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.

A090909 Terms a(k) of A073869 for which a(k-1) = a(k), and a(k) and a(k+1) are distinct.

Original entry on oeis.org

2, 5, 7, 10, 13, 15, 18, 20, 23, 26, 28, 31, 34, 36, 39, 41, 44, 47, 49, 52, 54, 57, 60, 62, 65, 68, 70, 73, 75, 78, 81, 83, 86, 89, 91, 94, 96, 99, 102, 104, 107, 109, 112, 115, 117, 120, 123, 125, 128, 130, 133, 136, 138, 141, 143, 146, 149, 151, 154, 157, 159, 162
Offset: 1

Views

Author

Amarnath Murthy, Dec 14 2003

Keywords

Comments

Is this the same as A001950? - Alec Mihailovs (alec(AT)mihailovs.com), Jul 23 2007
Identical to n + A066096(n)? - Ed Russell (times145(AT)hotmail.com), May 09 2009
From Michel Dekking, Dec 18 2024: (Start)
Proof of Mihailovs's conjecture: This follows immediately from the result in my 2023 paper in JIS that A073869 is equal to Hofstadter’s G-sequence A005206, and my recent comment in A005206 on the pairs of duplicate values in A005206.
The answer to Russell’s question is well-known, and also Detlef’s formula is well-known.
Originally, this sequence was given the name “Terms a(k) of A073869 for which a(k-1), a(k) and a(k+1) are distinct.’’ These are the triples (1,2,3),(4,5,6),(6,7,8), (9,10,11), ... occurring at k = 3, k = 8, k = 11, k = 16,... in A005206. Note that if a duplicate pair (a(m-1), a(m)) is followed directly by another duplicate pair, then a(m-3), a(m-2) and a(m-1) are distinct, and only so. This corresponds to the block 00 occurring in the Fibonacci word obtained by projecting A005206 on the Fibonacci word (see Corollary in my recent comment in A005206). These occurrences are at the Wythoff AB numbers A003623 according to Wolfdieter Lang’s comment in A003623. Conclusion: the sequence of terms a(k) of A073869 for which a(k-1), a(k), and a(k+1) are distinct is given by the Wythoff AB-numbers. (End)

Examples

			A073869 = A005206 = 0,1,1,2,3,3,4,4,5,6,6,... The pair (1,1) occurs at k = 2.
		

Crossrefs

Programs

  • Magma
    [Floor((3+Sqrt(5))*n/2): n in [0..80]]; // G. C. Greubel, Sep 12 2023
    
  • Mathematica
    (* First program *)
    A002251= Fold[Append[#1, #2 Ceiling[#2/GoldenRatio] -Total[#1]] &, {1}, Range[2, 500]] - 1; (* Birkas Gyorgy's code of A019444, modified *)
    A090909= Join[{0}, Select[Partition[A002251, 2, 1], #[[2]] > #[[1]] &][[All, 2]]] (* G. C. Greubel, Sep 12 2023 *)
    (* Second program *)
    Floor[GoldenRatio^2*Range[0,80]] (* G. C. Greubel, Sep 12 2023 *)
  • SageMath
    [floor(golden_ratio^2*n) for n in range(81)] # G. C. Greubel, Sep 12 2023

Formula

a(n) = floor(phi^2*n), where phi = (1+sqrt(5))/2. - Gary Detlefs, Mar 10 2011

Extensions

More terms from R. J. Mathar, Sep 29 2017
Name corrected by Michel Dekking, Dec 13 2024

A090908 Terms a(k) of A073869 for which a(k) is distinct from a(k-1).

Original entry on oeis.org

1, 3, 4, 6, 8, 9, 11, 12, 14, 16, 17, 19, 21, 22, 24, 25, 27, 29, 30, 32, 33, 35, 37, 38, 40, 42, 43, 45, 46, 48, 50, 51, 53, 55, 56, 58, 59, 61, 63, 64, 66, 67, 69, 71, 72, 74, 76, 77, 79, 80, 82, 84, 85, 87, 88, 90, 92, 93, 95, 97, 98, 100, 101, 103, 105, 106, 108, 110, 111
Offset: 1

Views

Author

Amarnath Murthy, Dec 14 2003

Keywords

Comments

Is this the same as A000201?
The answer is yes: see A090909. - Michel Dekking, Dec 19 2024

Examples

			A073869 = A005206 = 0,1,1,2,3,3,4,4,5,6,6,... The pair (1,2) occurs at k = 3.
		

Crossrefs

Formula

See A000201.

Extensions

More terms from R. J. Mathar, Sep 29 2017
Name corrected by Michel Dekking, Dec 19 2024

A090638 Duplicate of A073869.

Original entry on oeis.org

0, 1, 1, 2, 3, 3, 4, 4, 5, 6, 6, 7, 8, 8, 9, 9, 10, 11, 11, 12, 12, 13, 14, 14, 15, 16, 16, 17, 17
Offset: 1

Views

Author

Keywords

A005206 Hofstadter G-sequence: a(0) = 0; a(n) = n - a(a(n-1)) for n > 0.

Original entry on oeis.org

0, 1, 1, 2, 3, 3, 4, 4, 5, 6, 6, 7, 8, 8, 9, 9, 10, 11, 11, 12, 12, 13, 14, 14, 15, 16, 16, 17, 17, 18, 19, 19, 20, 21, 21, 22, 22, 23, 24, 24, 25, 25, 26, 27, 27, 28, 29, 29, 30, 30, 31, 32, 32, 33, 33, 34, 35, 35, 36, 37, 37, 38, 38, 39, 40, 40, 41, 42, 42, 43, 43, 44, 45, 45, 46, 46, 47
Offset: 0

Views

Author

Keywords

Comments

Rule for finding n-th term: a(n) = An, where An denotes the Fibonacci antecedent to (or right shift of) n, which is found by replacing each F(i) in the Zeckendorf expansion (obtained by repeatedly subtracting the largest Fibonacci number you can until nothing remains) by F(i-1) (A1=1). For example: 58 = 55 + 3, so a(58) = 34 + 2 = 36. - Diego Torres (torresvillarroel(AT)hotmail.com), Nov 24 2002
From Albert Neumueller (albert.neu(AT)gmail.com), Sep 28 2006: (Start)
A recursively built tree structure can be obtained from the sequence (see Hofstadter, p. 137):
14 15 16 17 18 19 20 21
\ / / \ / \ / /
9 10 11 12 13
\ / / \ /
6 7 8
\ / /
\ / /
\ / /
4 5
\ /
\ /
\ /
\ /
\ /
3
/
2
\ /
1
To construct the tree: node n is connected with the node a(n) below
n
/
a(n)
For example, since a(7) = 4:
7
/
4
If the nodes of the tree are read from bottom to top, left to right, one obtains the positive integers: 1, 2, 3, 4, 5, 6, ... The tree has a recursive structure, since the construct
/
x
\ /
x
can be repeatedly added on top of its own ends, to construct the tree from its root: e.g.,
/
x
/ \ /
x x
\ / /
x x
\ /
\ /
x
When moving from a node to a lower connected node, one is moving to the parent. Parent node of n: floor((n+1)/tau). Left child of n: floor(tau*n). Right child of n: floor(tau*(n+1))-1 where tau=(1+sqrt(5))/2. (See the Sillke link.)
(End)
The number n appears A001468(n) times; A001468(n) = floor((n+1)*Phi) - floor(n*Phi) with Phi = (1 + sqrt 5)/2. - Philippe Deléham, Sep 22 2005
Number of positive Wythoff A-numbers A000201 not exceeding n. - N. J. A. Sloane, Oct 09 2006
Number of positive Wythoff B-numbers < A000201(n+1). - N. J. A. Sloane, Oct 09 2006
From Bernard Schott, Apr 23 2022: (Start)
Properties coming from the 1st problem proposed during the 45th Czech and Slovak Mathematical Olympiad in 1996 (see IMO Compendium link):
-> a(n) >= a(n-1) for any positive integer n,
-> a(n) - a(n-1) belongs to {0,1},
-> No integer n exists such that a(n-1) = a(n) = a(n+1). (End)
For n >= 1, find n in the Wythoff array (A035513). a(n) is the number that precedes n in its row, using the preceding column of the extended Wythoff array (A287870) if n is at the start of the (unextended) row. - Peter Munn, Sep 17 2022
See my 2023 publication on Hofstadter's G-sequence for a proof of the equality of (a(n)) with the sequence A073869. - Michel Dekking, Apr 28 2024
From Michel Dekking, Dec 16 2024: (Start)
Focus on the pairs of duplicate values, i.e., the pairs (a(n-1),a(n)) with a(n-1) = a(n). Directly from Theorem 1 in Kimberling and Stolarsky (2016) one derives that the m-th pair of duplicate values (a(n-1),a(n)) occurs at n = U(m), where U = 2,5,7,10,... is the upper Wythoff sequence. For example, (3,3) is the second pair, and occurs at U(2) = 5.
This property can be used to give a simple construction for (a(n)) -- ignoring the superfluous a(0) = 0.
Let 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,... be the sequence of positive natural numbers. Double all the elements of the lower Wythoff sequence (L(n)) = 1,3,4,6,8,9,11,....:
(x(n)) := 1,1, 2, 3,3, 4,4, 5, 6,6, 7, 8,8, 9,9, 10,....
Claim: the result is (a(n)). This follows since because of the doubling, the m-th pair of duplicate values (a(n-1),a(n)) occurs in x at n = L(m) + m = U(m). The second equality by a well-known formula.
It follows from this by Theorem 1 of K&S, that a(n-1) = x(n-1), and a(n) = x(n) if n = U(m), for all m. But since L and U are complementary sequences, a(n) = x(n) will also hold if n = L(k), for all k. For example, L(4) = 6, and a(6) = x(6) = 4.
Corollary: for n >= 2 replace every pair of duplicate values (a(n-1),a(n)) by 0, and all the remaining elements of (a(n)) by 1. Then the result is the Fibonacci word 0,1,0,0,1,0,1,0,0... This is implied by the fact that L gives the positions of the 0s, and U the position of the 1's in the Fibonacci word. (End)
For all n >= 0, a(n) <= A005374(n), as proved in Letouzey-Li-Steiner link. Last equality occurs at n = 12, while a(n) < A005374(n) afterwards. - Pierre Letouzey, Feb 20 2025

References

  • D. R. Hofstadter, Goedel, Escher, Bach: an Eternal Golden Braid, Random House, 1980, p. 137.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Apart from initial terms, same as A060143. Cf. A123070.
a(n):=Sum{k=1..n} h(k), n >= 1, with h(k):= A005614(k-1) and a(0):=0.

Programs

  • Haskell
    a005206 n = a005206_list !! n
    a005206_list = 0 : zipWith (-) [1..] (map a005206 a005206_list)
    -- Reinhard Zumkeller, Feb 02 2012, Aug 07 2011
    
  • Haskell
    a005206 = sum . zipWith (*) a000045_list . a213676_row . a000201 . (+ 1)
    -- Reinhard Zumkeller, Mar 10 2013
    
  • Magma
    [Floor((n+1)*(1+Sqrt(5))/2)-n-1: n in [0..80]]; // Vincenzo Librandi, Nov 19 2016
    
  • Maple
    H:=proc(n) option remember; if n=0 then 0 elif n=1 then 1 else n-H(H(n-1)); fi; end proc: seq(H(n),n=0..76);
  • Mathematica
    a[0] = 0; a[n_] := a[n] = n - a[a[n - 1]]; Array[a, 77, 0]
    (* Second program: *)
    Fold[Append[#1, #2 - #1[[#1[[#2]] + 1 ]] ] &, {0}, Range@ 76] (* Michael De Vlieger, Nov 13 2017 *)
  • PARI
    first(n)=my(v=vector(n)); v[1]=1; for(k=2,n, v[k]=k-v[v[k-1]]); concat(0,v) \\ Charles R Greathouse IV, Sep 02 2015
    
  • Python
    from math import isqrt
    def A005206(n): return (n+1+isqrt(5*(n+1)**2)>>1)-n-1 # Chai Wah Wu, Aug 09 2022

Formula

a(n) = floor((n+1)*tau) - n - 1 = A000201(n+1)-n-1, where tau = (1+sqrt(5))/2; or a(n) = floor(sigma*(n+1)) where sigma = (sqrt(5)-1)/2.
a(0)=0, a(1)=1, a(n) = n - a(floor(n/tau)). - Benoit Cloitre, Nov 27 2002
a(n) = A019446(n) - 1. - Reinhard Zumkeller, Feb 02 2012
a(n) = n - A060144(n+1). - Reinhard Zumkeller, Apr 07 2012
a(n) = Sum_{k=1..A072649(m)} A000045(m)*A213676(m,k): m=A000201(n+1). - Reinhard Zumkeller, Mar 10 2013
From Pierre Letouzey, Sep 09 2015: (Start)
a(n + a(n)) = n.
a(n) + a(a(n+1) - 1) = n.
a(0) = 0, a(n+1) = a(n) + d(n) and d(0) = 1, d(n+1)=1-d(n)*d(a(n)). (End)
a(n) = A293688(n)/(n+1) for n >= 0 (conjectured). - Enrique Navarrete, Oct 15 2017
A generalization of Diego Torres's 2002 comment as a formula: if n = Sum_{i in S} A000045(i+1), where S is a set of positive integers, then a(n) = Sum_{i in S} A000045(i). - Peter Munn, Sep 28 2022
Conjectures from Chunqing Liu, Aug 01 2023: (Start)
a(A000201(n)-1) = n-1.
a(A001950(n)-1) = a(A001950(n)) = A000201(n). (End)

Extensions

a(0) = 0 added in the Name by Bernard Schott, Apr 23 2022

A002251 Start with the nonnegative integers; then swap L(k) and U(k) for all k >= 1, where L = A000201, U = A001950 (lower and upper Wythoff sequences).

Original entry on oeis.org

0, 2, 1, 5, 7, 3, 10, 4, 13, 15, 6, 18, 20, 8, 23, 9, 26, 28, 11, 31, 12, 34, 36, 14, 39, 41, 16, 44, 17, 47, 49, 19, 52, 54, 21, 57, 22, 60, 62, 24, 65, 25, 68, 70, 27, 73, 75, 29, 78, 30, 81, 83, 32, 86, 33, 89, 91, 35, 94, 96, 37, 99, 38, 102, 104, 40, 107, 109
Offset: 0

Views

Author

Keywords

Comments

(n,a(n)) are Wythoff pairs: (0,0), (1,2), (3,5), (4,7), ..., where each difference occurs once.
Self-inverse when considered as a permutation or function, i.e., a(a(n)) = n. - Howard A. Landman, Sep 25 2001
If the offset is 1, the sequence can also be obtained by rearranging the natural numbers so that sum of n terms is a multiple of n, or equivalently so that the arithmetic mean of the first n terms is an integer. - Amarnath Murthy, Aug 16 2002
For n = 1, 2, 3, ..., let p(n)=least natural number not already an a(k), q(n) = n + p(n); then a(p(n)) = q(n), a(q(n)) = p(n). - Clark Kimberling
Also, indices of powers of 2 in A086482. - Amarnath Murthy, Jul 26 2003
There is a 7-state Fibonacci automaton (see a002251_1.pdf) that accepts, in parallel, the Zeckendorf representations of n and a(n). - Jeffrey Shallit, Jul 14 2023

References

  • E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 1982, see p. 76.

Crossrefs

The sequence maps between A000201 and A001950, in that a(A000201(n)) = A001950(n), a(A001950(n)) = A000201(n).
Row 0 of A018219.

Programs

  • Mathematica
    With[{n = 42}, {0}~Join~Take[Values@ #, LengthWhile[#, # == 1 &] &@ Differences@ Keys@ #] &@ Sort@ Flatten@ Map[{#1 -> #2, #2 -> #1} & @@ # &, Transpose@ {Array[Floor[# GoldenRatio] &, n], Array[Floor[# GoldenRatio^2] &, n]}]] (* Michael De Vlieger, Nov 14 2017 *)
  • PARI
    A002251_upto(N,c=0,A=Vec(0,N))={for(n=1,N, A[n]||(#AA002251[1]=2, a(0)=0 is not included. - M. F. Hasler, Nov 27 2019, replacing earlier code from Sep 17 2014

Formula

a(n) = A019444(n+1) - 1.

Extensions

Edited by Christian G. Bower, Oct 29 2002

A367068 a(n) = ((Sum_{i=1..n} A340510(i))-1)/(n+1).

Original entry on oeis.org

0, 1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 7, 8, 9, 9, 10, 11, 11, 12, 12, 13, 14, 14, 15, 16, 16, 17, 17, 18, 19, 19, 20, 20, 21, 22, 22, 23, 24, 24, 25, 25, 26, 27, 27, 28, 29, 29, 30, 30, 31, 32, 32, 33, 33, 34, 35, 35, 36, 37, 37, 38, 38, 39, 40, 40, 41, 42, 42
Offset: 1

Views

Author

Zenan Sabanac, Dec 17 2023

Keywords

Comments

For a positive integer k define the Avdispahić-Zejnulahi sequence AZ(k) by b(1)=k, and thereafter b(n) is the least positive integer not yet in the sequence such that Sum_{i=1..n} b(i) == k (mod n+k).
Define the Avdispahić-Zejnulahi means sequence AZM(k) by a(n) = ((Sum_{i=1..n} b(i))-k)/(n+k). This is the AZM(1) sequence.
Is this a duplicate of A005379? For n<=1300 at least a(n)=A005379(n). - R. J. Mathar, Jan 30 2024

Crossrefs

Cf. A340510.
Cf. A073869 (AZM(0)), A367066 (AZM(2)).

Programs

  • Maple
    A367068 := proc(n)
        add(A340510(i),i=1..n)-1 ;
        %/(n+1) ;
    end proc:
    seq(A367068(n),n=1..50) ; # R. J. Mathar, Jan 30 2024
  • Mathematica
    zlist = {-1, 1, 3};
    mlist = {-1, 0, 1};
    For[n = 3, n <= 101, n++,
      If[MemberQ[zlist, mlist[[n]]], AppendTo[mlist, mlist[[n]] + 1];
        AppendTo[zlist, mlist[[n + 1]] + n];, AppendTo[mlist, mlist[[n]]];
         AppendTo[zlist, mlist[[n + 1]]];];];
    mlist = Drop[mlist, 1]; mlist
  • Python
    z_list=[-1, 1, 3]
    m_list=[-1, 0, 1]
    n=2
    for n in range(2, 100):
        if m_list[n] in z_list:
            m_list.append(m_list[n] + 1)
            z_list.append(m_list[n+1] + n+1)
        else:
            m_list.append(m_list[n])
            z_list.append(m_list[n+1])
    print(m_list[1:])

Formula

For n>2, a(n) = a(n-1) if a(n-1) <> A340510(k) (for k=1..n-1) and a(n) = a(n-1)+1=A340510(n)-n otherwise. (See Proposition 3.1. of Avdispahić and Zejnulahi in the link above).

A367069 a(n) = ((Sum_{i=1..n} A367067(i))-3)/(n+3).

Original entry on oeis.org

0, 1, 1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 7, 8, 9, 9, 10, 10, 11, 12, 12, 13, 14, 14, 15, 15, 16, 17, 17, 18, 19, 19, 20, 20, 21, 22, 22, 23, 23, 24, 25, 25, 26, 27, 27, 28, 28, 29, 30, 30, 31, 31, 32, 33, 33, 34, 35, 35, 36, 36, 37, 38, 38, 39, 40, 40, 41, 41, 42
Offset: 1

Views

Author

Zenan Sabanac, Dec 17 2023

Keywords

Comments

For a positive integer k define the Avdispahić-Zejnulahi sequence AZ(k) by b(1)=k, and thereafter b(n) is the least positive integer not yet in the sequence such that Sum_{i=1..n} b(i) == k (mod n+k). Define the Avdispahić-Zejnulahi means sequence AZM(k) by a(n) = ((Sum_{i=1..n} b(i))-k)/(n+k). This is the AZM(3) sequence.

Crossrefs

Cf. A367067.
Cf. A073869 (AZM(0)), A367068 (AZM(1)), A367066 (AZM(2)).

Programs

  • Mathematica
    zlist = {-1, 3, 5};
    mlist = {-1, 0, 1};
    For[n = 3, n <= 101, n++,
      If[MemberQ[zlist, mlist[[n]]], AppendTo[mlist, mlist[[n]] + 1];
        AppendTo[zlist, mlist[[n + 1]] + n + 2];,
        AppendTo[mlist, mlist[[n]]]; AppendTo[zlist, mlist[[n + 1]]];];];
    mlist = Drop[mlist, 1]; mlist
  • Python
    z_list=[-1, 3, 5]
    m_list=[-1, 0, 1]
    n=2
    for n in range(2, 100):
        if m_list[n] in z_list:
            m_list.append(m_list[n] + 1)
            z_list.append(m_list[n+1] + n+3)
        else:
            m_list.append(m_list[n])
            z_list.append(m_list[n+1])
    print(m_list[1:])
Showing 1-7 of 7 results.