Search: id:A084091
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%I A084091
%S A084091 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,
%T A084091 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,
%U A084091 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
%V A084091 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,
%W A084091 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,
%X A084091 -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
%N A084091 Expansion of sum(k>=0, x^2^k/(1+x^2^k+x^2^(k+1))).
%C A084091 Chances of values -1/0/+1 are ~2:5:2.
%C A084091 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
%C A084091 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
%F A084091 a(2n) = a(n) + 1 - (n+1 mod 3), a(2n+1) = 1 - (n mod 3). - Ralf Stephan, 
               Sep 27 2003
%F A084091 G.f.: Sum_{k>=0} f(x^2^k) where f(x)=x(1-x)/(1-x^3). - Michael Somos, 
               Jul 18 2004
%F A084091 max(sum(0<=k<=n,a(k))) = floor(log_4(n))+1. Proof by Nikolaus Meyberg.
%o A084091 (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 */
%Y A084091 Cf. A002487, A084091.
%Y A084091 Positions of 0 are in A084090, of 1 in A084089, of -1 in A084088, of 
               a(n)!=0 in A084087.
%Y A084091 Sequence in context: A079813 A078580 A059651 this_sequence A080846 A082401 
               A157238
%Y A084091 Adjacent sequences: A084088 A084089 A084090 this_sequence A084092 A084093 
               A084094
%K A084091 sign,mult
%O A084091 0,1
%A A084091 Ralf Stephan (ralf(AT)ark.in-berlin.de), May 11 2003

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