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 42 results. Next

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)

A278039 The tribonacci representation of a(n) is obtained by appending a 0 to the tribonacci representation of n (cf. A278038).

Original entry on oeis.org

0, 2, 4, 6, 7, 9, 11, 13, 15, 17, 19, 20, 22, 24, 26, 28, 30, 31, 33, 35, 37, 39, 41, 43, 44, 46, 48, 50, 51, 53, 55, 57, 59, 61, 63, 64, 66, 68, 70, 72, 74, 75, 77, 79, 81, 83, 85, 87, 88, 90, 92, 94, 96, 98, 100, 101, 103, 105, 107, 109, 111, 112, 114, 116, 118, 120, 122, 124, 125, 127, 129, 131, 132, 134, 136
Offset: 0

Views

Author

N. J. A. Sloane, Nov 18 2016

Keywords

Comments

This sequence records the indices for the 0 values of A080843, ordered increasingly. In the W. Lang link a(n) = B(n). - Wolfdieter Lang, Dec 06 2018
Sequence gives the positions of letter a in the tribonacci word generated by a->ab, b->ac, c->a, when given offset 0. - Michel Dekking, Apr 03 2019

Examples

			The tribonacci representation of 7 is 1000 (see A278038), so a(7) has tribonacci representation 10000, which is 13, so a(7) = 13.
		

Crossrefs

Partial sums of A276788.

Formula

a(n) = A003144(n+1) - 1 = Sum_{k=1..n} A276788(k), n >= 0 (an empty sum is 0).
a(n) = 2*n - (A276798(n) - 1), n >= 0. For a proof see the link, Proposition 6 B). - Wolfdieter Lang, Dec 04 2018

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)

A352089 Tribonacci-Niven numbers: numbers that are divisible by the number of terms in their minimal (or greedy) representation in terms of the tribonacci numbers (A278038).

Original entry on oeis.org

1, 2, 4, 6, 7, 8, 12, 13, 14, 18, 20, 21, 24, 26, 27, 28, 30, 33, 36, 39, 40, 44, 46, 48, 56, 60, 68, 69, 72, 75, 76, 80, 81, 82, 84, 87, 88, 90, 94, 96, 100, 108, 115, 116, 120, 126, 128, 129, 132, 135, 136, 138, 140, 149, 150, 156, 162, 168, 174, 176, 177, 180
Offset: 1

Views

Author

Amiram Eldar, Mar 04 2022

Keywords

Comments

Numbers k such that A278043(k) | k.
The positive tribonacci numbers (A000073) are all terms.
If k = A000073(A042964(m)) is an odd tribonacci number, then k+1 is a term.
Ray (2005) and Ray and Cooper (2006) called these numbers "3-Zeckendorf Niven numbers" and proved that their asymptotic density is 0. - Amiram Eldar, Sep 06 2024

Examples

			6 is a term since its minimal tribonacci representation, A278038(6) = 110, has A278043(6) = 2 1's and 6 is divisible by 2.
		

References

  • Andrew B. Ray, On the natural density of the k-Zeckendorf Niven numbers, Ph.D. dissertation, Central Missouri State University, 2005.

Crossrefs

Programs

  • Mathematica
    t[1] = 1; t[2] = 2; t[3] = 4; t[n_] := t[n] = t[n - 1] + t[n - 2] + t[n - 3]; q[n_] := Module[{s = {}, m = n, k}, While[m > 0, k = 1; While[t[k] <= m, k++]; k--; AppendTo[s, k]; m -= t[k]; k = 1]; Divisible[n, DigitCount[Total[2^(s - 1)], 2, 1]]]; Select[Range[180], q]

A278043 Number of 1's in tribonacci representation of n (cf. A278038).

Original entry on oeis.org

0, 1, 1, 2, 1, 2, 2, 1, 2, 2, 3, 2, 3, 1, 2, 2, 3, 2, 3, 3, 2, 3, 3, 4, 1, 2, 2, 3, 2, 3, 3, 2, 3, 3, 4, 3, 4, 2, 3, 3, 4, 3, 4, 4, 1, 2, 2, 3, 2, 3, 3, 2, 3, 3, 4, 3, 4, 2, 3, 3, 4, 3, 4, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 3, 4, 4, 5, 4, 5, 1, 2, 2, 3, 2, 3, 3, 2, 3, 3, 4, 3, 4, 2, 3, 3, 4, 3, 4
Offset: 0

Views

Author

N. J. A. Sloane, Nov 18 2016

Keywords

Crossrefs

Programs

  • Mathematica
    t[1] = 1; t[2] = 2; t[3] = 4; t[n_] := t[n] = t[n - 1] + t[n - 2] + t[n - 3]; a[0] = 0; a[n_] := Module[{s = {}, m = n, k}, While[m > 0, k = 1; While[t[k] <= m, k++]; k--; AppendTo[s, k]; m -= t[k]; k = 1]; DigitCount[Total[2^(s - 1)], 2, 1]]; Array[a, 100, 0] (* Amiram Eldar, Mar 04 2022 *)

Formula

a(n) = A000120(A003726(n+1)). - John Keith, May 23 2022

A352087 Numbers whose minimal (or greedy) tribonacci representation (A278038) is palindromic.

Original entry on oeis.org

0, 1, 3, 5, 8, 14, 18, 23, 25, 36, 40, 45, 52, 62, 71, 78, 82, 102, 110, 128, 148, 150, 163, 181, 198, 211, 229, 233, 246, 264, 275, 312, 326, 360, 397, 411, 426, 463, 477, 505, 529, 562, 593, 617, 650, 658, 682, 715, 746, 770, 781, 805, 838, 869, 893, 926, 928
Offset: 1

Views

Author

Amiram Eldar, Mar 04 2022

Keywords

Comments

A000073(n) + 1 is a term for n>=4, since its minimal tribonacci representation is 10...01 with n-4 0's between the two 1's.

Examples

			The first 10 terms are:
   n  a(n)  A278038(a(n))
  -----------------------
   1   0                0
   2   1                1
   3   3               11
   4   5              101
   5   8             1001
   6  14            10001
   7  18            10101
   8  23            11011
   9  25           100001
  10  36           101101
		

Crossrefs

Programs

  • Mathematica
    t[1] = 1; t[2] = 2; t[3] = 4; t[n_] := t[n] = t[n - 1] + t[n - 2] + t[n - 3]; q[n_] := Module[{s = {}, m = n, k}, While[m > 0, k = 1; While[t[k] <= m, k++]; k--; AppendTo[s, k]; m -= t[k]; k = 1]; PalindromeQ[FromDigits @ IntegerDigits[Total[2^(s - 1)], 2]]]; Select[Range[0, 1000], q]

A278045 Number of trailing 0's in tribonacci representation of n (cf. A278038).

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane, Nov 18 2016

Keywords

Comments

The number mod 3 of trailing 0's in the tribonacci representation of n >= 1 (this sequence mod 3) is the tribonacci word itself (A080843). - N. J. A. Sloane, Oct 04 2018
The number of trailing 1's in the tribonacci representation of n >= 0 (cf. A278038) is also the tribonacci word itself (A080843).
From Amiram Eldar, Mar 04 2022: (Start)
The asymptotic density of the occurrences of k = 0, 1, 2, ... is (c-1)/c^(k+1), where c = 1.839286... (A058265) is the tribonacci constant.
The asymptotic mean of this sequence is 1/(c-1) = 1.191487... (End)

Crossrefs

Programs

  • Mathematica
    t[1] = 1; t[2] = 2; t[3] = 4; t[n_] := t[n] = t[n - 1] + t[n - 2] + t[n - 3]; a[0] = 1; a[n_] := Module[{s = {}, m = n, k}, While[m > 0, k = 1; While[t[k] <= m, k++]; k--; AppendTo[s, k]; m -= t[k]; k = 1]; Min[s] - 1]; Array[a, 100, 0] (* Amiram Eldar, Mar 04 2022 *)

A308199 The tribonacci representation of a(n) is obtained by appending 0,0 to the tribonacci representation of n (cf. A278038).

Original entry on oeis.org

0, 4, 7, 11, 13, 17, 20, 24, 28, 31, 35, 37, 41, 44, 48, 51, 55, 57, 61, 64, 68, 72, 75, 79, 81, 85, 88, 92, 94, 98, 101, 105, 109, 112, 116, 118, 122, 125, 129, 132, 136, 138, 142, 145, 149, 153, 156, 160, 162, 166, 169, 173, 177, 180, 184, 186, 190, 193, 197, 200, 204, 206, 210, 213, 217, 221, 224, 228
Offset: 0

Views

Author

N. J. A. Sloane, Jun 23 2019

Keywords

Comments

From Michel Dekking, Oct 06 2019: (Start)
If w is a binary vector not containing 111, then w00 and w01 are also binary vectors not containing 111. So a(n) = A278040(n) - 1.
This sequence gives the positions of the word ab in the tribonacci word t, when t is given offset 0.
This sequence is the compound sequence A278039(A278039) of the three sequences A278039, A278040, A278041, which are the building blocks of the tribonacci world with offset 0. (End)

Examples

			u = abacabaabacaba.., then u(0)u(1) = ab, u(4)u(5) = ab, u(7)u(8) = ab, u(11)u(12) = ab.
		

Crossrefs

Essentially partial sums of A276789.

Formula

From Michel Dekking, Oct 06 2019: (Start)
a(n) = Sum_{k=1..n-1} d(k), where d is the tribonacci word on the alphabet {4,3,2}.
a(n) = A003144(A003144(n)) - 1. (End)

A278042 Number of 0's in tribonacci representation of n (cf. A278038).

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane, Nov 18 2016

Keywords

Crossrefs

Programs

  • Mathematica
    t[1] = 1; t[2] = 2; t[3] = 4; t[n_] := t[n] = t[n - 1] + t[n - 2] + t[n - 3]; a[n_] := Module[{s = {}, m = n, k}, While[m > 0, k = 1; While[t[k] <= m, k++]; k--; AppendTo[s, k]; m -= t[k]; k = 1]; DigitCount[Total[2^(s - 1)], 2, 0]]; Array[a, 100, 0] (* Amiram Eldar, Mar 04 2022 *)

A278044 Length of tribonacci representation of n (cf. A278038).

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane, Nov 18 2016

Keywords

Comments

For n>=2, n appears A001590(n+2) times. - John Keith, May 23 2022

Crossrefs

Cf. A001590.
Similar to, but strictly different from, A201052.

Programs

  • Mathematica
    t[1] = 1; t[2] = 2; t[3] = 4; t[n_] := t[n] = t[n - 1] + t[n - 2] + t[n - 3]; a[0] = 1; a[n_] := Module[{k = 1}, While[t[k] <= n, k++]; k - 1]; Array[a, 100, 0] (* Amiram Eldar, Mar 04 2022 *)

Formula

a(n) = A278042(n) + A278043(n).
Showing 1-10 of 42 results. Next