|
|
A082507
|
|
Generated by a 3rd-order formal recursion with suitable initial values as follows: a(n) = n - a(n-1) - a(n-2) - a(n-3); a(0)=a(1)=a(2)=0.
|
|
0
|
|
|
2, 0, 0, 0, 3, 1, 1, 1, 4, 2, 2, 2, 5, 3, 3, 3, 6, 4, 4, 4, 7, 5, 5, 5, 8, 6, 6, 6, 9, 7, 7, 7, 10, 8, 8, 8, 11, 9, 9, 9, 12, 10, 10, 10
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
-1,1
|
|
LINKS
|
|
|
FORMULA
|
O.g.f.: 2/x+x^3(3-2x)/((1-x)^2*(1+x)(1+x^2)).
a(n) = 3/8 -5*(-1)^n/8 + n/4 + (1/4)*cos(Pi*n/2) - (5/4)*sin(Pi*n/2), n > -1. (End)
|
|
EXAMPLE
|
Sum of 4 successive terms gives n for n > 2:
n = 2 = a(-1) + a(0) + a(1) + a(2) = 2 + 0 + 0 + 0;
n = 3 = a(3) = a(0) + a(1) + a(2) + a(3) = 0 + 0 + 0 + 3;
n = 4 = a(1) + a(2) + a(3) + a(4) = 0 + 0 + 3 + 1;
Value of a(-1)=2 is arbitrary but provides a suitable extension.
|
|
MATHEMATICA
|
f[x_] := x-f[x-1]-f[x-2]-f[x-3]; {f[0]=0, f[1]=0, f[2]=0}; Table[f[w], {w, 1, 25}]
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|