|
Search: id:A084091
|
|
|
| A084091 |
|
Expansion of sum(k>=0, x^2^k/(1+x^2^k+x^2^(k+1))). |
|
+0 6
|
|
| 0, 1, 0, 0, 1, -1, 0, 1, 0, 0, 0, -1, 0, 1, 0, 0, 1, -1, 0, 1, -1, 0, 0, -1, 0, 1, 0, 0, 1, -1, 0, 1, 0, 0, 0, -1, 0, 1, 0, 0, 0, -1, 0, 1, -1, 0, 0, -1, 0, 1, 0, 0, 1, -1, 0, 1, 0, 0, 0, -1, 0, 1, 0, 0, 1, -1, 0, 1, -1, 0, 0, -1, 0, 1, 0, 0, 1, -1, 0, 1, -1, 0, 0, -1, 0, 1, 0, 0, 0, -1, 0, 1, -1, 0, 0, -1, 0, 1, 0, 0, 1, -1, 0, 1, 0
(list; graph; listen)
|
|
|
OFFSET
|
0,1
|
|
|
COMMENT
|
Chances of values -1/0/+1 are ~2:5:2.
G.f. A(x) satisfies 0=f(A(x),A(x^2),A(x^4)) where f(u,v,w)=u^2-v^2+2w(v-u)+w-v - Michael Somos, Jul 18 2004
Multiplicative with a(2^e)=(1+(-1)^e)/2, a(3^e)=0, a(p^e)=1 if p=1 mod 6, a(p^e)=(-1)^e if p=5 mod 6. - Michael Somos, Jul 18 2004
|
|
FORMULA
|
a(2n) = a(n) + 1 - (n+1 mod 3), a(2n+1) = 1 - (n mod 3). - Ralf Stephan, Sep 27 2003
G.f.: Sum_{k>=0} f(x^2^k) where f(x)=x(1-x)/(1-x^3). - Michael Somos, Jul 18 2004
max(sum(0<=k<=n,a(k))) = floor(log_4(n))+1. Proof by Nikolaus Meyberg.
|
|
PROGRAM
|
(PARI) a(n)=local(A, m); if(n<1, 0, A=O(x); m=1; while(m<=n, m*=2; A=x/(1+x+x^2) +subst(A, x, x^2)); polcoeff(A, n)) /* Michael Somos, Jul 18 2004 */
|
|
CROSSREFS
|
Cf. A002487, A084091.
Positions of 0 are in A084090, of 1 in A084089, of -1 in A084088, of a(n)!=0 in A084087.
Sequence in context: A079813 A078580 A059651 this_sequence A080846 A082401 A059448
Adjacent sequences: A084088 A084089 A084090 this_sequence A084092 A084093 A084094
|
|
KEYWORD
|
sign,mult
|
|
AUTHOR
|
Ralf Stephan (ralf(AT)ark.in-berlin.de), May 11 2003
|
|
|
Search completed in 0.002 seconds
|