login
A079944
A run of 2^n 0's followed by a run of 2^n 1's, for n=0, 1, 2, ...
119
0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 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, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1
OFFSET
0,1
COMMENTS
With offset 2, this is the second bit in the binary expansion of n. - Franklin T. Adams-Watters, Feb 13 2009
a(n) = A173920(n+2,2); in the sequence of nonnegative integers (cf. A001477) substitute all n by 2^floor(n/2) occurrences of (n mod 2). - Reinhard Zumkeller, Mar 04 2010
REFERENCES
Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. See Example 1.34.
FORMULA
a(n) = floor(log[2](4*(n+2)/3)) - floor(log[2](n+2)). - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 22 2003
For n >= 2, a(n-2)=1+floor(log[2](n/3))-floor(log[2](n/2)) - Benoit Cloitre, Mar 03 2003
G.f.: 1/x^2/(1-x) * (1/x + sum(k>=0, x^(3*2^k)-x^2^(k+1))). - Ralf Stephan, Jun 04 2003
a(n) = A000035(A004526(A030101(n+2))). - Reinhard Zumkeller, Mar 04 2010
MATHEMATICA
Table[IntegerDigits[n + 2, 2][[2]], {n, 0, 100}] (* Jean-François Alcover, Jul 26 2019 *)
PROG
(Haskell)
a079944 n = a079944_list !! n
a079944_list = f [0, 1] where f (x:xs) = x : f (xs ++ [x, x])
-- Reinhard Zumkeller, Oct 14 2010, Mar 28 2011
(PARI) a(n)=binary(n+2)[2] \\ Charles R Greathouse IV, Nov 07 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, Feb 21 2003
STATUS
approved