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 1-6 of 6 results.

A384947 Positive integers m for which A183136(m) != f(m), where f(m) = floor( (m*(m+1)/2)/phi - m/2 + 1/(2*phi) ) and phi = (1+sqrt(5))/2 is the golden ratio.

Original entry on oeis.org

15, 18, 36, 39, 41, 47, 49, 52, 91, 94, 96, 102, 103, 104, 107, 109, 123, 125, 128, 130, 136, 138, 141, 235, 238, 240, 246, 247, 248, 251, 252, 253, 267, 268, 269, 272, 273, 274, 277, 280, 281, 282, 285, 287, 303, 306, 322, 324, 327, 328
Offset: 1

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Author

Hoang Xuan Thanh, Jun 13 2025

Keywords

Comments

f(m) is an approximation to A183136(m) = Sum_{k=1..m} floor(k/phi) based on assuming the floor in each term decreases it by 1/2 from what is otherwise a triangular sum; and further offset + 1/(2*phi) in f(m) chosen to improve the accuracy of this approximation.
The actual values of frac(k/phi) can differ from 1/2 each by a net amount which is enough to make m a term of this sequence.

Examples

			41 is term, because A183136(41) = 512 != 511 = floor(((41^2+1)*phi - 41) / (2*phi^2)).
		

Crossrefs

Programs

  • Mathematica
    PositionIndex[MapIndexed[# != Floor[PolygonalNumber[#2[[1]]]/GoldenRatio - #2[[1]]/2 + 1/(2*GoldenRatio)] &, Accumulate[Floor[Range[500]/GoldenRatio]]]][True] (* Paolo Xausa, Jun 20 2025 *)

A005614 The binary complement of the infinite Fibonacci word A003849. Start with 1, apply 0->1, 1->10, iterate, take limit.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0
Offset: 0

Views

Author

Keywords

Comments

Previous name was: The infinite Fibonacci word (start with 1, apply 0->1, 1->10, iterate, take limit).
Characteristic function of A022342. - Philippe Deléham, May 03 2004
a(n) = number of 0's between successive 1's (see also A003589 and A007538). - Eric Angelini, Jul 06 2005
With offset 1 this is the characteristic sequence for Wythoff A-numbers A000201=[1,3,4,6,...].
Eric Angelini's comment made me think that if 1 is defined to be the number of 0's between successive 1's in a string of 0's and 1's, then this string is 101. Applying the same operation to the digits of 101 leads to 101101, the iteration leads to successive palindromes of lengths given by A001911, up to a(n). - Rémi Schulz, Jul 06 2010
For generalized Fibonacci words see A221150, A221151, A221152, ... - Peter Bala, Nov 11 2013
The limiting mean of the first n terms is phi - 1; the limiting variance is phi (A001622). - Clark Kimberling, Mar 12 2014
Apply the difference operator to every column of the Wythoff difference array, A080164, to get an array of Fibonacci numbers, F(h). Replace each F(h) with h, and apply the difference operator to every column. In the resulting array, every column is A005614. - Clark Kimberling, Mar 02 2015
Binary expansion of the rabbit constant A014565. - M. F. Hasler, Nov 10 2018

Examples

			The infinite word is 101101011011010110101101101011...
		

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003.
  • G. Melançon, Factorizing infinite words using Maple, MapleTech journal, vol. 4, no. 1, 1997, pp. 34-42, esp. p. 36.

Crossrefs

Binary complement of A003849, which is the standard form of this sequence.
Two other essentially identical sequences are A096270, A114986.
Subwords: A178992, A171676.
Cf. A000045 (Fibonacci numbers), A001468, A001911, A005206 (partial sums), A014565, A014675, A022342, A036299, A044432, A221150, A221151, A221152.
Cf. A339051 (odd bisection), A339052 (even bisection).
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
    a005614 n = a005614_list !! n
    a005614_list = map (1 -) a003849_list
    -- Reinhard Zumkeller, Apr 07 2012
    
  • Magma
    [Floor((n+1)*(-1+Sqrt(5))/2)-Floor(n*(-1+Sqrt(5))/2): n in [1..100]]; // Vincenzo Librandi, Jan 17 2019
    
  • Maple
    Digits := 50; u := evalf((1-sqrt(5))/2); A005614 := n->floor((n+1)*u)-floor(n*u);
  • Mathematica
    Nest[ Flatten[ # /. {0 -> {1}, 1 -> {1, 0}}] &, {1}, 10] (* Robert G. Wilson v, Jan 30 2005 *)
    Flatten[Nest[{#, #[[1]]} &, {1, 0}, 9]] (* IWABUCHI Yu(u)ki, Oct 23 2013 *)
    SubstitutionSystem[{0 -> {1}, 1 -> {1, 0}}, {1, 0}, 9] // Last (* Jean-François Alcover, Feb 06 2020 *)
  • PARI
    a(n,w1,s0,s1)=local(w2); for(i=2,n,w2=[ ]; for(k=1,length(w1),w2=concat(w2, if(w1[ k ],s1,s0))); w1=w2); w2
    for(n=2,10,print(n" "a(n,[ 0 ],[ 1 ],[ 1,0 ]))) \\ Gives successive convergents to sequence
    
  • PARI
    /* for m>=1 compute exactly A183136(m+1)+1 terms of the sequence */
    r=(1+sqrt(5))/2;v=[1,0];for(n=2,m,v=concat(v,vector(floor((n+1)/r),i,v[i]));a(n)=v[n];) /* Benoit Cloitre, Jan 16 2013 */
    
  • Python
    from math import isqrt
    def A005614(n): return (n+isqrt(m:=5*(n+2)**2)>>1)-(n+1+isqrt(m-10*n-15)>>1) # Chai Wah Wu, Aug 17 2022

Formula

Define strings S(0)=1, S(1)=10, thereafter S(n)=S(n-1)S(n-2); iterate. Sequence is S(oo). The individual S(n)'s are given in A036299.
a(n) = floor((n+2)*u) - floor((n+1)*u), where u = (-1 + sqrt(5))/2.
Sum_{n>=0} a(n)/2^(n+1) = A014565. - R. J. Mathar, Jul 19 2013
From Peter Bala, Nov 11 2013: (Start)
If we read the present sequence as the digits of a decimal constant c = 0.101101011011010 ... then we have the series representation c = Sum_{n >= 1} 1/10^floor(n*phi). An alternative representation is c = Sum_{n >= 1} 1/10^floor(n/phi) - 10/9.
The constant 9*c has the simple continued fraction representation [0; 1, 10, 10, 100, 1000, ..., 10^Fibonacci(n), ...]. See A010100.
Using this result we can find the alternating series representation c = 1/9 - 9*Sum_{n >= 1} (-1)^(n+1)*(1 + 10^Fibonacci(3*n+1))/( (10^(Fibonacci(3*n - 1)) - 1)*(10^(Fibonacci(3*n + 2)) - 1) ). The series converges very rapidly: for example, the first 10 terms of the series give a value for c accurate to more than 5.7 million decimal places. Cf. A014565. (End)
a(n) = A005206(n+1) - A005206(n). a(2*n) = A339052(n); a(2*n+1) = A339051(n+1). - Peter Bala, Aug 09 2022

Extensions

Corrected by Clark Kimberling, Oct 04 2000
Name corrected by Michel Dekking, Apr 02 2019

A060143 a(n) = floor(n/tau), where tau = (1 + sqrt(5))/2.

Original entry on oeis.org

0, 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
Offset: 0

Views

Author

Clark Kimberling, Mar 05 2001

Keywords

Comments

Fibonacci base shift right: for n >= 0, a(n+1) = Sum_{k in A_n} F_{k-1}, where n = Sum_{k in A_n} F_k (unique) expression of n as a sum of "noncontiguous" Fibonacci numbers (with index >=2). - Michele Dondi (bik.mido(AT)tiscalenet.it), Dec 30 2001 [corrected, and aligned with sequence offset by Peter Munn, Jan 10 2018]
Numerators a(n) of fractions slowly converging to phi, the golden ratio: let a(1) = 0, b(n) = n - a(n); if (a(n) + 1) / b(n) < (1 + sqrt(5))/2, then a(n+1) = a(n) + 1, else a(n+1) = a(n). a(n) + b(n) = n and as n -> +infinity, a(n) / b(n) converges to (1 + sqrt(5))/2. For all n, a(n) / b(n) < (1 + sqrt(5))/2. - Robert A. Stump (bee_ess107(AT)msn.com), Sep 22 2002
a(10^n) gives the first few digits of phi=(sqrt(5)-1)/2.
Comment corrected, two alternative ways, by Peter Munn, Jan 10 2018: (Start)
(a(n) = a(n+1) or a(n) = a(n-1)) if and only if a(n) is in A066096.
a(n+1) = a(n+2) if and only if n is in A003622.
(End)
From Wolfdieter Lang, Jun 28 2011: (Start)
a(n+1) counts for n >= 1 the number of Wythoff A-numbers not exceeding n.
a(n+1) counts also the number of Wythoff B-numbers smaller than A(n+2), with the Wythoff A- and B-sequences A000201 and A001950, respectively.
a(n+1) = Sum_{j=1..n} A005614(j-1) for n >= 1 (no rounding problems like in the above definition, because the rabbit sequence A005614(n-1) for n >= 1, can be defined by a substitution rule).
a(n+1) = A(n+1)-(n+1) (serving, together with the last equation, as definition for A(n+1), given the rabbit sequence).
a(n+1) = A005206(n), n >= 0.
(End)
Let b(n) = floor((n+1)/phi). Then b(n) + b(b(n-1)) = n [Granville and Rasson]. - N. J. A. Sloane, Jun 13 2014

Examples

			a(6)= 3 so b(6) = 6 - 3 = 3. a(7) = 4 because (a(6) + 1) / b(6) = 4/3 which is < (1 + sqrt(5))/2. So b(7) = 7 - 4 = 3. a(8) = 4 because (a(7) + 1) / b(7) = 5/3 which is > (1 + sqrt(5))/2. - Robert A. Stump (bee_ess107(AT)msn.com), Sep 22 2002
From _Wolfdieter Lang_, Jun 28 2011: (Start)
There are a(4) = 2 (positive) Wythoff A-numbers <= 3, namely 1 and 3.
There are a(4) = 2 (positive) Wythoff B-numbers < A(4) = 6, namely 2 and 5.
a(4) = 2 = A(4) - 4 = 6 - 4.
(End)
		

Crossrefs

Cf. A000045 (Fibonacci numbers), A003622, A022342, A035336.
Terms that occur only once: A001950.
Terms that occur twice: A066096 (a version of A000201).
Numerator sequences for other values, as described in Robert A. Stump's 2002 comment: A074065 (sqrt(3)), A074840 (sqrt(2)).
Apart from initial terms, same as A005206.
First differences: A096270 (a version of A005614).
Partial sums: A183136.

Programs

  • Magma
    [Floor(2*n/(1+Sqrt(5))): n in [0..80]]; // Vincenzo Librandi, Mar 29 2015
    
  • Mathematica
    Floor[Range[0,80]/GoldenRatio] (* Harvey P. Dale, May 09 2013 *)
  • PARI
    { default(realprecision, 10); p=(sqrt(5) - 1)/2; for (n=0, 1000, write("b060143.txt", n, " ", floor(n*p)); ) } \\ Harry J. Smith, Jul 02 2009
    
  • Python
    from math import isqrt
    def A060143(n): return (n+isqrt(5*n**2)>>1)-n # Chai Wah Wu, Aug 10 2022

Formula

a(n) = floor(phi(n)), where phi=(sqrt(5)-1)/2. [corrected by Casey Mongoven, Jul 18 2008]
a(F_n + 1) = F_{n-1} if F_n is the n-th Fibonacci number. [aligned with sequence offset by Peter Munn, Jan 10 2018]
a(1) = 0. b(n) = n - a(n). If (a(n) + 1) / b(n) < (1 + sqrt(5))/2, then a(n+1) = a(n) + 1, else a(n+1) = a(n). - Robert A. Stump (bee_ess107(AT)msn.com), Sep 22 2002 [corrected by Peter Munn, Jan 07 2018]
A006336(n) = A006336(n-1) + A006336(a(n)) for n>1. - Reinhard Zumkeller, Oct 24 2007
a(n) = floor(n*phi) - n, where phi = (1+sqrt(5))/2. - William A. Tedeschi, Mar 06 2008
Celaya and Ruskey give an interesting formula for a(n). - N. J. A. Sloane, Jun 13 2014

Extensions

I merged three identical sequences to create this entry. Some of the formulas may need their initial terms adjusting now. - N. J. A. Sloane, Mar 05 2003
More terms from William A. Tedeschi, Mar 06 2008

A183143 [1/r]+[2/r]+...+[n/r], where r=sqrt(3) and []=floor.

Original entry on oeis.org

0, 1, 2, 4, 6, 9, 13, 17, 22, 27, 33, 39, 46, 54, 62, 71, 80, 90, 100, 111, 123, 135, 148, 161, 175, 190, 205, 221, 237, 254, 271, 289, 308, 327, 347, 367, 388, 409, 431, 454, 477, 501, 525, 550, 575, 601, 628, 655, 683, 711, 740, 770, 800, 831
Offset: 1

Views

Author

Clark Kimberling, Dec 26 2010

Keywords

Comments

A183143 + A183144 = A000217 (the triangular numbers).

Crossrefs

Programs

  • Maple
    A183143 := proc(n)
        local r;
        r := sqrt(3) ;
        add( floor(i/r),i=1..n) ;
    end proc:
    seq( A183143(n),n=1..200) ; # R. J. Mathar, Aug 28 2025
  • Mathematica
    Accumulate[Floor[Range[60]/Sqrt[3]]] (* Harvey P. Dale, Sep 10 2019 *)
  • PARI
    default(realprecision,100); r=sqrt(3); for(n=1,99,print1(sum(k=1,n,floor(k/r)),", "))

Formula

[1/r]+[2/r]+...+[n/r], where r=sqrt(3) and []=floor.

A183137 a(n) = [1/s] + [2/s] + ... + [n/s], where s = (golden ratio)^2 = (3+sqrt(5))/2 and [] = floor.

Original entry on oeis.org

0, 0, 1, 2, 3, 5, 7, 10, 13, 16, 20, 24, 28, 33, 38, 44, 50, 56, 63, 70, 78, 86, 94, 103, 112, 121, 131, 141, 152, 163, 174, 186, 198, 210, 223, 236, 250, 264, 278, 293, 308, 324, 340, 356, 373, 390, 407, 425, 443, 462, 481, 500, 520, 540, 561, 582, 603, 625
Offset: 1

Views

Author

Clark Kimberling, Dec 26 2010

Keywords

Comments

A183136(n) + a(n) = A000217(n+1) (the triangular numbers).

Examples

			a(7) = 7 = 0+0+1+1+1+2+2.
		

Crossrefs

Programs

  • Mathematica
    Accumulate[With[{c=GoldenRatio^2},Floor[Range[60]/c]]]  (* Harvey P. Dale, Apr 20 2011 *)

Formula

a(n+1) = a(n) + n - A005206(n). - John Furey, Jun 03 2015

A385177 a(n) = Sum_{k=1..n} ceiling(k/phi), where phi is the golden ratio (A001622).

Original entry on oeis.org

1, 3, 5, 8, 12, 16, 21, 26, 32, 39, 46, 54, 63, 72, 82, 92, 103, 115, 127, 140, 153, 167, 182, 197, 213, 230, 247, 265, 283, 302, 322, 342, 363, 385, 407, 430, 453, 477, 502, 527, 553, 579, 606, 634, 662, 691, 721, 751, 782, 813, 845, 878, 911, 945, 979, 1014, 1050, 1086, 1123
Offset: 1

Views

Author

Paolo Xausa, Jun 20 2025

Keywords

Crossrefs

Partial sums of A019446.

Programs

  • Mathematica
    Accumulate[Ceiling[Range[100]/GoldenRatio]]
  • Python
    from math import isqrt
    def A385177(n): return n+sum(isqrt(5*i**2)-i>>1 for i in range(1,n+1)) # Chai Wah Wu, Jun 20 2025

Formula

a(n) = A183136(n) + n.
Showing 1-6 of 6 results.