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 1-7 of 7 results.

A266122 a(n) = index of n in A266121 or 0 if n is not present in that sequence.

Original entry on oeis.org

1, 5, 2, 4, 3, 10, 8, 6, 7, 17, 13, 9, 14, 11, 21, 16, 20, 35, 24, 15, 41, 29, 27, 12, 26, 18, 25, 23, 37, 30, 32, 34, 33, 50, 53, 36, 52, 77, 22, 19, 60, 65, 46, 31, 39, 95, 38, 28, 43, 63, 47, 40, 48, 78, 42, 76, 59, 71, 54, 81, 45, 86, 74, 51, 73, 117, 109, 111, 110, 83, 80, 70, 113, 145, 55, 79, 151, 82, 101, 49
Offset: 1

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Author

Antti Karttunen, Dec 23 2015

Keywords

Comments

If A266121 is really a permutation of natural numbers, then this sequence is also, and no hypothetical zero-values are ever needed.

Links

Crossrefs

Inverse: A266121 (with provisions, see comment section).

A266191 Sequence beginning with a(1)=1, a(2)=2, a(3)=3 and then after each new term a(n) is selected as the least unused number for which a(n)*a(n-1)*a(n-2) is a fibbinary number (A003714), i.e., has no adjacent 1's in its base-2 representation.

Original entry on oeis.org

1, 2, 3, 6, 4, 7, 12, 8, 11, 15, 16, 22, 24, 5, 39, 14, 10, 59, 28, 20, 67, 31, 9, 63, 19, 18, 27, 38, 17, 54, 47, 34, 44, 51, 33, 88, 30, 32, 43, 48, 21, 86, 23, 42, 35, 46, 84, 70, 91, 168, 71, 55, 36, 75, 99, 40, 135, 110, 41, 60, 111, 80, 120, 62, 160, 134, 56, 141, 76, 112, 64, 78, 119, 113, 103, 183, 37, 167, 366, 73
Offset: 1

Views

Author

Antti Karttunen, Dec 23 2015

Keywords

Comments

It is conjectured that this sequence is not only injective, but also surjective on N, i.e., that it is a true permutation of natural numbers.
See also comments in related A266121.

Examples

			For n=4, we start testing from the least so far unused number, which is 4, by multiplying it by a(3)*a(2) = 6. Because 6*4 = 24, which has two adjacent 1's in its binary representation "11000", 6 is disqualified. Next we try 5, and 6*5 = "11110", and 5 is also disqualified. Next we try 6, and 6*6 = "100100", with no adjacent 1's, and we have found the least unused number satisfying the required condition, thus we set a(4) = 6.

Links

Crossrefs

Left inverse: A266192 (also the right inverse if this sequence is a permutation of natural numbers).
Cf. A003714, A003987.
Cf. also A266117, A266121 and A266195 for similar permutations.

A269361 Lexicographically first injection of natural numbers beginning with a(1)=1 such that a(n)*a(n+1) is a fibbinary number (A003714), i.e., has no adjacent 1's in its base-2 representation.

Original entry on oeis.org

1, 2, 4, 5, 8, 9, 16, 10, 13, 20, 17, 32, 18, 29, 36, 33, 40, 26, 21, 49, 42, 52, 25, 41, 50, 82, 57, 45, 53, 80, 34, 64, 37, 113, 74, 125, 84, 61, 72, 58, 73, 65, 68, 69, 128, 66, 129, 132, 133, 77, 114, 81, 117, 90, 104, 85, 98, 89, 93, 100, 105, 161, 106, 97, 169, 122, 137, 157, 138, 136, 121, 141, 149, 221, 153, 109, 160, 116, 144
Offset: 1

Views

Author

Antti Karttunen, Feb 25 2016

Keywords

Comments

The sequence is conjectured to be a permutation of A091072.

Links

Crossrefs

Cf. A003714, A091072, A266121, A266191, A269363, A269365, A269366.

A269363 Lexicographically first injection of natural numbers beginning with a(1)=3 such that for all n >= 1, a(n)*a(n+1) is a fibbinary number (A003714), i.e., has no adjacent 1's in its base-2 representation.

Original entry on oeis.org

3, 6, 7, 12, 11, 15, 22, 24, 14, 19, 27, 38, 28, 23, 46, 48, 43, 30, 39, 35, 59, 44, 31, 75, 62, 87, 51, 83, 56, 47, 88, 54, 76, 55, 96, 86, 60, 71, 67, 70, 78, 112, 79, 107, 102, 91, 120, 139, 118, 140, 119, 142, 131, 134, 155, 240, 156, 135, 152, 108, 95, 92, 103, 179, 184, 115, 147, 224, 94, 175, 123, 150, 111, 158, 214, 163, 203
Offset: 1

Views

Author

Antti Karttunen, Feb 25 2016

Keywords

Comments

The sequence is conjectured to be a permutation of A091067.
The scatter plot is quite interesting (essentially the same as A269367). Compare also to the graph of A269361.

Links

Crossrefs

Cf. A003714, A003987, A091067, A266121, A266191, A269361, A269365, A269366.
Cf. A269367 (the terms ranked with A255070).

Programs

  • Mathematica
    fibbinaryQ[n_] := BitAnd[n, 2 n]==0; a[1]=3; a[n_] := a[n] = For[k=1, True, k++, If[Mod[k, 4] != 1, If[fibbinaryQ[a[n-1] k], If[FreeQ[Array[a, n-1], k], Return[k]]]]]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Mar 02 2016 *)

A266117 Lexicographically first injection of positive integers beginning with a(1) such that a(n)*a(n+1) is a term of A265349, i.e., has no multiple occurrences of any nonzero digit when viewed in factorial base (A007623).

Original entry on oeis.org

1, 2, 3, 4, 5, 10, 11, 6, 7, 12, 8, 9, 24, 13, 18, 19, 26, 14, 17, 22, 21, 16, 15, 32, 30, 20, 23, 29, 42, 28, 25, 48, 34, 53, 41, 54, 27, 40, 33, 36, 37, 61, 65, 44, 49, 72, 39, 38, 51, 52, 55, 59, 47, 46, 58, 50, 60, 62, 31, 66, 56, 57, 64, 45, 80, 63, 74, 69, 68, 35, 79, 100, 73, 43, 67, 70, 71, 78, 84, 87, 92, 90, 76, 96, 75
Offset: 1

Views

Author

Antti Karttunen, Dec 22 2015

Keywords

Comments

After a(1) = 1, always choose for a(n+1) the least unused k such that in factorial base representation (A007623), the product a(n)*a(n+1) will not show any of the nonzero digits present twice (or more times), regardless of the positions of the digits.

Examples

			For n = 6, we start searching from the least not yet used number in range a(1) .. a(5) [which is 6, because all the previous terms are fixed] for the first number whose product with a(5) = 5 results a number in A265349.
Multiplying 5 (in factorial base "21") with 6 (in factorial base "100") results 30, which in factorial base is "1100", containing digit "1" twice, thus 6 is disqualified.
Similarly, products 5*7, 5*8 and 5*9 result 35 = "1121", 40 = "1220" and 45 = "1311", where in all cases one of the nonzero digits occur more than once, so 7, 8 and 9 are also all disqualified.
But 5*10 = 50, which has a factorial base representation ("2010") that matches the criterion, thus a(6) = 10.

Links

Crossrefs

Left inverse: A266118 (also the right inverse if this sequence is a permutation of the positive integers).
Cf. A264990, A265349.
Cf. also A266121, A266191 and A266195 for similar permutations.

A269367 Suspected permutation of natural numbers: a(n) = A255070(A269363(n)).

Original entry on oeis.org

1, 2, 3, 5, 4, 7, 9, 11, 6, 8, 12, 17, 13, 10, 21, 23, 19, 14, 18, 16, 28, 20, 15, 35, 30, 41, 24, 39, 27, 22, 42, 25, 36, 26, 47, 40, 29, 34, 32, 33, 37, 55, 38, 51, 49, 43, 59, 67, 57, 68, 58, 69, 64, 65, 75, 119, 76, 66, 74, 52, 46, 44, 50, 87, 90, 56, 71, 111, 45, 85, 60, 72, 54, 77, 104, 79, 99, 88, 95, 48, 53, 78, 102, 82, 31
Offset: 1

Views

Author

Antti Karttunen, Feb 27 2016

Keywords

Comments

Lexicographically first injection of natural numbers beginning with a(1)=1 such that for all n >= 1, A091067(a(n))*A091067(a(n+1)) is a fibbinary number (A003714), i.e., has no adjacent 1's in its base-2 representation.

Links

Crossrefs

Left inverse: A269368 (also the right inverse if this is a permutation of natural numbers).
Cf. A003714, A091067, A266121, A269361, A269363, A255070.

Formula

a(n) = A255070(A269363(n)).

A269366 a(1) = 1, a(2n) = A269361(1+a(n)), a(2n+1) = A269363(a(n)).

Original entry on oeis.org

1, 2, 3, 4, 6, 5, 7, 8, 12, 16, 15, 9, 11, 10, 22, 13, 24, 18, 38, 40, 48, 33, 46, 20, 14, 32, 27, 17, 19, 25, 44, 29, 28, 50, 75, 21, 30, 72, 71, 73, 70, 133, 139, 113, 76, 129, 91, 42, 35, 36, 23, 37, 54, 45, 51, 26, 43, 49, 39, 82, 62, 128, 107, 80, 56, 53, 83, 114, 140, 109, 214, 52, 59, 34, 47, 149, 150, 141, 123, 221, 111, 121
Offset: 1

Views

Author

Antti Karttunen, Feb 25 2016

Keywords

Comments

This sequence can be represented as a binary tree. Each left hand child is produced as A269361(1+n), and each right hand child as A269363(n), when the parent node contains n:
|
...................1...................
2 3
4......../ \........6 5......../ \........7
/ \ / \ / \ / \
/ \ / \ / \ / \
/ \ / \ / \ / \
8 12 16 15 9 11 10 22
13 24 18 38 40 48 33 46 20 14 32 27 17 19 25 44
etc.
An example of (suspected) "entanglement permutation" where the other pair of complementary sequences is generated by a greedy algorithm.
Sequence is not only injective, but also surjective on N (thus a permutation of natural numbers) provided that A269361 is surjective on A091072 and A269363 is surjective on A091067.

Links

Crossrefs

Left inverse: A269365 (also right inverse, if this sequence is a permutation of natural numbers).
Cf. A091067, A091072, A269361, A269363, A269367, A266121.
Showing 1-7 of 7 results.

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