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A188376 Positions of 1 in A188374; complement of A188375.

%I A188376 #23 Jan 05 2025 19:51:39
%S A188376 1,4,7,8,11,14,15,18,21,24,25,28,31,32,35,38,41,42,45,48,49,52,55,56,
%T A188376 59,62,65,66,69,72,73,76,79,82,83,86,89,90,93,96,97,100,103,106,107,
%U A188376 110,113,114,117,120,123,124,127,130,131,134,137,140,141,144,147,148,151,154,155,158,161,164,165,168,171,172,175,178
%N A188376 Positions of 1 in A188374; complement of A188375.
%C A188376 See A187950.
%C A188376 From _Michel Dekking_, Feb 27 2018: (Start)
%C A188376 Let d = 3,3,1,3,3,1,3,3,3,1,3,3,1,3,3,3,1,3, ... be the sequence of first differences: d(n):=a(n+1)-a(n).
%C A188376 CLAIM: d equals the Pell word A171588 on the alphabet {3,1}, i.e., d is the unique fixed point of the morphism 3->331, 1->3.
%C A188376 Proof: recall that
%C A188376     A188374 = [nr+2r]-[nr]-[2r] = 1,0,0,1,0,0,1,1,0,0,1,0,0,1,1,0,... where r=1/sqrt(2).
%C A188376 It was shown in the comments of A294180 that (a(n)) gives the positions of 1 in the 3-symbol Pell word b = A294180 , which is the unique fixed point of the morphism
%C A188376      beta: 1->123, 2->123, 3->1.
%C A188376 The letter 1 occurs in b if and only if it appears as the first letter of a beta(1), beta(2) or beta(3). The differences between the occurrences of 1's are therefore equal to 3, 3, or 1, and moreover, these differences occur exactly as the sequence of beta(1)'s, beta(2)'s and beta(3)'s. After projecting 1->3, 2->3, 3->1 this yields the morphism 3->331, 1->3.
%C A188376 COROLLARY: a(n) = 2[nr] +n.
%C A188376 Proof:  We know that the Pell word A171588 = 0010010001001... has a Sturmian representation
%C A188376      A289001(n) = [(n+1)(1-r)]- [n(1-r)] = [nr]-[(n+1)r]-1.
%C A188376 Mapping 0 to 3, and 1 to 1, we find that d = 3313313331331 has a representation  d(n) = 2[(n+1)r]-2[nr] +1. This leads to
%C A188376      a(n+1) = 1+d(1)+...+d(n) = n+1+2[(n+1)r].
%C A188376 CLAIM: (a(n)) equals the sequence ad' in the paper "Pellian representations", defined by ad'(n) = [2r[n(1+r)]], for n=1,2,...
%C A188376 Proof: The double floor in the definition of ad' can be reduced to a single floor by Theorem 7.10 of "Pellian representations":
%C A188376     ad'(n) = 2d'(n)-n, for n=1,2,...
%C A188376 Here d' is defined as d'(n) = [n(1+r)]. It follows that
%C A188376       ad'(n) = [n(1+r)]+[nr] = 2[nr]+n = a(n).
%C A188376 (End)
%H A188376 Vincenzo Librandi, <a href="/A188376/b188376.txt">Table of n, a(n) for n = 1..2000</a>
%H A188376 L. Carlitz, R. Scoville and V. E. Hoggatt, Jr.,<a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/10-5/carlitz1.pdf"> Pellian Representations</a>, Fib. Quart. Vol. 10, No. 5, (1972), pp. 449-488.
%F A188376 a(n) = 2[nr]+n, where r = 1/sqrt(2). - _Michel Dekking_, Feb 27 2018
%t A188376 (See A188374.)
%t A188376 Table[(2 Floor[n (1/Sqrt[2])] + n), {n, 100}] (* _Vincenzo Librandi_, Mar 01 2018 *)
%o A188376 (Magma) [2*Floor(n*(1/Sqrt(2)))+n: n in [1..80]]; // _Vincenzo Librandi_, Mar 01 2018
%Y A188376 Cf. A187950, A188374, A188375.
%K A188376 nonn
%O A188376 1,2
%A A188376 _Clark Kimberling_, Mar 29 2011