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 1-5 of 5 results.

A316713 Unique representation of nonnegative numbers by iterated tribonacci A, B and C sequences.

Original entry on oeis.org

1, 21, 121, 31, 1121, 221, 131, 11121, 2121, 1221, 321, 1131, 231, 111121, 21121, 12121, 3121, 11221, 2221, 1321, 11131, 2131, 1231, 331, 1111121, 211121, 121121, 31121, 112121, 22121, 13121, 111221, 21221, 12221, 3221, 11321, 2321, 111131, 21131, 12131, 3131, 11231, 2231, 1331, 11111121, 2111121, 1211121, 311121, 1121121, 221121, 131121, 1112121, 212121, 122121, 32121, 113121, 23121, 1111221, 211221, 121221, 31221, 112221, 22221, 13221, 111321, 21321, 12321
Offset: 0

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Author

Wolfdieter Lang, Sep 11 2018

Keywords

Comments

This representation is the tribonacci A000073 analog of the Wythoff representation of numbers (A189921 or A317208) for the Fibonacci case.
The complementary and disjoint sets A, B and C are given by the sequences A278040, A278039, and A278041, respectively.
The present representation uses 1 for B, 2 for A and 3 for C numbers. The brackets for sequence iteration and the final argument 0 have to be added. E.g.: a(0) = 1 for B(1), a(1) = 21 for A(B(0)), a(2) = 121 for B(A(B(0))), a(3) = 31 for C(B(0)), ...
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 length of the string a(n) is A316714(n). The number of B, A and C sequences used for the ABC-representation of n (that is the number of 1s, 2s and 3s of a(n)) is A316715, A316716 and A316717, respectively.

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 representations begin:
                                              #(1)   #(2)    #(3)   L(a(n))
           a(n)                             A316715 A316716 A316717 A316714
n = 0:       1                  B(0) =  0      1      0       0       1
n = 1:      21               A(B(0)) =  1      1      1       0       2
n = 2:     121            B(A(B(0))) =  2      2      1       0       3
n = 3:      31               C(B(0)) =  3      1      0       1       2
n = 4:    1121         B(B(A(B(0)))) =  4      3      1       0       4
n = 5:     221            A(A(B(0))) =  5      1      2       0       3
n = 6:     131            B(C(B(0))) =  6      2      0       1       3
n = 7:   11121      B(B(B(A(B(0))))) =  7      4      1       0       5
n = 8:    2121         A(B(A(B(0)))) =  8      2      2       0       4
n = 9:    1221         B(A(A(B(0)))) =  9      2      2       0       4
n = 10:    321            C(A(B(0))) = 10      1      1       1       3
n = 11:   1131         B(B(C(B(0)))) = 11      3      0       1       4
n = 12:    231            A(C(B(0))) = 12      1      1       1       3
n = 13: 111121   B(B(B(B(A(B(0)))))) = 13      5      1       0       6
n = 14:  21121      A(B(B(A(B(0))))) = 14      3      2       0       5
n = 15:  12121      B(A(B(A(B(0))))) = 15      3      2       0       5
n = 16:   3121         C(B(A(B(0)))) = 16      2      1       1       4
n = 17:  11221      B(B(A(A(B(0))))) = 17      3      2       0       5
n = 18:   2221         A(A(A(B(0)))) = 18      1      3       0       4
n = 19:   1321         B(C(A(B(0)))) = 19      2      1       1       4
n = 20:  11131      B(B(B(C(B(0))))) = 20      4      0       1       5
...
----------------------------------------------------------------------------
		

Crossrefs

A316714 a(n) is the number of digits of A316713(n). This is the number of A, B and C sequences used in the tribonacci ABC-representation of n >= 0.

Original entry on oeis.org

1, 2, 3, 2, 4, 3, 3, 5, 4, 4, 3, 4, 3, 6, 5, 5, 4, 5, 4, 4, 5, 4, 4, 3, 7, 6, 6, 5, 6, 5, 5, 6, 5, 5, 4, 5, 4, 6, 5, 5, 4, 5, 4, 4, 8, 7, 7, 6, 7, 6, 6, 7, 6, 6, 5, 6, 5, 7, 6, 6, 5, 6, 5, 5, 6, 5, 5, 4, 7, 6, 6, 5, 6, 5, 5, 6, 5, 5, 4, 5, 4, 9, 8, 8, 7, 8, 7, 7, 8, 7, 7, 6, 7, 6, 8, 7, 7, 6, 7, 6, 6
Offset: 0

Views

Author

Wolfdieter Lang, Sep 11 2018

Keywords

Comments

The number of 1's, 2's and 3's in the representation of n, given in A316713(n), is given in A316715(n), A316716(n) and A316717(n).

Examples

			See A316713, the column Length(a(n)).
		

Crossrefs

Formula

a(n) = length(A316713(n)), n >= 0.
a(n) = A316715(n) + A316716(n) + A316717(n), n >= 0.

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

Views

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.

A316716 a(n) is the number of 2s in A316713(n). That is, a(n) is the number of A-sequences (A278040) used in the tribonacci ABC-representation of n >= 0.

Original entry on oeis.org

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

Views

Author

Wolfdieter Lang, Sep 11 2018

Keywords

Comments

The number of 1s and 3s in A316713(n) is given in A316715 and A316717, respectively.

Examples

			See column #(2) of A316713.
		

Crossrefs

Formula

a(n) = number of 2s in A316713(n), that is the number of As in the tribonacci ABC-representation of n >= 0.

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

Showing 1-5 of 5 results.