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.

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A247461 Subsequence obtained from A026242 by applying an Eratosthenes-type sieve: strike out every second number after the first "2", then if m is the next number not yet struck out, strike out every m-th number following this one, etc.

Original entry on oeis.org

1, 1, 2, 3, 4, 5, 8, 15, 20, 35, 50, 37, 40, 46, 109, 124, 134, 92, 183, 198, 223, 159, 272, 282, 205, 214, 356, 371, 406, 445, 480, 495, 312, 321, 569, 579, 367, 628, 653, 434, 742, 801, 816, 851, 535, 925, 940, 587, 596, 999, 1014, 1024, 709, 755, 1261
Offset: 1

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Author

M. F. Hasler and Eric Angelini, Sep 17 2014

Keywords

Comments

The first two terms a(1)=a(2)=1 are included here but are not considered to be part of what one might call A026242-primes, in analogy of the usual primes A000040 obtained by applying the very same procedure to the set of all positive integers.
Since A026242 is not increasing, this sequence isn't, either. However, since the remaining "primes" > 1 are exactly the numbers used during the sieve, and for all m, the second m occurs m places after the first m in A026242, no number can occur twice here.
There is a stronger version of the sieve, which consists of considering all numbers "m", whether or not they have been crossed out earlier. When this is applied, then the result is the finite subsequence [1, 1, 2, 3, 4, 5, 8, 15, 50]. Eric Angelini calls these numbers, {2, 3, 4, 5, 8, 15, 50}, "Biprimes of K = A026242".

Programs

  • PARI
    /* first compute A026242 to a sufficient number of terms, then: */ for(k=3,#K=A026242, K[k] && forstep(i=k+K[k],#K,K[k],K[i]=0)); A247461=select(x->x,K)
    /* to apply the stronger sieve */ for(k=3,#K=A026242, forstep(i=k+A026242[k],#K,A026242[k],K[i]=0)); select(x->x,K)

A026271 a(n) = sum of the numbers between the two n's in A026242.

Original entry on oeis.org

0, 3, 6, 14, 25, 33, 49, 60, 81, 105, 121, 150, 182, 203, 240, 264, 306, 351, 380, 430, 462, 517, 575, 612, 675, 741, 783, 854, 899, 975, 1054, 1104, 1188, 1275, 1330, 1422, 1480, 1577, 1677, 1740, 1845, 1911, 2021, 2134, 2205
Offset: 1

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Author

Keywords

A000201 Lower Wythoff sequence (a Beatty sequence): a(n) = floor(n*phi), where phi = (1+sqrt(5))/2 = A001622.

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
Offset: 1

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Author

Keywords

Comments

This is the unique sequence a satisfying a'(n)=a(a(n))+1 for all n in the set N of natural numbers, where a' denotes the ordered complement (in N) of a. - Clark Kimberling, Feb 17 2003
This sequence and A001950 may be defined as follows. Consider the maps a -> ab, b -> a, starting from a(1) = a; then A000201 gives the indices of a, A001950 gives the indices of b. The sequence of letters in the infinite word begins a, b, a, a, b, a, b, a, a, b, a, ... Setting a = 0, b = 1 gives A003849 (offset 0); setting a = 1, b = 0 gives A005614 (offset 0). - Philippe Deléham, Feb 20 2004
These are the numbers whose lazy Fibonacci representation (see A095791) includes 1; the complementary sequence (the upper Wythoff sequence, A001950) are the numbers whose lazy Fibonacci representation includes 2 but not 1.
a(n) is the unique monotonic sequence satisfying a(1)=1 and the condition "if n is in the sequence then n+(rank of n) is not in the sequence" (e.g. a(4)=6 so 6+4=10 and 10 is not in the sequence) - Benoit Cloitre, Mar 31 2006
Write A for A000201 and B for A001950 (the upper Wythoff sequence, complement of A). Then the composite sequences AA, AB, BA, BB, AAA, AAB,...,BBB,... appear in many complementary equations having solution A000201 (or equivalently, A001950). Typical complementary equations: AB=A+B (=A003623), BB=A+2B (=A101864), BBB=3A+5B (=A134864). - Clark Kimberling, Nov 14 2007
Cumulative sum of A001468 terms. - Eric Angelini, Aug 19 2008
The lower Wythoff sequence also can be constructed by playing the so-called Mancala-game: n piles of total d(n) chips are standing in a row. The piles are numbered from left to right by 1, 2, 3, ... . The number of chips in a pile at the beginning of the game is equal to the number of the pile. One step of the game is described as follows: Distribute the pile on the very left one by one to the piles right of it. If chips are remaining, build piles out of one chip subsequently to the right. After f(n) steps the game ends in a constant row of piles. The lower Wythoff sequence is also given by n -> f(n). - Roland Schroeder (florola(AT)gmx.de), Jun 19 2010
With the exception of the first term, a(n) gives the number of iterations required to reverse the list {1,2,3,...,n} when using the mapping defined as follows: remove the first term of the list, z(1), and add 1 to each of the next z(1) terms (appending 1's if necessary) to get a new list. See A183110 where this mapping is used and other references given. This appears to be essentially the Mancala-type game interpretation given by R. Schroeder above. - John W. Layman, Feb 03 2011
Also row numbers of A213676 starting with an even number of zeros. - Reinhard Zumkeller, Mar 10 2013
From Jianing Song, Aug 18 2022: (Start)
Numbers k such that {k*phi} > phi^(-2), where {} denotes the fractional part.
Proof: Write m = floor(k*phi).
If {k*phi} > phi^(-2), take s = m-k+1. From m < k*phi < m+1 we have k < (m-k+1)*phi < k + phi, so floor(s*phi) = k or k+1. If floor(s*phi) = k+1, then (see A003622) floor((k+1)*phi) = floor(floor(s*phi)*phi) = floor(s*phi^2)-1 = s+floor(s*phi)-1 = m+1, but actually we have (k+1)*phi > m+phi+phi^(-2) = m+2, a contradiction. Hence floor(s*phi) = k.
If floor(s*phi) = k, suppose otherwise that k*phi - m <= phi^(-2), then m < (k+1)*phi <= m+2, so floor((k+1)*phi) = m+1. Suppose that A035513(p,q) = k for p,q >= 1, then A035513(p,q+1) = floor((k+1)*phi) - 1 = m = A035513(s,1). But it is impossible for one number (m) to occur twice in A035513. (End)
The formula from Jianing Song above is a direct consequence of an old result by Carlitz et al. (1972). Their Theorem 11 states that (a(n)) consists of the numbers k such that {k*phi^(-2)} < phi^(-1). One has {k*phi^(-2)} = {k*(2-phi)} = {-k*phi}. Using that 1-phi^(-1) = phi^(-2), the Jianing Song formula follows. - Michel Dekking, Oct 14 2023
In the Fokkink-Joshi paper, this sequence is the Cloitre (1,1,2,1)-hiccup sequence, i.e., a(1) = 1; for m < n, a(n) = a(n-1)+2 if a(m) = n-1, else a(n) = a(n-1)+1. - Michael De Vlieger, Jul 28 2025

Examples

			From Roland Schroeder (florola(AT)gmx.de), Jul 13 2010: (Start)
Example for n = 5; a(5) = 8;
(Start: [1,2,3,4,5]; 8 steps until [5,4,3,2,1]):
[1,2,3,4,5]; [3,3,4,5]; [4,5,6]; [6,7,1,1]; [8,2,2,1,1,1]: [3,3,2,2,2,1,1,1]; [4,3,3,2,1,1,1]; [4,4,3,2,1,1]; [5,4,3,2,1]. (End)
		

References

  • Eric Friedman, Scott M. Garrabrant, Ilona K. Phipps-Morgan, A. S. Landsberg and Urban Larsson, Geometric analysis of a generalized Wythoff game, in Games of no Chance 5, MSRI publ. Cambridge University Press, date?
  • M. Gardner, Penrose Tiles to Trapdoor Ciphers, W. H. Freeman, 1989; see p. 107.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • I. M. Yaglom, Two games with matchsticks, pp. 1-7 of Qvant Selecta: Combinatorics I, Amer Math. Soc., 2001.

Crossrefs

a(n) = least k such that s(k) = n, where s = A026242. Complement of A001950. See also A058066.
The permutation A002251 maps between this sequence and A001950, in that A002251(a(n)) = A001950(n), A002251(A001950(n)) = a(n).
First differences give A014675. a(n) = A022342(n) + 1 = A005206(n) + n + 1. a(2n)-a(n)=A007067(n). a(a(a(n)))-a(n) = A026274(n-1). - Benoit Cloitre, Mar 08 2003
A185615 gives values n such that n divides A000201(n)^m for some integer m>0.
Let A = A000201, B = A001950. Then AA = A003622, AB = A003623, BA = A035336, BB = A101864.
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A000201 as the parent: A000201, A001030, A001468, A001950, A003622, A003842, A003849, A004641, A005614, A014675, A022342, A088462, A096270, A114986, A124841. - N. J. A. Sloane, Mar 11 2021
Bisections: A276854, A342279.

Programs

  • Haskell
    a000201 n = a000201_list !! (n-1)
    a000201_list = f [1..] [1..] where
       f (x:xs) (y:ys) = y : f xs (delete (x + y) ys)
    -- Reinhard Zumkeller, Jul 02 2015, Mar 10 2013
    
  • Maple
    Digits := 100; t := evalf((1+sqrt(5))/2); A000201 := n->floor(t*n);
  • Mathematica
    Table[Floor[N[n*(1+Sqrt[5])/2]], {n, 1, 75}]
    Array[ Floor[ #*GoldenRatio] &, 68] (* Robert G. Wilson v, Apr 17 2010 *)
  • Maxima
    makelist(floor(n*(1+sqrt(5))/2),n,1,60); /* Martin Ettl, Oct 17 2012 */
    
  • PARI
    a(n)=floor(n*(sqrt(5)+1)/2)
    
  • PARI
    a(n)=(n+sqrtint(5*n^2))\2 \\ Charles R Greathouse IV, Feb 07 2013
    
  • Python
    def aupton(terms):
      alst, aset = [None, 1], {1}
      for n in range(1, terms):
        an = alst[n] + (1 if n not in aset else 2)
        alst.append(an); aset.add(an)
      return alst[1:]
    print(aupton(68)) # Michael S. Branicky, May 14 2021
    
  • Python
    from math import isqrt
    def A000201(n): return (n+isqrt(5*n**2))//2 # Chai Wah Wu, Jan 11 2022

Formula

Zeckendorf expansion of n (cf. A035517) ends with an even number of 0's.
Other properties: a(1)=1; for n>1, a(n) is taken to be the smallest integer greater than a(n-1) which is consistent with the condition "n is in the sequence if and only if a(n)+1 is not in the sequence".
a(1) = 1; for n>0, a(n+1) = a(n)+1 if n is not in the sequence, a(n+1) = a(n)+2 if n is in the sequence.
a(a(n)) = floor(n*phi^2) - 1 = A003622(n).
{a(k)} union {a(k)+1} = {1, 2, 3, 4, ...}. Hence a(1) = 1; for n>1, a(a(n)) = a(a(n)-1)+2, a(a(n)+1) = a(a(n))+1. - Benoit Cloitre, Mar 08 2003
{a(n)} is a solution to the recurrence a(a(n)+n) = 2*a(n)+n, a(1)=1 (see Barbeau et al.).
a(n) = A001950(n) - n. - Philippe Deléham, May 02 2004
a(0) = 0; a(n) = n + Max_{k : a(k) < n}. - Vladeta Jovovic, Jun 11 2004
a(Fibonacci(r-1)+j) = Fibonacci(r)+a(j) for 0 < j <= Fibonacci(r-2); 2 < r. - Paul Weisenhorn, Aug 18 2012
With 1 < k and A001950(k-1) < n <= A001950(k): a(n) = 2*n-k; A001950(n) = 3*n-k. - Paul Weisenhorn, Aug 21 2012

A001950 Upper Wythoff sequence (a Beatty sequence): a(n) = floor(n*phi^2), where phi = (1+sqrt(5))/2.

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
Offset: 1

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Author

Keywords

Comments

Indices at which blocks (1;0) occur in infinite Fibonacci word; i.e., n such that A005614(n-2) = 0 and A005614(n-1) = 1. - Benoit Cloitre, Nov 15 2003
A000201 and this sequence may be defined as follows: Consider the maps a -> ab, b -> a, starting from a(1) = a; then A000201 gives the indices of a, A001950 gives the indices of b. The sequence of letters in the infinite word begins a, b, a, a, b, a, b, a, a, b, a, ... Setting a = 0, b = 1 gives A003849 (offset 0); setting a = 1, b = 0 gives A005614 (offset 0). - Philippe Deléham, Feb 20 2004
a(n) = n-th integer which is not equal to the floor of any multiple of phi, where phi = (1+sqrt(5))/2 = golden number. - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), May 09 2007
Write A for A000201 and B for the present sequence (the upper Wythoff sequence, complement of A). Then the composite sequences AA, AB, BA, BB, AAA, AAB, ..., BBB, ... appear in many complementary equations having solution A000201 (or equivalently, the present sequence). Typical complementary equations: AB=A+B (=A003623), BB=A+2B (=A101864), BBB=3A+5B (=A134864). - Clark Kimberling, Nov 14 2007
Apart from the initial 0 in A090909, is this the same as that sequence? - Alec Mihailovs (alec(AT)mihailovs.com), Jul 23 2007
If we define a base-phi integer as a positive number whose representation in the golden ratio base consists only of nonnegative powers of phi, and if these base-phi integers are ordered in increasing order (beginning 1, phi, ...), then it appears that the difference between the n-th and (n-1)-th base-phi integer is phi-1 if and only if n belongs to this sequence, and the difference is 1 otherwise. Further, if each base-phi integer is written in linear form as a + b*phi (for example, phi^2 is written as 1 + phi), then it appears that there are exactly two base-phi integers with b=n if and only if n belongs to this sequence, and exactly three base-phi integers with b=n otherwise. - Geoffrey Caveney, Apr 17 2014
Numbers with an odd number of trailing zeros in their Zeckendorf representation (A014417). - Amiram Eldar, Feb 26 2021
Numbers missing from A066096. - Philippe Deléham, Jan 19 2023

Examples

			From _Paul Weisenhorn_, Aug 18 2012 and Aug 21 2012: (Start)
a(14) = floor(14*phi^2) = 36; a'(14) = floor(14*phi)=22;
with r=9 and j=1: a(13+1) = 34 + 2 = 36;
with r=8 and j=1: a'(13+1) = 21 + 1 = 22.
k=6 and a(5)=13 < n <= a(6)=15
a(14) = 3*14 - 6 = 36; a'(14) = 2*14 - 6 = 22;
a(15) = 3*15 - 6 = 39; a'(15) = 2*15 - 6 = 24. (End)
		

References

  • Claude Berge, Graphs and Hypergraphs, North-Holland, 1973; p. 324, Problem 2.
  • Eric Friedman, Scott M. Garrabrant, Ilona K. Phipps-Morgan, A. S. Landsberg and Urban Larsson, Geometric analysis of a generalized Wythoff game, in Games of no Chance 5, MSRI publ. Cambridge University Press, 2019.
  • Martin Gardner, Penrose Tiles to Trapdoor Ciphers, W. H. Freeman, 1989; see p. 107.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • I. M. Yaglom, Two games with matchsticks, pp. 1-7 of Qvant Selecta: Combinatorics I, Amer Math. Soc., 2001.

Crossrefs

a(n) = greatest k such that s(k) = n, where s = A026242.
Complement of A000201 or A066096.
A002251 maps between A000201 and A001950, in that A002251(A000201(n)) = A001950(n), A002251(A001950(n)) = A000201(n).
Let A = A000201, B = A001950. Then AA = A003622, AB = A003623, BA = A035336, BB = A101864.
First differences give (essentially) A076662.
Bisections: A001962, A001966.
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A000201 as the parent: A000201, A001030, A001468, A001950, A003622, A003842, A003849, A004641, A005614, A014675, A022342, A088462, A096270, A114986, A124841. - N. J. A. Sloane, Mar 11 2021

Programs

  • Haskell
    a001950 n = a000201 n + n  -- Reinhard Zumkeller, Mar 10 2013
    
  • Magma
    [Floor(n*((1+Sqrt(5))/2)^2): n in [1..80]]; // Vincenzo Librandi, Nov 19 2016
    
  • Maple
    A001950 := proc(n)
        floor(n*(3+sqrt(5))/2) ;
    end proc:
    seq(A001950(n),n=0..40) ; # R. J. Mathar, Jul 16 2024
  • Mathematica
    Table[Floor[N[n*(1+Sqrt[5])^2/4]], {n, 1, 75}]
    Array[ Floor[ #*GoldenRatio^2] &, 60] (* Robert G. Wilson v, Apr 17 2010 *)
  • PARI
    a(n)=floor(n*(sqrt(5)+3)/2)
    
  • PARI
    A001950(n)=(sqrtint(n^2*5)+n*3)\2 \\ M. F. Hasler, Sep 17 2014
    
  • Python
    from math import isqrt
    def A001950(n): return (n+isqrt(5*n**2)>>1)+n # Chai Wah Wu, Aug 10 2022

Formula

a(n) = n + floor(n*phi). In general, floor(n*phi^m) = Fibonacci(m-1)*n + floor(Fibonacci(m)*n*phi). - Benoit Cloitre, Mar 18 2003
a(n) = n + floor(n*phi) = n + A000201(n). - Paul Weisenhorn and Philippe Deléham
Append a 0 to the Zeckendorf expansion (cf. A035517) of n-th term of A000201.
a(n) = A003622(n) + 1. - Philippe Deléham, Apr 30 2004
a(n) = Min(m: A134409(m) = A006336(n)). - Reinhard Zumkeller, Oct 24 2007
If a'=A000201 is the ordered complement (in N) of {a(n)}, then a(Fib(r-2) + j) = Fib(r) + a(j) for 0 < j <= Fib(r-2), 3 < r; and a'(Fib(r-1) + j) = Fib(r) + a'(j) for 0 < j <= Fib(r-2), 2 < r. - Paul Weisenhorn, Aug 18 2012
With a(1)=2, a(2)=5, a'(1)=1, a'(2)=3 and 1 < k and a(k-1) < n <= a(k) one gets a(n)=3*n-k, a'(n)=2*n-k. - Paul Weisenhorn, Aug 21 2012

Extensions

Corrected by Michael Somos, Jun 07 2000

A026272 a(n) = smallest k such that k=a(n-k-1) is the only appearance of k so far; if there is no such k, then a(n) = least positive integer that has not yet appeared.

Original entry on oeis.org

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

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Author

Keywords

Comments

From Daniel Joyce, Apr 13 2001: (Start)
This sequence displays every positive integer exactly twice and the gap between the two occurrences of n contains exactly n other values. The first occurrence of n precedes the first occurrence of n+1.
Also related to the Wythoff array (A035513) and the Para-Fibonacci sequence (A035513) where every positive integer is displayed exactly once in the whole array. Take any integer n in A026272 and let C = number of terms from the beginning of the sequence to the second occurrence of n. Then C = (2nd term after n in the applicable sequence for n in A035513).
Also in the second occurrence of n in A026272, let N=n ( - one term) = (first term value after n in the applicable sequence for n in A035513). In this format the second occurrence of n in A026272 will produce in A035513, n itself and two of the succeeding terms of n in the Wythoff array where every positive integer can only be displayed once.
In A026272 if |a(n)-a(n+1)| > 10 then phi ~ a(n)/|a(n)-a(n+1)|. When n -> infinity it will converge to phi. (End)
Or, put a copy of n in A000027 n places further along! - Zak Seidov, May 24 2008
Another version would prefix this sequence with two leading 0's (see the Angelini reference). If we use this form and write down the indices of the two 0's, the two 1's, the two 2's, the two 3's, etc., then we get A072061. - Jacques ALARDET, Jul 26 2008

References

  • Eric Angelini, "Jeux de suites", in Dossier Pour La Science, pp. 32-35, Volume 59 (Jeux math'), April/June 2008, Paris.

Crossrefs

Programs

  • Mathematica
    s=Range[1000];n=0;Do[n++;s=Insert[s,n,Position[s,n][[1]]+n+1],{500}];A026272=Take[s,1000] (* Zak Seidov, May 24 2008 *)
  • PARI
    A026272=apply(t->t-1,A026242[3..-1]) \\ Use vecextract(A026242,"3..") in PARI versions < 2.7. - M. F. Hasler, Sep 17 2014
    
  • Python
    from collections import Counter
    from itertools import count, islice
    def agen(): # generator of terms
        aset, alst, k, mink, counts = set(), [0], 0, 1, Counter()
        for n in count(1):
            for k in range(1, len(alst)-1):
                if k == alst[n-k-1] and counts[alst[n-k-1]] == 1: an = k; break
            else: an = mink
            yield an; aset.add(an); alst.append(an); counts.update([an])
            while mink in aset: mink += 1
    print(list(islice(agen(), 66))) # Michael S. Branicky, Jun 27 2022

Formula

a(n) = A026242(n+2) - 1 = A026350(n+3) - 2 = A026354(n+4) - 3.

Extensions

Edited by Max Alekseyev, May 31 2011

A026350 a(n) = a(m) if a(m) has already occurred exactly once and n = a(m)+m-1, else a(n) = least positive integer that has not yet occurred.

Original entry on oeis.org

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

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Keywords

Crossrefs

Cf. A026272.

Programs

Formula

a(n+1)=A026242(n)+1, for all n>0. - M. F. Hasler, Sep 17 2014

A026354 a(n) = a(m) if a(m) has already occurred exactly once and n = a(m)+m-2, else a(n) = least positive integer that has not yet occurred.

Original entry on oeis.org

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

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Keywords

Crossrefs

Cf. A026272.

Programs

Formula

a(n)=n for n<4, a(n)=A026242(n-2)+2 for all n>2, a(n)=A026272(n-4)+3 for n>4. - M. F. Hasler, Sep 17 2014
There is a 35-state automaton that accepts, in parallel, the Fibonacci (Zeckendorf) representation of n and a(n). See the file a026354.pdf. - Jeffrey Shallit, Dec 21 2023

A307664 Lexicographically earliest sequence of positive terms, such that each value, say v, appears three times at indices k, k + v and k + 2*v for some k.

Original entry on oeis.org

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

Views

Author

Rémy Sigrist, Apr 20 2019

Keywords

Comments

This sequence is a variant of A026242 where each value, say v, appears twice at indices k and k + v for some k.
Graphically, we have three lines.
Apparently every positive integer appears in the sequence.

Examples

			For n = 1:
- we can set a(1) = a(2) = a(3) = 1.
For n = 4:
- we can set a(4) = a(6) = a(8) = 2.
For n = 5:
- a(8) is already known, hence a(5) <> 3,
- we can set a(5) = a(9) = a(13) = 4.
		

Crossrefs

See A307667 and A307669 for other variants.
Cf. A026242.

Programs

  • PARI
    See Links section.

A130526 A permutation of the integers induced by the lower and upper Wythoff sequences.

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Jun 02 2007

Keywords

Comments

Another permutation of the integers is -A130526.

Examples

			6=L(4), so a(6)=4.
7=U(3), so a(7)=-3.
		

Crossrefs

Formula

a(0) = 0; a(n)=k if n=L(k); a(n)=-k if n=U(k), where L(k) = A000201(k)=Floor(n*phi) = Lower Wythoff sequence, U(k) = A001950(k) = k+L(k) = Upper Wythoff sequence.

A307667 Lexicographically earliest sequence of positive terms, such that each value, say v, appears four times at indices k, k + v, k + 2*v and k + 3*v for some k.

Original entry on oeis.org

1, 1, 1, 1, 2, 4, 2, 7, 2, 4, 2, 8, 6, 4, 7, 11, 9, 4, 6, 8, 12, 7, 14, 15, 6, 9, 11, 8, 7, 10, 6, 16, 12, 18, 9, 8, 14, 11, 15, 10, 17, 13, 20, 9, 12, 25, 19, 16, 11, 10, 14, 18, 3, 15, 13, 3, 12, 17, 3, 10, 23, 3, 20, 16, 14, 19, 5, 13, 15, 18, 25, 5, 21, 24
Offset: 1

Views

Author

Rémy Sigrist, Apr 20 2019

Keywords

Comments

This sequence is a variant of A307664.
Graphically, we have four wavy lines.
Apparently every positive integer appears in the sequence.

Examples

			For n = 1:
- we can set a(1) = a(2) = a(3) = a(4) = 1.
For n = 5:
- we can set a(5) = a(7) = a(9) = a(11) = 2.
For n = 6:
- a(9) is already known, hence a(6) <> 3,
- we can set a(6) = a(10) = a(14) = a(18) = 4.
		

Crossrefs

Programs

  • PARI
    See Links section.
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