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 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. 4
 -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 (list; graph; refs; listen; history; text; internal format)
 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: A178225 A264739 A257170 * A117569 A135528 A163805 Adjacent sequences:  A073094 A073095 A073096 * A073098 A073099 A073100 KEYWORD sign AUTHOR Benoit Cloitre, Aug 18 2002 STATUS approved

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Last modified June 5 12:47 EDT 2020. Contains 334840 sequences. (Running on oeis4.)