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First differences of A049472, floor(n/sqrt(2)).
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%I #60 Jul 05 2023 04:00:02

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

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

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

%N First differences of A049472, floor(n/sqrt(2)).

%C Fixed point of the morphism 0->1, 1->110. - _Benoit Cloitre_, May 31 2004

%C As binary constant 0.1101101110110... = 0.85826765646... (A119812), see Fxtbook link. - _Joerg Arndt_, May 15 2011

%C Characteristic word with slope 1/sqrt(2) [see J. L. Ramirez et al.]. - _R. J. Mathar_, Jul 09 2013

%C From _Peter Bala_, Nov 22 2013: (Start)

%C Sturmian word: equals the limit word S(infinity) where S(0) = 0, S(1) = 1 and for n >= 1, S(n+1) = S(n)S(n)S(n-1).

%C More generally, for k = 0,1,2,..., we can define a sequence of words S_k(n) by S_k(0) = 0, S_k(1) = 0...01 (k 0's) and for n >= 1, S_k(n+1) = S_k(n)S_k(n)S_k(n-1). Then the limit word S_k(infinity) is a Sturmian word whose terms are given by a(n) = floor((n + 2)/(k + sqrt(2))) - floor((n + 1)/(k + sqrt(2))).

%C This sequence corresponds to the case k = 0. See A159684 (case k = 1) and A171588 (case k = 2). Compare with the Fibonacci words A005614, A221150, A221151 and A221152. See also A230901. (End)

%C For n > 0: a(A001951(n)) = 1, a(A001952(n)) = 0. - _Reinhard Zumkeller_, Jul 03 2015

%C Binary complement of the Pell word A171588. - _Michel Dekking_, Feb 22 2018

%H Reinhard Zumkeller, <a href="/A080764/b080764.txt">Table of n, a(n) for n = 0..10000</a>

%H Joerg Arndt, <a href="http://www.jjj.de/fxt/#fxtbook">Matters Computational (The Fxtbook)</a>, section 38.12, pp. 757-758.

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Sturmian_word">Sturmian word</a>

%H <a href="/index/Fi#FIXEDPOINTS">Index entries for sequences that are fixed points of mappings</a>

%F a(n) = floor((n+2)*sqrt(2)/2) - floor((n+1)*sqrt(2)/2).

%F a(n) = A188295(n+2) for all n in Z. - _Michael Somos_, Aug 19 2018

%e From _Peter Bala_, Nov 22 2013: (Start)

%e The first few Sturmian words S(n) are

%e S(0) = 0

%e S(1) = 1

%e S(2) = 110

%e S(3) = 110 110 1

%e S(4) = 1101101 1101101 110

%e S(5) = 11011011101101110 11011011101101110 1101101

%e The lengths of the words are [1, 1, 3, 7, 17, 41, ...] = A001333. (End)

%p A080764 := proc(n)

%p alpha := 1/sqrt(2) ;

%p floor((n+2)*alpha)-floor((n+1)*alpha) ;

%p end proc: # _R. J. Mathar_, Jul 09 2013

%t Nest[ Flatten[ # /. {0 -> 1, 1 -> {1, 1, 0}}] &, {1}, 7] (* _Robert G. Wilson v_, Apr 16 2005 *)

%t NestList[ Flatten[ # /. {0 -> {1}, 1 -> {1, 0, 1}}] &, {1}, 5] // Flatten (* or *)

%t t = Table[Floor[n/Sqrt[2]], {n, 111}]; Drop[t, 1] - Drop[t, -1] (* _Robert G. Wilson v_, Nov 03 2005 *)

%t a[ n_] := With[{m = n + 1}, Floor[(m + 1) / Sqrt[2]] - Floor[m / Sqrt[2]]]; (* _Michael Somos_, Aug 19 2018 *)

%o (Haskell)

%o a080764 n = a080764_list !! n

%o a080764_list = tail $ zipWith (-) (tail a049472_list) a049472_list

%o -- _Reinhard Zumkeller_, Jul 03 2015

%o (PARI) {a(n) = n++; my(k = sqrtint(n*n\2)); n*(n+2) > 2*k*(k+2)}; /* _Michael Somos_, Aug 19 2018 */

%Y Cf. A005614, A159684, A171588, A221150, A221151, A221152, A230901.

%Y Cf. A049472, A001951, A001952, A188295.

%K nonn,easy

%O 0,1

%A _Matthew Vandermast_, Mar 25 2003