

A285683


Positions of 2 in A285680.


5



5, 10, 12, 17, 22, 24, 29, 31, 36, 41, 43, 48, 53, 55, 60, 62, 67, 72, 74, 79, 81, 86, 91, 93, 98, 103, 105, 110, 112, 117, 122, 124, 129, 134, 136, 141, 143, 148, 153, 155, 160, 162, 167, 172, 174, 179, 184, 186, 191, 193, 198, 203, 205, 210, 212, 217, 222
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OFFSET

1,1


COMMENTS

A 3way partition of the positive integers, by positions of 0, 1, 2 in A285680:A285681: positions of 0; slope t = (15+sqrt(5))/10;
A285682: positions of 1; slope u = (4+sqrt(5);
A285683: positions of 2; slope v = (1+3*sqrt(5))/2;
where 1/t + 1/u + 1/v = 1.
Conjecture: a(n)  a(n1) is in {5,7} for n >= 2.
Proof of the correct version of Kimberling's conjecture: This conjecture is false; it should be replaced by a(n)  a(n1) is in {2,5} for n >= 2. The truth of that conjecture follows from Lemma 8 in Allouche and Dekking, since (a(n)) is a generalized Beatty sequence (see Formula below).  Michel Dekking, Oct 09 2018
Apparently a(n) = A285679(n+1).  R. J. Mathar, May 14 2017
Proof of Mathar's conjecture: Let xF:=A003849 = 01001010010010100... be the Fibonacci sequence. The two return words of 1010 are r1=10100100, and r2=10100. [See Justin & Vuillon (2000) for definition of return word.  N. J. A. Sloane, Sep 23 2019] If these occur at index i, then in BOTH cases there is an occurrence of 0010 in xF at index i+3 in xF. These will be called typeA occurrences. Since r1 has suffix 00, there will also be occurrences of 0010 at index i+6, if r1 occurs at index i. These will be called typeB occurrences. It is clear that there are no other occurrences of 0010 in xF. Now if we perform the [0010>2] transform T on xF, then the typeB occurrences of 0010 will not yield the occurrence of a 2 in T(xF), since the StringReplace procedure acts from left to right, destroying the 0010 occurrences of typeB. The conclusion is that the positions of 2 in the [1010>2] transform and the [0010>2] transform are coupled, occurring at a fixed shift of each other, except that there is an extra first occurrence of 0010 at index 3. By coincidence, the first occurrence of 1010 in xF is at index 5, which equals the index of the second occurrence of 0010, with 3 subtracted (that has to be done because of the shrinking of a block of length 4 to the letter 2). It follows that the two occurrence sequences are equal, except for the extra 3.  Michel Dekking, Oct 09 2018


LINKS

Clark Kimberling, Table of n, a(n) for n = 1..10000
J.P. Allouche, F. M. Dekking, Generalized Beatty sequences and complementary triples, arXiv:1809.03424v3 [math.NT], 20182019.
Jacques Justin and Laurent Vuillon, Return words in Sturmian and episturmian words, RAIROTheoretical Informatics and Applications 34.5 (2000): 343356.


FORMULA

a(n) = 3*floor(n*phi)  n + 3 (this follows from Theorem 29 in Allouche and Dekking, since the overlap word 101010 that contains 1010 does not occur in the Fibonacci word).  Michel Dekking, Oct 09 2018


MATHEMATICA

s = Nest[Flatten[# /. {0 > {0, 1}, 1 > {0}}] &, {0}, 13] ; (* A003849 *)
w = StringJoin[Map[ToString, s]]
w1 = StringReplace[w, {"1010" > "2"}]
st = ToCharacterCode[w1]  48; (* A285680 *)
Flatten[Position[st, 0]]; (* A285681 *)
Flatten[Position[st, 1]]; (* A285682 *)
Flatten[Position[st, 2]]; (* A285683 *)


CROSSREFS

Cf. A003849, A285680, A258681, A285682.
Sequence in context: A313363 A313364 A313365 * A313366 A313367 A313368
Adjacent sequences: A285680 A285681 A285682 * A285684 A285685 A285686


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, May 11 2017


STATUS

approved



