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.

A087686 Elements of A004001 that repeat consecutively.

Original entry on oeis.org

1, 2, 4, 7, 8, 12, 14, 15, 16, 21, 24, 26, 27, 29, 30, 31, 32, 38, 42, 45, 47, 48, 51, 53, 54, 56, 57, 58, 60, 61, 62, 63, 64, 71, 76, 80, 83, 85, 86, 90, 93, 95, 96, 99, 101, 102, 104, 105, 106, 109, 111, 112, 114, 115, 116, 118, 119, 120, 121, 123, 124, 125, 126, 127
Offset: 1

Views

Author

Roger L. Bagula, Sep 27 2003

Keywords

Comments

Complement of A088359; A051135(a(n)) > 1. [Reinhard Zumkeller, Jun 03 2011]
From Antti Karttunen, Jan 18 2016: (Start)
This set of numbers is closed with respect to A004001, see A266188.
After 1, one more than the positions of zeros in A093879.
(End)

Links

Crossrefs

Cf. A004001, A051135, A093879.
Cf. A088359 (complement), A188163 (almost complement).
Cf. A080677 (the least monotonic left inverse).
Cf. also A266188, A265901, A265903, A267110, A267111, A267112.

Programs

  • Haskell
    import Data.List (findIndices)
a087686 n = a087686_list !! (n-1)
a087686_list = map succ $ findIndices (> 1) a051135_list
-- Reinhard Zumkeller, Jun 03 2011
(Scheme, with Antti Karttunen's IntSeq-library)
(define A087686 (MATCHING-POS 1 1 (lambda (n) (> (A051135 n) 1))))
;; Antti Karttunen, Jan 18 2016
  • Mathematica
    Conway[n_Integer?Positive] := Conway[n] =Conway[Conway[n-1]] + Conway[n - Conway[n-1]] Conway[1] = Conway[2] = 1 digits=1000 a=Table[Conway[n], {n, 1, digits}]; b=Table[If[a[[n]]-a[[n-1]]==0, a[[n]], 0], {n, 2, digits}]; c=Delete[Union[b], 1]

Formula

Other identities. For all n >= 1:
A080677(a(n)) = n. [See comments in A080677.]

A088359 Numbers which occur only once in A004001.

Original entry on oeis.org

3, 5, 6, 9, 10, 11, 13, 17, 18, 19, 20, 22, 23, 25, 28, 33, 34, 35, 36, 37, 39, 40, 41, 43, 44, 46, 49, 50, 52, 55, 59, 65, 66, 67, 68, 69, 70, 72, 73, 74, 75, 77, 78, 79, 81, 82, 84, 87, 88, 89, 91, 92, 94, 97, 98, 100, 103, 107, 108, 110, 113, 117, 122, 129, 130, 131, 132
Offset: 1

Views

Author

Robert G. Wilson v, Sep 26 2003

Keywords

Comments

Out of the first one million terms (a(10^6) = 510403), 258661 occur only once.
Complement of A087686; A051135(a(n)) = 1. - Reinhard Zumkeller, Jun 03 2011
From Antti Karttunen, Jan 18 2016: (Start)
In general, out of the first 2^(n+1) terms of A004001, 2^(n-1) - 1 terms (a quarter) occur only once. See also illustration in A265332.
One more than the positions of ones in A093879.
(End)

Links

Crossrefs

Positions of ones in A051135.
Cf. A188163 (same sequence with prepended 1).
Cf. A087686 (complement).
Cf. A004001, A093879, A265332, A266399.
Cf. also A267110, A267111, A267112.

Programs

  • Haskell
    import Data.List (elemIndices)
a088359 n = a088359_list !! (n-1)
a088359_list = map succ $ elemIndices 1 a051135_list
-- Reinhard Zumkeller, Jun 03 2011
(Scheme, with Antti Karttunen's IntSeq-library)
(define A088359 (ZERO-POS 1 1 (COMPOSE -1+ A051135)))
;; Antti Karttunen, Jan 18 2016
  • Mathematica
    a[1] = 1; a[2] = 1; a[n_] := a[n] = a[ a[n - 1]] + a[n - a[n - 1]]; hc = Table[ a[n], {n, 1, 261}]; RunLengthEncodeOne[x_List] := Length[ # ] == 1 & /@ Split[x]; r = RunLengthEncodeOne[hc]; Select[ Range[ Length[r]], r[[ # ]] == True &]

Formula

From Antti Karttunen, Jan 18 2016: (Start)
Other identities.
For all n >= 0, a(A000079(n)) = A000051(n+1), that is, a(2^n) = 2^(n+1) + 1.
For all n >= 1:
a(n) = A004001(A266399(n)).
(End)

A265332 a(n) is the index of the column in A265901 where n appears; also the index of the row in A265903 where n appears.

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Jan 09 2016

Keywords

Comments

If all 1's are deleted, the remaining terms are the sequence incremented. - after Franklin T. Adams-Watters Oct 05 2006 comment in A051135.
Ordinal transform of A162598.

Examples

			Illustration how the sequence can be constructed by concatenating the frequency counts Q_n of each successive level n of A004001-tree:
--
             1                                      Q_0 = (1)
             |
            _2__                                    Q_1 = (2)
           /    \
         _3    __4_____                             Q_2 = (1,3)
        /     /  |     \
      _5    _6  _7    __8___________                Q_3 = (1,1,2,4)
     /     /   / |   /  |  \        \
   _9    10  11 12  13  14  15___    16_________    Q_4 = (1,1,1,2,1,2,3,5)
  /     /   /  / |  /  / |   |\  \   | \  \  \  \
17    18  19 20 21 22 23 24 25 26 27 28 29 30 31 32
--
The above illustration copied from the page 229 of Kubo and Vakil paper (page 5 in PDF).

Links

Crossrefs

Essentially same as A051135 apart from the initial term, which here is set as a(1)=1.
Cf. A004001, A265901, A265903.
Cf. A162598 (corresponding other index).
Cf. A265754.
Cf. also A267108, A267109, A267110.

Programs

  • Mathematica
    terms = 120;
h[1] = 1; h[2] = 1;
h[n_] := h[n] = h[h[n - 1]] + h[n - h[n - 1]];
seq[nmax_] := seq[nmax] = (Length /@ Split[Sort @ Array[h, nmax, 2]])[[;; terms]];
seq[nmax = 2 terms];
seq[nmax += terms];
While[seq[nmax] != seq[nmax - terms], nmax += terms];
seq[nmax] (* Jean-François Alcover, Dec 19 2021 *)
  • Scheme
    (define (A265332 n) (if (= 1 n) 1 (A051135 n)))

Formula

a(1) = 1; for n > 1, a(n) = A051135(n).

A267112 Permutation of natural numbers: a(1) = 1; a(2n) = A087686(1+a(n)), a(2n+1) = A088359(a(n)), where A088359 and A087686 = numbers that occur only once (resp. more than once) in A004001.

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 6, 8, 9, 12, 10, 15, 13, 14, 11, 16, 17, 21, 18, 27, 22, 24, 19, 31, 28, 29, 23, 30, 25, 26, 20, 32, 33, 38, 34, 48, 39, 42, 35, 58, 49, 51, 40, 54, 43, 45, 36, 63, 59, 60, 50, 61, 52, 53, 41, 62, 55, 56, 44, 57, 46, 47, 37, 64, 65, 71, 66, 86, 72, 76, 67, 106, 87, 90, 73, 96, 77, 80, 68, 121, 107, 109, 88
Offset: 1

Views

Author

Antti Karttunen, Jan 10 2016

Keywords

Comments

This sequence can be represented as a binary tree. Each left hand child is produced as A087686(1+n), and each right hand child as A088359(n), when their parent contains n:
|
...................1...................
2 3
4......../ \........5 7......../ \........6
/ \ / \ / \ / \
/ \ / \ / \ / \
/ \ / \ / \ / \
8 9 12 10 15 13 14 11
16 17 21 18 27 22 24 19 31 28 29 23 30 25 26 20
etc.
The level k of the tree contains all numbers of binary width k, like many base-2 related permutations (A003188, A054429, etc). For a proof, see A267110, which gives the contents of each parent node (for node containing n).
A276442 shows the mirror-image of the same tree.

Links

Crossrefs

Inverse: A267111.
Cf. A000079, A000225, A004001, A006127, A087686, A088359, A267110.
Similar or related permutations: A003188, A054429, A276442, A233276, A233278, A276344, A276346, A276446.
Cf. also permutations A266411, A266412 and arrays A265901, A265903.

Formula

a(1) = 1; after which, a(2n) = A087686(1+a(n)), a(2n+1) = A088359(a(n)).
As a composition of other permutations:
a(n) = A276442(A054429(n)).
a(n) = A276344(A233276(n)).
a(n) = A276346(A233278(n)).
a(n) = A276446(A003188(n)).
Other identities. For all n >= 0:
a(2^n) = 2^n. [Follows from the properties (3) and (4) of A004001 given on page 227 of Kubo & Vakil paper.]
a(A000225(n)) = A006127(n), i.e., a((2^(n+1)) - 1) = 2^n + n. [Numbers at the right edge.]

A276442 Permutation of natural numbers: a(1) = 1; a(2n) = A088359(a(n)), a(2n+1) = A087686(1+a(n)), where A088359 & A087686 = numbers that occur only once & more than once in A004001.

Original entry on oeis.org

1, 3, 2, 6, 7, 5, 4, 11, 14, 13, 15, 10, 12, 9, 8, 20, 26, 25, 30, 23, 29, 28, 31, 19, 24, 22, 27, 18, 21, 17, 16, 37, 47, 46, 57, 44, 56, 55, 62, 41, 53, 52, 61, 50, 60, 59, 63, 36, 45, 43, 54, 40, 51, 49, 58, 35, 42, 39, 48, 34, 38, 33, 32, 70, 85, 84, 105, 82, 104, 103, 120, 79, 101, 100, 119, 98, 118, 117, 126, 75, 95, 94
Offset: 1

Views

Author

Antti Karttunen, Sep 03 2016

Keywords

Comments

This sequence can be represented as a binary tree. Each left hand child is produced as A088359(n), and each right hand child as A087686(1+n), when their parent contains n:
|
...................1...................
3 2
6......../ \........7 5......../ \........4
/ \ / \ / \ / \
/ \ / \ / \ / \
/ \ / \ / \ / \
11 14 13 15 10 12 9 8
20 26 25 30 23 29 28 31 19 24 22 27 18 21 17 16
etc.
As in the mirror image permutation A267112, the level k of the tree contains all numbers of binary width k like many other base-2 related permutations (A003188, A054429, A233278, etc). For a proof, see A267110, which gives the contents of each parent node (for a node containing n > 1).

Links

Crossrefs

Inverse: A276441.
Cf. A004001, A087686, A088359, A267110.
Related or similar permutations: A003188, A054429, A233276, A233278, A267112, A276344, A276346, A276444.

Programs

Formula

a(1) = 1; after which, a(2n) = A088359(a(n)), a(2n+1) = A087686(1+a(n)).
As a composition of other permutations:
a(n) = A267112(A054429(n)).
a(n) = A276344(A233278(n)).
a(n) = A276346(A233276(n)).
a(n) = A276444(A003188(n)).

A267108 a(n) = A000120(A267111(n)).

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Jan 16 2016

Keywords

Links

Crossrefs

Cf. A000120, A004001, A070939, A088359, A265332, A267109, A267110, A267111.

Formula

a(1) = 1; for n > 1, if A265332(n) = 1 [when n is one of the terms of A088359], a(n) = 1 + a(A004001(n)-1), otherwise a(n) = a(n-A004001(n)).
a(n) = A000120(A267111(n)).
Other identities. For all n >= 1:
a(n) = A070939(n) - A267109(n).

A267109 a(n) = A080791(A267111(n)).

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Jan 16 2016

Keywords

Links

Crossrefs

Cf. A004001, A070939, A080791, A088359, A265332, A267108, A267110, A267111.

Formula

a(1) = 0; for n > 1, if A265332(n) = 1 [when n is one of the terms of A088359], a(n) = a(A004001(n)-1), otherwise 1 + a(n) = a(n-A004001(n)).
a(n) = A080791(A267111(n)).
Other identities. For all n >= 1:
a(n) = A070939(n) - A267108(n).
Showing 1-7 of 7 results.

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