A073097 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.

-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, 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, 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

Offset

0,1

Comments

The positive sequence has a(n) = mod(A000120(A047849(n)),2) = mod(A000120(A078008(2n)),2) - Paul Barry, Jan 13 2005

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

Links

Antti Karttunen, Table of n, a(n) for n = 0..65537

Formula

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

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

Prog

(PARI)
up_to = 65537;
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
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); };
v073097 = A073097list(up_to);
A073097(n) = if(!n, -1, v073097[n]); \\ Antti Karttunen, Jan 12 2019

Crossrefs

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

Sequence in context: A264739 A373252 A257170 * A117569 A135528 A163805

Adjacent sequences: A073094 A073095 A073096 * A073098 A073099 A073100

Keyword

sign

Author

Benoit Cloitre, Aug 18 2002

Status

approved