A086694 A run of 2^n 1's followed by a run of 2^n 0's, for n=0, 1, 2, ...
1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0
Offset: 1
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
- Ralf Stephan, Some divide-and-conquer sequences ...
- Ralf Stephan, Table of generating functions
Programs
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Maple
seq(op([1$(2^n),0$(2^n)]),n=0..6); # Robert Israel, Jul 27 2017
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Mathematica
Table[{PadRight[{},2^n,1],PadRight[{},2^n,0]},{n,0,5}]//Flatten (* Harvey P. Dale, May 29 2017 *) Table[{Array[1&,2^n],Array[0&,2^n]},{n,0,5}]//Flatten (* Wolfgang Hintze, Jul 27 2017 *)
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PARI
a(n)=if(n<3,if(n<2,1,0),if(n%2==0,a(n/2-1),a((n-1)/2)))
Formula
a(n) = 1 - floor(log_2(4*(n+1)/3)) + floor(log_2(n+1)).
a(1) = 1, a(2) = 0, a(2n+1) = a(n), a(2n) = a(n-1).
G.f.: Sum_{k>=1} (x^(2^k)-x^(3*2^(k-1)))/(x-x^2). - Robert Israel, Jul 27 2017
G.f.: g(x) = (1/(1 - x))*( Sum_{n >= 1} x^(2^n-1)*(1 - x^2^(n-1)) ). Functional equation: g(x) = x + x*(1+x)*g(x^2). - Wolfgang Hintze, Aug 05 2017
Comments