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.

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A322414 Compound tribonacci sequence with a(n) = A278041(A278041(n)), for n >= 0.

Original entry on oeis.org

23, 67, 104, 148, 172, 216, 253, 297, 341, 378, 422, 446, 490, 527, 571, 608, 652, 676, 720, 757, 801, 845, 882, 926, 950, 994, 1031, 1075, 1099, 1143, 1180, 1224, 1268, 1305, 1349, 1373, 1417, 1454, 1498, 1535, 1579, 1603, 1647, 1684, 1728, 1772, 1809, 1853, 1877, 1921, 1958, 2002, 2046, 2083, 2127, 2151, 2195, 2232, 2276, 2313, 2357
Offset: 0

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Author

Wolfdieter Lang, Jan 02 2019

Keywords

Comments

(a(n+1)) = A319972(n)-1 = A003146(A003146(n))-1, the corresponding classical compound tribonacci sequence. - Michel Dekking, Apr 04 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(C(n)) = C(C(n) + 1) - 4 = 7*A(n) + 6*B(n) + 4*(n + 4), for n >= 0, where A = A278040, B = A278039 and C = A278041. For a proof see the W. Lang link in A278040, Proposition 9, eq. (56).

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

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
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