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.

Previous Showing 21-29 of 29 results.

A322408 Compound sequence with a(n) = A319198(A278041(n)), for n >= 0.

Original entry on oeis.org

3, 7, 11, 15, 18, 22, 26, 30, 34, 38, 42, 45, 49, 53, 57, 61, 65, 68, 72, 76, 80, 84, 88, 92, 95, 99, 103, 107, 110, 114, 118, 122, 126, 130, 134, 137, 141, 145, 149, 153, 157, 160, 164, 168, 172, 176, 180, 184, 187, 191, 195, 199, 203, 207, 211, 214, 218, 222, 226, 230, 234
Offset: 0

Views

Author

Wolfdieter Lang, Jan 02 2019

Keywords

Comments

Old name was: Compound tribonacci sequence a(n) = A319198(A278041(n)), for n >= 0.
a(n) gives the sum of the entries of the tribonacci word sequence t = A080843 not exceeding t(C(n)), with C(n) = A278041(n).
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
The difference sequence (a(n+1)-a(n)) is equal to a change of alphabet of the tribonacci word t = A092782. The alphabet is {4,4,3}. This follows from the formula a(n) = A278039(n) + 2*n + 3. - Michel Dekking, Oct 05 2019

Examples

			n = 2: C(2) = 16, t = {0, 1, 0, 2, 0, 1, 0, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, ...} which sums to 11 = a(2) = 4 + 7, because B(2) = 4.
		

Crossrefs

Formula

a(n) = z(C(n)) = Sum_{j=0..C(n)} t(j), n >= 0, with z = A319198, C = A278041 and t = A080843.
a(n) = B(n) + 2*n + 3, where B(n) = A278039(n). For a proof see the W. Lang link in A080843, Proposition 8, eq. (47).
a(n) = 3 + Sum_{k=1..n-1} d(k), where d is the tribonacci sequence on the alphabet {4,4,3}. - Michel Dekking, Oct 05 2019

Extensions

Name changed by Michel Dekking, Oct 08 2019

A319969 a(n) = A003145(A003146(n)).

Original entry on oeis.org

13, 37, 57, 81, 94, 118, 138, 162, 186, 206, 230, 243, 267, 287, 311, 331, 355, 368, 392, 412, 436, 460, 480, 504, 517, 541, 561, 585, 598, 622, 642, 666, 690, 710, 734, 747, 771, 791, 815, 835, 859, 872, 896, 916, 940, 964, 984, 1008, 1021, 1045, 1065, 1089
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 bab 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 baa, bab and bac give a splitting of the positional sequence of the word ba, and the three sets BA(N), BB(N) and BC(N), give a splitting of the set B(N), where A := A003144, B := A003145, C := A003146. Here N denotes the set of positive integers. - Michel Dekking, Apr 09 2019

Crossrefs

Formula

a(n) = A003145(A003146(n)).
a(n) = 3*A003144(n) + 4*A003145(n) + 2*n = 4*A276040(n-1) + 3*A278039(n-1) + 2*n + 7, n >= 1. For a proof see the W. Lang link, Proposition 9, eq. (50). - Wolfdieter Lang, Apr 11 2019

Extensions

More terms from Rémy Sigrist, Oct 16 2018

A276383 Complement of A158919: complementary Beatty sequence to the Beatty sequence defined by the tribonacci constant tau = A058265.

Original entry on oeis.org

2, 4, 6, 8, 10, 13, 15, 17, 19, 21, 24, 26, 28, 30, 32, 35, 37, 39, 41, 43, 46, 48, 50, 52, 54, 56, 59, 61, 63, 65, 67, 70, 72, 74, 76, 78, 81, 83, 85, 87, 89, 92, 94, 96, 98, 100, 102, 105, 107, 109, 111, 113, 116, 118, 120, 122, 124, 127, 129, 131, 133, 135, 138, 140, 142, 144, 146, 149, 151, 153, 155, 157, 159, 162, 164, 166, 168, 170, 173, 175, 177, 179, 181, 184, 186, 188, 190, 192, 195, 197, 199, 201, 203, 205, 208, 210, 212, 214, 216, 219, 221, 223, 225, 227, 230, 232, 234, 236, 238, 241, 243
Offset: 1

Views

Author

N. J. A. Sloane, Sep 02 2016

Keywords

Comments

This is the Beatty sequence for tau_prime = 2.191487883953118747061354268227517294...,
defined by 1/tau + 1/tau_prime = 1.
Differs from A172278 at n = 162, 209, 256, 303, 324, ...
Note that Beatty sequences do not normally include 0 - see the classic pair A000201, A001950. - N. J. A. Sloane, Oct 19 2018
Note that the tribonacci numbers T = A000073 related to the ternary sequence A080843 lead to the three complementary sequences for the nonnegative integers AT(n) = A278040(n), BT(n) = A278039(n) and CT(n) = A278041(n). - Wolfdieter Lang, Sep 08 2018

Examples

			Comments from _Wolfdieter Lang_, Sep 08 2018 (Start):
The complementary sequences A158919 and A276383 begin:
n:       1 2 3 4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 ...
A158919: 1 3 5 7  9 11 12 14 16 18 20 22 23 25 27 29 31 33 34 36 ...
A276383: 2 4 6 8 10 13 15 17 19 21 24 26 28 30 32 35 37 39 41 43 ...
--------------------------------------------------------------------
The complementary sequences AT, BT and CT begin:
n:  0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15  16  17  18  19 ...
AT: 1  5  8 12 14 18 21 25 29 32 36 38 42 45 49 52  56  58  62  65 ...
BT: 0  2  4  6  7  9 11 13 15 17 19 20 22 24 26 28  30  31  33  35 ...
CT: 3 10 16 23 27 34 40 47 54 60 67 71 78 84 91 97 104 108 115 121 ...
(End)
		

Crossrefs

Similar to but strictly different from A172278.

Programs

  • Maple
    A276383 := proc(n)
        Tau := (1/3)*(1+(19+3*sqrt(33))^(1/3)+(19-3*sqrt(33))^(1/3));
        taupr := 1/(1-1/Tau) ;
        floor(n*taupr) ;
    end proc: # R. J. Mathar, Sep 04 2016
    a:=proc(n) local s,t; t:=evalf(solve(x^3-x^2-x-1=0,x),120)[1]; s:=t/(t-1); floor(n*s) end; seq(a(n),n=0..70); # Muniru A Asiru, Oct 16 2018

Formula

a(n) = floor(n*tau_prime), with tau_prime = tau/(tau - 1), where tau is the tribonacci constant A058265.
tau_prime = (1 + (19 + 3*sqrt(33))^(1/3) + (19 - 3*sqrt(33))^(1/3)) / (-2 + (19 + 3*sqrt(33))^(1/3) + (19 - 3*sqrt(33))^(1/3)). - Wolfdieter Lang, Sep 08 2018

Extensions

Edited by N. J. A. Sloane, Oct 19 2018 at the suggestion of Georg Fischer

A319970 a(n) = A003146(A003144(n)).

Original entry on oeis.org

4, 17, 28, 41, 48, 61, 72, 85, 98, 109, 122, 129, 142, 153, 166, 177, 190, 197, 210, 221, 234, 247, 258, 271, 278, 291, 302, 315, 322, 335, 346, 359, 372, 383, 396, 403, 416, 427, 440, 451, 464, 471, 484, 495, 508, 521, 532, 545, 552, 565, 576, 589, 602, 613
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 cabaa 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 baa is always preceded in t by the word ca, and the formula CA = BB-2, where A := A003144, B := A003145, C := A003146. See A319968 for BB. - Michel Dekking, Apr 09 2019
The fact that this sequence is the positional sequence of cabaa in the tribonacci word permits to apply Theorem 5.1. in the paper by Huang and Wen. This gives that the sequence (a(n+1)-a(n)) equals the tribonacci word on the alphabet {a(2)-a(1), a(3)-a(2), a(5)-a(4)} = {13, 11, 7}. - Michel Dekking, Oct 04 2019

Crossrefs

Formula

a(n) = A003146(A003144(n)).
a(n) = 2*(A003144(n) + A003145(n)) + n - 3 = 2*(A278040(n-1) + A278039(n-1)) + n + 1, n >= 1. For a proof see the W. Lang link in A278040, Proposition 9, eq. (55). Wolfdieter Lang, Apr 11 2019
a(1) = 4, a(n+1) = 4 + Sum_{k=1..n} d(k), where d is the tribonacci sequence on the alphabet {13,11,7}. - Michel Dekking, Oct 04 2019

Extensions

More terms from Rémy Sigrist, Oct 16 2018

A319971 a(n) = A003146(A003145(n)).

Original entry on oeis.org

11, 35, 55, 79, 92, 116, 136, 160, 184, 204, 228, 241, 265, 285, 309, 329, 353, 366, 390, 410, 434, 458, 478, 502, 515, 539, 559, 583, 596, 620, 640, 664, 688, 708, 732, 745, 769, 789, 813, 833, 857, 870, 894, 914, 938, 962, 982, 1006, 1019, 1043, 1063, 1087
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 cabab 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 bab is always preceded in t by the word ca, and the formula CB = BC-2, where A := A003144, B := A003145, C := A003146. See A319969 for BC, the positional sequence of the word bab. - Michel Dekking, Apr 09 2019

Crossrefs

Formula

a(n) = A003146(A003145(n)).
a(n) = 3*A003144(n) + 4*A003145(n) + 2*(n-1) = 4*A278040(n-1) + 3*A278039(A27n-1) + 2*n + 5, n >= 1. For a proof see the W. Lang link in A278040, Proposition 9, eq. (54). Wolfdieter Lang, Apr 11 2019

Extensions

More terms from Rémy Sigrist, Oct 16 2018

A321333 Compound sequence with a(n) = A319198(A278040(n)), for n >= 0.

Original entry on oeis.org

1, 4, 5, 8, 9, 12, 13, 16, 19, 20, 23, 24, 27, 28, 31, 32, 35, 36, 39, 40, 43, 46, 47, 50, 51, 54, 55, 58, 59, 62, 63, 66, 69, 70, 73, 74, 77, 78, 81, 82, 85, 86, 89, 90, 93, 96, 97, 100, 101, 104, 105, 108, 111, 112, 115, 116, 119, 120, 123, 124, 127
Offset: 0

Views

Author

Wolfdieter Lang, Dec 27 2018

Keywords

Comments

Old name was: Compound tribonacci sequence a(n) = A319198(A278040(n)), for n >= 0.
a(n) gives the sum of the entries of the tribonacci word sequence t = A080843 not exceeding t(A(n)), with A(n) = A278040(n).

Examples

			n = 4, A(4) = 14, t = {0, 1, 0, 2, 0, 1, 0, 0, 1, 0, 2, 0, 1, 0, 1, ...}, which sums to  9  = a(4) = 2*(14 - 7) - 5, because B(4) = 7.
		

Crossrefs

Formula

a(n) = z(A(n)) = Sum_{j=0..A(n)} t(j), n >= 0, with z = A319198, A = A278040 and t = A080843.
a(n) = 2*(A(n) - B(n)) - (n + 1), where B(n) = A278039(n). For a proof see the W. Lang link in A080843, Proposition 8, eq. (45).
a(n)= 1 + Sum_{k=1..n-1} d(k), where d is the tribonacci sequence on the alphabet {3,1,1}. - Michel Dekking, Oct 08 2019

Extensions

Name changed by Michel Dekking, Oct 08 2019

A385532 Second prepended column of the tribonacci array of the second kind, A385436.

Original entry on oeis.org

-1, 0, 0, 1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 7, 8, 9, 9, 10, 11, 11, 12, 13, 13, 14, 15, 15, 16, 17, 17, 18, 19, 19, 20, 20, 21, 22, 22, 23, 24, 24, 25, 26, 26, 27, 28, 28, 29, 30, 30, 31, 31, 32, 33, 33, 34, 35, 35, 36, 37, 37, 38, 39, 39, 40, 41, 41, 42, 43, 43, 44
Offset: 1

Views

Author

A.H.M. Smeets, Jul 02 2025

Keywords

Comments

The numbers that occur twice in this sequence are the numbers in A278039.

Crossrefs

A308200 The tribonacci representation of a(n) is obtained by appending 0,0,0 to the tribonacci representation of n (cf. A278038).

Original entry on oeis.org

0, 7, 13, 20, 24, 31, 37, 44, 51, 57, 64, 68, 75, 81, 88, 94, 101, 105, 112, 118, 125, 132, 138, 145, 149, 156, 162, 169, 173, 180, 186, 193, 200, 206, 213, 217, 224, 230, 237, 243, 250, 254, 261, 267, 274, 281, 287, 294, 298, 305, 311, 318, 325, 331, 338, 342, 349, 355, 362, 368, 375, 379, 386
Offset: 0

Views

Author

N. J. A. Sloane, Jun 23 2019

Keywords

Crossrefs

Essentially partial sums of A276792.

A319195 Irregular triangle with the unique representation of positive integers in the tribonacci ABC-representation.

Original entry on oeis.org

1, 0, 0, 1, 0, 2, 0, 0, 0, 1, 0, 1, 1, 0, 0, 2, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 2, 1, 0, 0, 0, 2, 0, 1, 2, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 2, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 2, 1, 0, 0, 0, 0, 2, 0, 1, 0, 2, 0, 0, 1, 2, 0, 2, 2, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0
Offset: 1

Views

Author

Wolfdieter Lang, Sep 13 2018

Keywords

Comments

The row length L(n) of this irregular triangle is A316714(n), n >= 1.
This representation is based on the complementary and disjoint sets A, B and C given by the sequences A278040, A278039 and A278041, respectively. In the present notation sequences A, B and C are denoted by 1, 0 and 2.
The numbers are represented by iterations of these sequences always starting with B(0) = 0 (in analogy to the Wythoff B sequence in the Fibonacci case). Uniqueness requires that the representations end in A(B(0)) or C(B(0)).
B^[k](0) (k-fold iterations) for k >= 2 are forbidden. One could represent the number 0 by B(0), but this is not done here, because it is found that the ABC-representations of positive numbers is equivalent to the tribonacci representation of positive numbers given in A278038 for n >= 1 (n = 0 is not represented by T(1) = A000073(1) = 0. This representation uses the tribonacci numbers {T(k)}_{k >= 3} = {1, 2, 4, 7, 13, ...} for uniqueness reason).
For this table the operation of sequences A, B and C is denoted by 1, 0 and 2, respectively, and the brackets and the final argument (0) of B(0) are not recorded. E.g., A(B(C(B(0)))) is written as 1020.
Another form of this table is given in A316713 where A, B and C are denoted 2, 1 and 3, respectively.
An equivalent such representation is given by A317206 using different complementary sequences A, B and C, related to our B = A278039, A = A278040, and C = A278041: A(n) = A003144(n) = A278039(n-1) + 1, B(n) = A003145(n) = A278040(n-1) + 1, C(n) = A003146(n) = A278041(n-1) + 1 with n >= 1.
The present representation is the analog to the Wythoff representation of positive numbers (A189921 or A317208) using the Wythoff A and B sequences A000201 and A001950, respectively.
The number length of the ABC-representation of n >= 1 is L(n) = A316714(n). The number of 0's (B's), 1's (A's) and 2's (C's) of the representation of n is A316715, A316716, A316717.

Examples

			The complementary and disjoint sequences A, B, C begin, for n >= 0:
n: 0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15  16  17  18  19  20  21  22 ...
A: 1  5  8 12 14 18 21 25 29 32 36 38 42 45 49 52  56  58  62  65  69  73  76 ...
B: 0  2  4  6  7  9 11 13 15 17 19 20 22 24 26 28  30  31  33  35  37  39  41 ...
C: 3 10 16 23 27 34 40 47 54 60 67 71 78 84 91 97 104 108 115 121 128 135 141 ...
---------------------------------------------------------------------------------
The ABC-representation of the positive integers begins:
                                              #(1)   #(2)    #(3)    L(n)
                                           A316715 A316716 A316717 A316714
n = 1:      10               A(B(0)) =  1      1      1       0       2
n = 2:     010            B(A(B(0))) =  2      2      1       0       3
n = 3:      20               C(B(0)) =  3      1      0       1       2
n = 4:    0010         B(B(A(B(0)))) =  4      3      1       0       4
n = 5:     110            A(A(B(0))) =  5      1      2       0       3
n = 6:     020            B(C(B(0))) =  6      2      0       1       3
n = 7:   00010      B(B(B(A(B(0))))) =  7      4      1       0       5
n = 8:    1010         A(B(A(B(0)))) =  8      2      2       0       4
n = 9:    0110         B(A(A(B(0)))) =  9      2      2       0       4
n = 10:    210            C(A(B(0))) = 10      1      1       1       3
n = 11:   0020         B(B(C(B(0)))) = 11      3      0       1       4
n = 12:    120            A(C(B(0))) = 12      1      1       1       3
n = 13: 000010   B(B(B(B(A(B(0)))))) = 13      5      1       0       6
n = 14:  10010      A(B(B(A(B(0))))) = 14      3      2       0       5
n = 15:  01010      B(A(B(A(B(0))))) = 15      3      2       0       5
n = 16:   2010         C(B(A(B(0)))) = 16      2      1       1       4
n = 17:  00110      B(B(A(A(B(0))))) = 17      3      2       0       5
n = 18:   1110         A(A(A(B(0)))) = 18      1      3       0       4
n = 19:   0210         B(C(A(B(0)))) = 19      2      1       1       4
n = 20:  00020      B(B(B(C(B(0))))) = 20      4      0       1       5
...
		

Crossrefs

Previous Showing 21-29 of 29 results.