2/5) = floor(8/5) + floor(2/5) = 1 + 0 = 1. - Wesley Ivan Hurt, Mar 27 2014
G.f. = 1 + x + 2*x^2 + x^3 + x^4 + x^5 + x^6 + 2*x^7 + x^8 + x^9 + ...
G.f. = q + q^3 + 2*q^5 + q^7 + q^9 + q^11 + q^13 + 2*q^15 + q^17 + ...
|
|
MAPLE
|
|
|
MATHEMATICA
|
Table[Floor[(n + 3)/5] - Ceiling[(n - 7)/5], {n, 0, 100}] (* Wesley Ivan Hurt, Mar 27 2014 *)
|
|
PROG
|
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|