cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 29 results. Next

A322410 Compound tribonacci sequence with a(n) = A278040(A278039(n)), for n >= 0.

Original entry on oeis.org

1, 8, 14, 21, 25, 32, 38, 45, 52, 58, 65, 69, 76, 82, 89, 95, 102, 106, 113, 119, 126, 133, 139, 146, 150, 157, 163, 170, 174, 181, 187, 194, 201, 207, 214, 218, 225, 231, 238, 244, 251, 255, 262, 268, 275, 282, 288, 295, 299, 306, 312, 319, 326, 332, 339, 343, 350, 356, 363, 369, 376
Offset: 0

Views

Author

Wolfdieter Lang, Jan 02 2019

Keywords

Comments

The nine sequences A308199, A319967, A319968, A322410, A322409, A322411, A322413, A322412, A322414 are based on defining the tribonacci ternary word to start with index 0 (in contrast to the usual definition, in A080843 and A092782, which starts with index 1). As a result these nine sequences differ from the compound tribonacci sequences defined in A278040, A278041, and A319966-A319972. - N. J. A. Sloane, Apr 05 2019

Crossrefs

Formula

A(B(n)) = A(B(n) + 1) - 4 = A(n) + B(n) + n, for n >= 0, with A = A278040 and B = A278039. For a proof see the W. Lang link in A278040, Proposition 9, eq. (49).
a(n+1) = A319967(n)-1 = A003145(A003144(n))-1, the corresponding classical compound tribonacci sequence. - Michel Dekking, Apr 04 2019

A322413 Compound tribonacci sequence with a(n) = A278041(A278039(n)), for n >= 0.

Original entry on oeis.org

3, 16, 27, 40, 47, 60, 71, 84, 97, 108, 121, 128, 141, 152, 165, 176, 189, 196, 209, 220, 233, 246, 257, 270, 277, 290, 301, 314, 321, 334, 345, 358, 371, 382, 395, 402, 415, 426, 439, 450, 463, 470, 483, 494, 507, 520, 531, 544, 551, 564, 575, 588, 601, 612, 625, 632, 645, 656, 669, 680, 693
Offset: 0

Views

Author

Wolfdieter Lang, Jan 02 2019

Keywords

Comments

(a(n+1)) = A319970(n)-1 = A003146(A003144(n))-1, the corresponding classical compound tribonacci sequence. - Michel Dekking, Apr 03 2019
The nine sequences A308199, A319967, A319968, A322410, A322409, A322411, A322413, A322412, A322414 are based on defining the tribonacci ternary word to start with index 0 (in contrast to the usual definition, in A080843 and A092782, which starts with index 1). As a result these nine sequences differ from the compound tribonacci sequences defined in A278040, A278041, and A319966-A319972. - N. J. A. Sloane, Apr 05 2019

Crossrefs

Formula

a(n) = C(B(n)) = C(B(n) + 1) - 7 = 2*(A(n) + B(n)) + n + 1, for n >= 0, where A = A278040, B = A278039 and C = A278041. For a proof see the W. Lang link in A278040, Proposition 9, eq. (55).

A316715 a(n) is the number of 1s in A316713(n). That is, a(n) is the number of B-sequences (A278039) used in the tribonacci ABC-representation of n >= 0.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 2, 4, 2, 2, 1, 3, 1, 5, 3, 3, 2, 3, 1, 2, 4, 2, 2, 1, 6, 4, 4, 3, 4, 2, 3, 4, 2, 2, 1, 3, 1, 5, 3, 3, 2, 3, 1, 2, 7, 5, 5, 4, 5, 3, 4, 5, 3, 3, 2, 4, 2, 5, 3, 3, 2, 3, 1, 2, 4, 2, 2, 1, 6, 4, 4, 3, 4, 2, 3, 4, 2, 2, 1, 3, 1, 8, 6, 6, 5, 6, 4, 5, 6, 4, 4, 3, 5, 3, 6, 4, 4, 3, 4, 2, 3
Offset: 0

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Author

Wolfdieter Lang, Sep 11 2018

Keywords

Comments

The number of 2s and 3s in A316713(n) is given in A316716 and A316717, respectively.

Examples

			See column #(1) in A316713.
		

Crossrefs

Formula

a(n) = number of 1s in A316713(n), that is the number of Bs in the tribonacci ABC-representation of n >= 0.

A322407 Compound sequence a(n) = A319198(A278039(n)), for n >= 0.

Original entry on oeis.org

0, 1, 3, 4, 4, 5, 7, 8, 9, 11, 12, 12, 13, 15, 16, 18, 19, 19, 20, 22, 23, 24, 26, 27, 27, 28, 30, 31, 31, 32, 34, 35, 36, 38, 39, 39, 40, 42, 43, 45, 46, 46, 47, 49, 50, 51, 53, 54, 54, 55, 57, 58, 59, 61, 62, 62, 63, 65, 66, 68, 69
Offset: 0

Views

Author

Wolfdieter Lang, Jan 02 2019

Keywords

Comments

Old name was: Compound tribonacci sequence a(n) = A319198(A278039(n)), for n >= 0.
a(n) gives the sum of the entries of the tribonacci word sequence t = A080843 not exceeding t(B(n)), with B(n) = A278039(n).

Examples

			n = 3: B(3) = 6, t = {0, 1, 0, 2, 0, 1, 0, ...} which sums to 4 = a(3) = -12 + 3*6 - 2, because A(3) = 12.
		

Crossrefs

Formula

a(n) = z(B(n)) = Sum_{j=0..B(n)} t(j), n >= 0, with z = A319198, B = A278039 and t = A080843.
a(n) = -A(n) + 3*B(n) - (n - 1), where A(n) = A278040(n). For a proof see the W. Lang link in A080843, Proposition 8, eq. (46).
a(n) = Sum_{k=1..n-1} d(k), where d is the tribonacci sequence on the alphabet {1,2,0}. - Michel Dekking, Oct 08 2019

Extensions

Name changed by Michel Dekking, Oct 07 2019

A003144 Positions of letter a in the tribonacci word abacabaabacababac... generated by a->ab, b->ac, c->a (cf. A092782).

Original entry on oeis.org

1, 3, 5, 7, 8, 10, 12, 14, 16, 18, 20, 21, 23, 25, 27, 29, 31, 32, 34, 36, 38, 40, 42, 44, 45, 47, 49, 51, 52, 54, 56, 58, 60, 62, 64, 65, 67, 69, 71, 73, 75, 76, 78, 80, 82, 84, 86, 88, 89, 91, 93, 95, 97, 99, 101, 102, 104, 106, 108, 110, 112, 113, 115, 117, 119, 121, 123, 125
Offset: 1

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Author

Keywords

Comments

From Philippe Deléham, Feb 27 2009: (Start)
A003144, A003145, A003146 may be defined as follows. Consider the morphism psi: a -> ab, b -> ac, c -> a. The image (or trajectory) of a under repeated application of this map is the infinite ternary tribonacci 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. (End) [For the word with a -> 0, b -> 1, c -> 2 with offset 0 see A080843. - Wolfdieter Lang, Aug 10 2018]
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 a 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 0. [Duchene and Rigo, Remark 2.5] - N. J. A. Sloane, Nov 18 2016; corrected 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).

Crossrefs

Cf. A003145, A003146, A080843, A092782, A058265, A275926, A276793, A276796, A278039 (subtract 1 from each term, and use offset 0).
First differences are A276788.
For tribonacci representations of numbers see A278038.

Programs

  • 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) = `a` then t1:=[op(t1),i]; fi; od: t1; # N. J. A. Sloane, Nov 01 2006
  • Mathematica
    A003144L = StringPosition[SubstitutionSystem[{"a" -> "ab", "b" -> "ac", "c" -> "a"}, "a", {#}][[1]], "a"][[All, 1]] &; A003144L[7] (* JungHwan Min, Dec 22 2016 *)

Formula

It appears that a(n) is always either floor(n*t) or floor(n*t)+1 for all n, where t is the tribonacci constant A058265. See A275926. - N. J. A. Sloane, Oct 28 2016. This is true - see the Dekking et al. paper. - N. J. A. Sloane, Jul 22 2019

Extensions

More terms from Philippe Deléham, Apr 16 2004
Entry revised by N. J. A. Sloane, Oct 13 2016

A278040 The tribonacci representation of a(n) is obtained by appending 0,1 to the tribonacci representation of n (cf. A278038).

Original entry on oeis.org

1, 5, 8, 12, 14, 18, 21, 25, 29, 32, 36, 38, 42, 45, 49, 52, 56, 58, 62, 65, 69, 73, 76, 80, 82, 86, 89, 93, 95, 99, 102, 106, 110, 113, 117, 119, 123, 126, 130, 133, 137, 139, 143, 146, 150, 154, 157, 161, 163, 167, 170, 174, 178, 181, 185, 187, 191, 194, 198, 201, 205, 207, 211, 214, 218, 222, 225, 229, 231, 235
Offset: 0

Views

Author

N. J. A. Sloane, Nov 18 2016

Keywords

Comments

This sequence gives the A(n) numbers of the W. Lang link. There the B(n) and C(n) numbers are A278039(n) and A278041(n), respectively. - Wolfdieter Lang, Dec 05 2018
Positions of letter b in the tribonacci word t generated by a->ab, b->ac, c->a, when given offset 0. - Michel Dekking, Apr 03 2019
This sequence gives the positions of the word ab in the tribonacci word t. This follows from the fact that the letter b is always preceded in t by the letter a, and the formula AA = B-1, where A := A003144, B := A003145, C := A003146. - Michel Dekking, Apr 09 2019

Examples

			The tribonacci representation of 7 is 1000 (see A278038), so a(7) has tribonacci representation 100001, which is 24+1 = 25, so a(7) = 25.
		

Crossrefs

By analogy with the Wythoff compound sequences A003622 etc., the nine compounds of A003144, A003145, A003146 might be called the tribonacci compound sequences. They are A278040, A278041, and A319966-A319972.

Formula

a(n) = A003145(n+1) - 1.
a(n) = A003144(A003144(n)). - N. J. A. Sloane, Oct 05 2018
See Theorem 13 in the Carlitz, Scoville and Hoggatt paper. - Michel Dekking, Mar 20 2019
From Wolfdieter Lang, Dec 13 2018: (Start)
This sequence gives the indices k with A080843(k) = 1, ordered increasingly with offset 0.
a(n) = 1 + 4*n - A319198(n-1), n >= 0, with A319198(-1) = 0.
a(n) = A276796(C(n)) - 1, with C(n) = A278041(n).
For a proof see the W. Lang link, Proposition 5, and eq. (58).
a(n) - 1 = B1(n), where B1-numbers are B-numbers from A278039 followed by an A-number from A278040. See a comment and example in A319968.
a(n) - 1 = B(B(n)) = B(B(n) + 1) - 2, for n > = 0, where B = A278039.
(End)

A278041 The tribonacci representation of a(n) is obtained by appending 0,1,1 to the tribonacci representation of n (cf. A278038).

Original entry on oeis.org

3, 10, 16, 23, 27, 34, 40, 47, 54, 60, 67, 71, 78, 84, 91, 97, 104, 108, 115, 121, 128, 135, 141, 148, 152, 159, 165, 172, 176, 183, 189, 196, 203, 209, 216, 220, 227, 233, 240, 246, 253, 257, 264, 270, 277, 284, 290, 297, 301, 308, 314, 321, 328, 334, 341, 345, 352, 358, 365, 371, 378, 382, 389, 395, 402, 409, 415
Offset: 0

Views

Author

N. J. A. Sloane, Nov 18 2016

Keywords

Comments

This sequence gives the indices k for which A080843(k) = 2, sorted increasingly with offset 0. In the W. Lang link a(n) = C(n). - Wolfdieter Lang, Dec 06 2018
Positions of letter c in the tribonacci word t generated by a->ab, b->ac, c->a, when given offset 0. - Michel Dekking, Apr 03 2019
This sequence gives the positions of the word ac in the tribonacci word t. This follows from the fact that the letter c is always preceded in t by the letter a, and the formula AB = C-1, where A := A003144, B := A003145, C := A003146. - Michel Dekking, Apr 09 2019

Examples

			The tribonacci representation of 7 is 1000 (see A278038), so a(7) has tribonacci representation 1000011, which is 44+2+1 = 47, so a(7) = 47.
		

Crossrefs

By analogy with the Wythoff compound sequences A003622 etc., the nine compounds of A003144, A003145, A003146 might be called the tribonacci compound sequences. They are A278040, A278041, and A319966-A319972.

Formula

a(n) = A003146(n+1) - 1.
a(n) = A003144(A003145(n)). - N. J. A. Sloane, Oct 05 2018
From Wolfdieter Lang, Dec 06 2018: (Start)
a(n) = n + 2 + A(n) + B(n), where A(n) = A278040(n) and B = A278039(n).
a(n) = 7*n + 3 - (z_A(n-1) + 3*z_C(n-1)), where z_A(n) = A276797(n+1) and z_C(n) = A276798(n+1) - 1, n >= 0.
For proofs see the W. Lang link in A080843, eqs. 37 and 40.
a(n) - 1 = B2(n), where B2-numbers are B-numbers from A278039 followed by a C-number from A278041. See a comment and example in A319968.
(End)

A319972 a(n) = A003146(A003146(n)).

Original entry on oeis.org

24, 68, 105, 149, 173, 217, 254, 298, 342, 379, 423, 447, 491, 528, 572, 609, 653, 677, 721, 758, 802, 846, 883, 927, 951, 995, 1032, 1076, 1100, 1144, 1181, 1225, 1269, 1306, 1350, 1374, 1418, 1455, 1499, 1536, 1580, 1604, 1648, 1685, 1729, 1773, 1810, 1854
Offset: 1

Views

Author

N. J. A. Sloane, Oct 05 2018

Keywords

Comments

By analogy with the Wythoff compound sequences A003622 etc., the nine compounds of A003144, A003145, A003146 might be called the tribonacci compound sequences. They are A278040, A278041, and A319966-A319972.
This sequence gives the positions of the word cabac in the tribonacci word t = abacabaa..., fixed point of the morphism a->ab, b->ac, c->a. This follows from the fact that the positional sequences of cabaa, cabab and cabac give a splitting of the positional sequence of the word caba (the unique word in t with prefix the letter c), and that the three sets CA(N), CB(N) and CC(N), give a splitting of the set C(N), where A := A003144, B := A003145, C := A003146. Here N is the set of positive integers. - Michel Dekking, Apr 09 2019

Crossrefs

Formula

a(n) = A003146(A003146(n)).
a(n) = 6*A003144(n) + 7*A003145(n) + 4*n = 7*A278040(n-1) + 6*A278039(n-1) + 4*n + 13, n >= 1. For a proof see the W. Lang link in A278040, Proposition 9, eq. (56). - Wolfdieter Lang, Apr 11 2019

Extensions

More terms from Rémy Sigrist, Oct 16 2018

A319968 a(n) = A003145(A003145(n)).

Original entry on oeis.org

6, 19, 30, 43, 50, 63, 74, 87, 100, 111, 124, 131, 144, 155, 168, 179, 192, 199, 212, 223, 236, 249, 260, 273, 280, 293, 304, 317, 324, 337, 348, 361, 374, 385, 398, 405, 418, 429, 442, 453, 466, 473, 486, 497, 510, 523, 534, 547, 554, 567, 578, 591, 604, 615, 628, 635, 648, 659, 672, 683, 696, 703, 716, 727, 740, 753
Offset: 1

Views

Author

N. J. A. Sloane, Oct 05 2018

Keywords

Comments

By analogy with the Wythoff compound sequences A003622 etc., the nine compounds of A003144, A003145, A003146 might be called the tribonacci compound sequences. They are A278040, A278041, and A319966-A319972.
From Wolfdieter Lang, Oct 19 2018: (Start)
In another version with the tribonacci word TriWord = A080843 (written as a sequence which has offset 0) and the positions of 0, 1 and 2 given by the B = A278039, A = A278040 and C = A278041 numbers, respectively, the present sequence (with offset 0) gives the smaller of the B-number pairs (B(k), B(k+1)) with B(k+1) = B(k) + 1 for some k >= 0 (named tribonacci B0-numbers), ordered increasingly.
The B-numbers A278039 come in three disjoint and complementary types, called B0-, B1- and B2-numbers. They are defined by the indices k of pairs of consecutive entries TriWord(k), Triword(k+1) depending on their values 0, 0 or 0, 1 or 0, 2 for the B0- or B1- or B2-numbers, respectively.
The B0-numbers are a(n+1) = 2*C(n) - n = A(A(n)) + 1; the B1-numbers are B1(n) = A(n) - 1; and the B2-numbers are B2(n) = C(n) - 1, all for n >= 0.
B0(n) + 1 = B1(A(n)+1), B1(n) + 1 = A(n) and B2(n) + 1 = C(n).
(End)
(a(n)) equals the positions of the word baa in the tribonacci word t = abacabaa..., fixed point of the morphism a->ab, b->ac, c->a. This follows from the fact that the word aa is always preceded in t by the letter b, and the formula BB = AC-1, where A := A003144, B := A003145, C := A003146. - Michel Dekking, Apr 09 2019

Examples

			From _Wolfdieter Lang_, Oct 19 2018: (Start)
The TriWord A080843 starts: 0, 1, 0, 2, 0, 1, 0, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 0, 1, 0, 2, ... (offset 0)
The trisection of the B-numbers A278039 (indices for 0 in TriWord) begins:
n :  0   1   2   3   4   5   6   7    8    9   10   11   12   13   14   15   16 ...
B0:  6  19  30  43  50  63  74  87  100  111  124  131  144  155  168  179  192 ...
B1:  0   4   7  11  13  17  20  24   28   31   35   37   41   44   48   51   55 ...
B2:  2   9  15  22  26  33  39  46   53   59   66   70   77   83   90   96  103 ...
------------------------------------------------------------------------------------
(End)
		

Crossrefs

Formula

a(n) = A003145(A003145(n)), for n >= 1.
a(n) = B0(n-1) = 2*A003146(n) - (n+1) = 2*A278041(n-1) - (n-1) = A278040(A278040(n-1)) + 1, for n >= 1. For B0 see a comment above and the example. - Wolfdieter Lang, Oct 19 2018
a(n+1) = B(C(n)) = B(C(n) + 1) - 1 = 2*(A(n) + B(n)) + n + 4, for n >= 0, where B = A278039, C = A278041 and A = A278040. For a proof see the W. Lang link in A278040, Proposition 9, eq. (53). - Wolfdieter Lang, Dec 13 2018
a(n) = 2*(A003144(n) + A003145(n)) + n - 1, n >= 1. [Rewriting a formula of the precedimg entry]. - Wolfdieter Lang, Apr 11 2019

Extensions

More terms from Joerg Arndt, Oct 15 2018

A276797 Partial sums of A276794.

Original entry on oeis.org

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
Offset: 0

Views

Author

N. J. A. Sloane, Oct 28 2016

Keywords

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
Conjecture: A140102(n) - n = a(n-1). - N. J. A. Sloane, Oct 26 2016 (added Mar 21 2019). This is true - see the Dekking et al. paper. - N. J. A. Sloane, Jul 22 2019

Crossrefs

A276793(n) + A276794(n) + A276791(n) = 1;
A276796(n) + A276797(n) + A276798(n) = n + 1.

Programs

  • 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:
    N0; N1; N2; # prints A276796, A276797, A276798 (except A276798 is off by 1 because it does not count the initial 0 in A003146). # N. J. A. Sloane, Jun 08 2018

Formula

a(n) = Sum_{k=0..n} A276794(k).
a(n) = n + 1 - (A276796(n) + A276798(n)).
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
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