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0, 0, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 5, 5, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 10, 10, 10, 10, 11, 11, 12, 12, 12, 12, 13, 13, 13, 14, 14, 14, 14, 15, 15, 15, 16, 16, 16, 16, 17, 17, 18, 18, 18, 18, 19, 19, 19, 20, 20, 20, 20, 21, 21, 21, 21, 22, 22, 22, 23, 23, 23, 23, 24, 24, 25
COMMENTS
a(n+1) = z_A(n), the number of entries of A278040 (called A number in the W. Lang given there) not exceeding n, for n >= 1 and z_A(-1) := 0. - Wolfdieter Lang, Dec 06 2018
FORMULA
a(n) = 2*B(n) - A(n) + 1, for n >= 0, where A(n) = A278040(n) and B(n) = A278039(n). For a proof see the W. Lang link in A278040, Proposition 7, eq. (41). - Wolfdieter Lang, Dec 06 2018
MAPLE
M:=12;
S[1]:=`0`; S[2]:=`01`; S[3]:=`0102`;
for n from 4 to M do S[n]:=cat(S[n-1], S[n-2], S[n-3]); od:
t0:=S[M]: # has 927 terms of tribonacci ternary word A080843# get numbers of 0's, 1's, 2's
N0:=[]: N1:=[]: N2:=[]: c0:=0: c1:=0: c2:=0:
L:=length(t0);
for i from 1 to L do
js := substring(t0, i..i);
j:=convert(js, decimal, 10);
if j=0 then c0:=c0+1; elif j=1 then c1:=c1+1; else c2:=c2+1; fi;
N0:=[op(N0), c0]; N1:=[op(N1), c1]; N2:=[op(N2), c2];
od:
Positions of letter b in the tribonacci word abacabaabacababac... generated by a->ab, b->ac, c->a (cf. A092782).
(Formerly M1571)
+10
56
2, 6, 9, 13, 15, 19, 22, 26, 30, 33, 37, 39, 43, 46, 50, 53, 57, 59, 63, 66, 70, 74, 77, 81, 83, 87, 90, 94, 96, 100, 103, 107, 111, 114, 118, 120, 124, 127, 131, 134, 138, 140, 144, 147, 151, 155, 158, 162, 164, 168, 171, 175, 179, 182, 186, 188, 192, 195, 199, 202, 206, 208
COMMENTS
A003144, A003145, A003146 may be defined as follows. Consider the map psi: a -> ab, b -> ac, c -> a. The image (or trajectory) of a under repeated application of this map is the infinite word a, b, a, c, a, b, a, a, b, a, c, a, b, a, b, a, c, ... (setting a = 1, b = 2, c = 3 gives A092782). The indices of a, b, c give respectively A003144, A003145, A003146. - Philippe Deléham, Feb 27 2009
The infinite word may also be defined as the limit S_oo where S_1 = a, S_n = psi(S_{n-1}). Or, by S_1 = a, S_2 = ab, S_3 = abac, and thereafter S_n = S_{n-1} S_{n-2} S_{n-3}. It is the unique word such that S_oo = psi(S_oo).
Also indices of b in the sequence closed under a -> abac, b -> aba, c -> ab; starting with a(1) = a. - Philippe Deléham, Apr 16 2004 Theorem: A number m is in this sequence iff the tribonacci representation of m-1 ends with 01. [Duchene and Rigo, Remark 2.5] - N. J. A. Sloane, Mar 02 2019
REFERENCES
Eric Duchêne, Aviezri S. Fraenkel, Vladimir Gurvich, Nhan Bao Ho, Clark Kimberling, Urban Larsson, Wythoff Visions, Games of No Chance, Vol. 5; MSRI Publications, Vol. 70 (2017), pages 101-153.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
FORMULA
It appears that a(n) = floor(n*t^2) + eps for all n, where t is the tribonacci constant A058265 and eps is 0, 1, or 2. See A276799. - N. J. A. Sloane, Oct 28 2016. This is true - see the Dekking et al. paper. - N. J. A. Sloane, Jul 22 2019
MAPLE
M:=17; S[1]:=`a`; S[2]:=`ab`; S[3]:=`abac`;
for n from 4 to M do S[n]:=cat(S[n-1], S[n-2], S[n-3]); od:
t0:=S[M]: l:=length(t0); t1:=[];
for i from 1 to l do if substring(t0, i..i) = `b` then t1:=[op(t1), i]; fi; od: # N. J. A. Sloane
MATHEMATICA
StringPosition[SubstitutionSystem[{"a" -> "ab", "b" -> "ac", "c" -> "a"}, "b", {#}][[1]], "b"][[All, 1]] &@9 (* Michael De Vlieger, Mar 30 2017, Version 10.2, after JungHwan Min at A003144 *)
CROSSREFS
First differences give A276789. A278040 (subtract 1 from each term, and use offset 1). For tribonacci representations of numbers see A278038.
1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15
COMMENTS
a(n+1) - 1 = z_C(n), where z_C(n) is the number of C numbers A276798 not exceeding n, for n >= 0, and z_C(-1) = 0. - Wolfdieter Lang, Dec 05 2018
FORMULA
a(n) = Sum_{k=0..n} A276791(k), for n >= 0. a(n) = 2*n + 1 - B(n), where B(n) = A278039(n), n >= 0. For a proof see the comment on z_C and Proposition 7, eq. 43, of the W. Lang link given in A080843. - Wolfdieter Lang, Dec 05 2018
MAPLE
M:=12;
S[1]:=`0`; S[2]:=`01`; S[3]:=`0102`;
for n from 4 to M do S[n]:=cat(S[n-1], S[n-2], S[n-3]); od:
t0:=S[M]: # has 927 terms of tribonacci ternary word A080843# get numbers of 0's, 1's, 2's
N0:=[]: N1:=[]: N2:=[]: c0:=0: c1:=0: c2:=0:
L:=length(t0);
for i from 1 to L do
js := substring(t0, i..i);
j:=convert(js, decimal, 10);
if j=0 then c0:=c0+1; elif j=1 then c1:=c1+1; else c2:=c2+1; fi;
N0:=[op(N0), c0]; N1:=[op(N1), c1]; N2:=[op(N2), c2];
od:
0, 1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10, 11, 12, 12, 13, 13, 14, 14, 15, 15, 16, 16, 17, 18, 18, 19, 19, 20, 20, 21, 21, 22, 22, 23, 23, 24, 25, 25, 26, 26, 27, 27, 28, 29, 29, 30, 30, 31, 31, 32, 32, 33, 33, 34, 34, 35, 36, 36, 37, 37, 38, 38, 39, 39, 40, 40, 41, 42, 42, 43, 43, 44
FORMULA
a(n) = Sum_{k=0..n} A276793(k), n >= 0. a(n) = A(n) - B(n) - (n + 1), where A(n) = A278040(n) and B(n) = A278039(n), n >= 0. For a proof see the W. Lang link in A278040, Proposition 7, eq. (42). - Wolfdieter Lang, Dec 05 2018
MAPLE
M:=12;
S[1]:=`0`; S[2]:=`01`; S[3]:=`0102`;
for n from 4 to M do S[n]:=cat(S[n-1], S[n-2], S[n-3]); od:
t0:=S[M]: # has 927 terms of tribonacci ternary word A080843# get numbers of 0's, 1's, 2's
N0:=[]: N1:=[]: N2:=[]: c0:=0: c1:=0: c2:=0:
L:=length(t0);
for i from 1 to L do
js := substring(t0, i..i);
j:=convert(js, decimal, 10);
if j=0 then c0:=c0+1; elif j=1 then c1:=c1+1; else c2:=c2+1; fi;
N0:=[op(N0), c0]; N1:=[op(N1), c1]; N2:=[op(N2), c2];
od:
Indicator function of ( A003146 prefixed with 0).
+10
8
1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0
COMMENTS
Also, this has 1's where A275925 has 3's (which is why the offset is 0). Also, apart from the initial 1, this has 1's where the tribonacci word A080843 has 2's.
1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0
COMMENTS
a(n) = 1 iff n is a term of A003144. The binary complement of (a(n)) is called the "binary Tribonacci word" in Mousavi and Shallit (see Theorem 23). It is defined to be the change of alphabet {0,1,2} -> {0,1,1} of the tribonacci word 0102010010... - Michel Dekking, Oct 12 2019
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