A073097 - OEIS

%I #14 Jan 12 2019 20:45:19

%S -1,-1,0,-1,0,1,0,1,0,1,0,-1,0,-1,0,1,0,1,0,-1,0,1,0,1,0,-1,0,-1,0,-1,

%T 0,1,0,1,0,-1,0,1,0,1,0,1,0,-1,0,-1,0,1,0,-1,0,-1,0,1,0,1,0,-1,0,-1,0,

%U -1,0,1,0,1,0,-1,0,1,0,1,0,1,0,-1,0,-1,0,1,0,1,0,-1,0,1,0,1,0,-1,0,-1,0,-1,0,1,0,-1,0,-1

%N Let x(n) denote the number of 4's among the n first elements of the continued fraction for sum k>=0 1/2^(2^k) (A007400), y(n) the number of 6's and z(n) the number of 2's. Then a(n)=x(n)-y(n)-z(n)-1.

%C The positive sequence has a(n) = mod(A000120(A047849(n)),2) = mod(A000120(A078008(2n)),2) - _Paul Barry_, Jan 13 2005

%C Cosh(1) in 'reflected factorial' base is 1.10101010101010101010101010101010101010101010... - see A091337 for Sinh(1) (from _Robert G. Wilson v_, May 04 2005)

%H Antti Karttunen, <a href="/A073097/b073097.txt">Table of n, a(n) for n = 0..65537</a>

%F It seems that a(2k+1) = 0 for k>=1.

%F The positive sequence (assuming the pattern continues) has g.f. (1+x-x^2)/((1-x)(1-x^2)), with a(n)=(1-(1)^n)/2+0^n = mod((1+A001045(n+1))/2, 2) = mod(A005578, 2). The partial sums are A008619(n+1). - _Paul Barry_, Apr 28 2004

%o (PARI)

%o up_to = 65537;

%o A007400(n) = if(n<3, [0, 1, 4][n+1], if(n%8==1, A007400((n+1)/2), if(n%8==2, A007400((n+2)/2), [2, 0, 0, 2, 4, 4, 6, 4, 2, 0, 0, 2, 4, 6, 4, 4][(n%16)+1]))); \\ From A007400

%o A073097list(up_to) = { my(v=vector(up_to), x4=0, y6=0, z2=0, k); for(n=1, up_to, k=A007400(n); if(2==k,z2++,if(4==k,x4++,if(6==k,y6++))); v[n] = (x4-y6-z2-1)); (v); };

%o v073097 = A073097list(up_to);

%o A073097(n) = if(!n,-1,v073097[n]); \\ _Antti Karttunen_, Jan 12 2019

%Y Cf. A000120, A005578, A007400, A008619, A047849, A078008, A091337.

%K sign

%O 0,1

%A _Benoit Cloitre_, Aug 18 2002