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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A101864 Wythoff BB numbers.

Original entry on oeis.org

5, 13, 18, 26, 34, 39, 47, 52, 60, 68, 73, 81, 89, 94, 102, 107, 115, 123, 128, 136, 141, 149, 157, 162, 170, 178, 183, 191, 196, 204, 212, 217, 225, 233, 238, 246, 251, 259, 267, 272, 280, 285, 293, 301, 306, 314, 322, 327, 335, 340, 348, 356, 361, 369, 374, 382, 390, 395
Offset: 1

Views

Author

N. J. A. Sloane, Jan 28 2005

Keywords

Comments

a(n)-3 are also the positions of 1 in A188436. - Federico Provvedi, Nov 22 2018
The asymptotic density of this sequence is 1/phi^4 = A094214^4 = 0.145898... . - Amiram Eldar, Mar 24 2025

Crossrefs

Second row of A101858.
Let A = A000201, B = A001950. Then AA = A003622, AB = A003623, BA = A035336, BB = A101864.

Programs

  • Maple
    b:=n->floor(n*((1+sqrt(5))/2)^2): seq(b(b(n)),n=1..60); # Muniru A Asiru, Dec 05 2018
  • Mathematica
    b[n_] := Floor[n * GoldenRatio^2]; a[n_] := b[b[n]]; Array[a, 60] (* Amiram Eldar, Nov 22 2018 *)
  • Python
    from sympy import S
    for n in range(1,60): print(int(S.GoldenRatio**2*(int(n*S.GoldenRatio**2))), end=', ') # Stefano Spezia, Dec 06 2018

Formula

a(n) = B(B(n)), n>=1, with B(k)=A001950(k) (Wythoff B-numbers). a(0)=0 with B(0)=0.

A035337 Third column of Wythoff array.

Original entry on oeis.org

3, 11, 16, 24, 32, 37, 45, 50, 58, 66, 71, 79, 87, 92, 100, 105, 113, 121, 126, 134, 139, 147, 155, 160, 168, 176, 181, 189, 194, 202, 210, 215, 223, 231, 236, 244, 249, 257, 265, 270, 278, 283, 291, 299, 304, 312
Offset: 0

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Author

Keywords

Comments

Also, positions of 3's in A139764, the smallest term in Zeckendorf representation of n. - John W. Layman, Aug 25 2011
The formula a(n) = 3*A003622(n)-n+1 = 3AA(n)-n+1 conjectured by Layman below is correct, since it is well known that AA(n)+1 = B(n) = A(n)+n, where B = A001950, and so 3AA(n)-n+1 = 3B(n)-n-2 = 3A(n)+2n-2. - Michel Dekking, Aug 31 2017
From Amiram Eldar, Mar 21 2022: (Start)
Numbers k for which the Zeckendorf representation A014417(k) ends with 1, 0, 0.
The asymptotic density of this sequence is 1/phi^4 = 2/(7+3*sqrt(5)), where phi is the golden ratio (A001622). (End)

Crossrefs

Let A = A000201, B = A001950. Then AA = A003622, AB = A003623, BA = A035336, BB = A101864. The eight triples AAA, AAB, ..., BBB are A134859, A134860, A035337, A134862, A134861, A134863, A035338, A134864, resp.

Programs

  • Maple
    t := (1+sqrt(5))/2 ; [ seq(3*floor((n+1)*t)+2*n,n=0..80) ];
  • Mathematica
    Table[3 Floor[n GoldenRatio] + 2 n - 2, {n, 46}] (* Michael De Vlieger, Aug 31 2017 *)
  • PARI
    a(n) = 2*n + 3*floor((1+sqrt(5))*(n+1)/2); \\ Altug Alkan, Sep 18 2017
  • Python
    from sympy import floor
    from mpmath import phi
    def a(n): return 3*floor((n + 1)*phi) + 2*n # Indranil Ghosh, Jun 10 2017
    
  • Python
    from math import isqrt
    def A035337(n): return 3*(n+isqrt(5*n**2)>>1)+(n-1<<1) # Chai Wah Wu, Aug 11 2022
    

Formula

a(n) = F(4)A(n)+F(3)(n-1) = 3A(n)+2n-2, where A = A000201 and F = A000045. - Michel Dekking, Aug 31 2017
It appears that a(n) = 3*A003622(n) - n + 1. - John W. Layman, Aug 25 2011

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

A035338 4th column of Wythoff array.

Original entry on oeis.org

5, 18, 26, 39, 52, 60, 73, 81, 94, 107, 115, 128, 141, 149, 162, 170, 183, 196, 204, 217, 225, 238, 251, 259, 272, 285, 293, 306, 314, 327, 340, 348, 361, 374, 382, 395, 403, 416, 429, 437, 450, 458, 471, 484, 492, 505, 518, 526, 539, 547, 560, 573, 581, 594
Offset: 0

Views

Author

Keywords

Comments

The asymptotic density of this sequence is 1/phi^5 = phi^5 - 11 = A244593 - 4 = 0.0901699... . - Amiram Eldar, Mar 24 2025

Crossrefs

Let A = A000201, B = A001950. Then AA = A003622, AB = A003623, BA = A035336, BB = A101864. The eight triples AAA, AAB, ..., BBB are A134859, A134860, A035337, A134862, A134861, A134863, A035338, A134864, resp.

Programs

  • Maple
    t := (1+sqrt(5))/2 ; [ seq(5*floor((n+1)*t)+3*n,n=0..80) ];
  • Mathematica
    f[n_] := 5 Floor[(n + 1) GoldenRatio] + 3n; Array[f, 54, 0] (* Robert G. Wilson v, Dec 11 2017 *)
  • Python
    from math import isqrt
    def A035338(n): return 5*(n+1+isqrt(5*(n+1)**2)>>1)+3*n # Chai Wah Wu, Aug 11 2022

A134859 Wythoff AAA numbers.

Original entry on oeis.org

1, 6, 9, 14, 19, 22, 27, 30, 35, 40, 43, 48, 53, 56, 61, 64, 69, 74, 77, 82, 85, 90, 95, 98, 103, 108, 111, 116, 119, 124, 129, 132, 137, 142, 145, 150, 153, 158, 163, 166, 171, 174, 179, 184, 187, 192, 197, 200, 205, 208, 213, 218, 221, 226, 229, 234, 239, 242
Offset: 1

Views

Author

Clark Kimberling, Nov 14 2007

Keywords

Comments

The lower and upper Wythoff sequences, A and B, satisfy the complementary equations AAA = AB - 2 and AAA = A + B - 2.
Also numbers with suffix string 001, when written in Zeckendorf representation (with leading zero for the first term). - A.H.M. Smeets, Mar 20 2024
The asymptotic density of this sequence is 1/phi^3 = phi^3 - 4 = A098317 - 4 = 0.236067... . - Amiram Eldar, Mar 24 2025

Examples

			Starting with A=(1,3,4,6,8,9,11,12,14,16,17,19,...), we have A(2)=3, so A(A(2))=4, so A(A(A(2)))=6.
		

Crossrefs

Let A = A000201, B = A001950. Then AA = A003622, AB = A003623, BA = A035336, BB = A101864. The eight triples AAA, AAB, ..., BBB are A134859, A134860, A035337, A134862, A134861, A134863, A035338, A134864, resp.
Essentially the same as A095098.

Programs

  • Maple
    # For Maple code for these Wythoff compound sequences see A003622. - N. J. A. Sloane, Mar 30 2016
  • Mathematica
    A[n_] := Floor[n GoldenRatio];
    a[n_] := A@ A@ A@ n;
    a /@ Range[100] (* Jean-François Alcover, Oct 28 2019 *)
  • Python
    from sympy import floor
    from mpmath import phi
    def A(n): return floor(n*phi)
    def a(n): return A(A(A(n))) # Indranil Ghosh, Jun 10 2017
    
  • Python
    from math import isqrt
    def A134859(n): return ((n+isqrt(5*n**2)>>1)-1<<1)+n # Chai Wah Wu, Aug 10 2022

Formula

a(n) = A(A(A(n))), n >= 1, with A=A000201, the lower Wythoff sequence.
a(n) = 2*floor(n*Phi^2) - n - 2 where Phi = (1+sqrt(5))/2. - Benoit Cloitre, Apr 12 2008; R. J. Mathar, Oct 16 2009
a(n) = A095098(n-1), n > 1. - R. J. Mathar, Oct 16 2009
From A.H.M. Smeets, Mar 23 2024: (Start)
a(n) = A(n) + B(n) - 2 (see Clark Kimberling 2008), with A=A000201, B=A001950, the lower and upper Wythoff sequences, respectively.
Equals {A003622}\{A134860} (= Wythoff AA \ Wythoff AAB). (End)

Extensions

Incorrect PARI program removed by R. J. Mathar, Oct 16 2009

A134860 Wythoff AAB numbers; also, Fib101 numbers: those n for which the Zeckendorf expansion A014417(n) ends with 1,0,1.

Original entry on oeis.org

4, 12, 17, 25, 33, 38, 46, 51, 59, 67, 72, 80, 88, 93, 101, 106, 114, 122, 127, 135, 140, 148, 156, 161, 169, 177, 182, 190, 195, 203, 211, 216, 224, 232, 237, 245, 250, 258, 266, 271, 279, 284, 292, 300, 305, 313, 321, 326, 334, 339, 347, 355, 360, 368, 373
Offset: 1

Views

Author

Antti Karttunen, Jun 01 2004 and Clark Kimberling, Nov 14 2007

Keywords

Comments

The lower and upper Wythoff sequences, A and B, satisfy the complementary equations AAB=AA+AB and AAB=A+2B-1.
The asymptotic density of this sequence is 1/phi^4 = 2/(7+3*sqrt(5)), where phi is the golden ratio (A001622). - Amiram Eldar, Mar 21 2022

Crossrefs

Let A = A000201, B = A001950. Then AA = A003622, AB = A003623, BA = A035336, BB = A101864. The eight triples AAA, AAB, ..., BBB are A134859, A134860, A035337, A134862, A134861, A134863, A035338, A134864, resp.
Set-wise difference A003622 \ A095098. Cf. A095089 (fib101 primes).

Programs

  • Mathematica
    With[{r = Map[Fibonacci, Range[2, 14]]}, Position[#, {1, 0, 1}][[All, 1]] &@ Table[If[Length@ # < 3, {}, Take[#, -3]] &@ IntegerDigits@ Total@ Map[FromDigits@ PadRight[{1}, Flatten@ #] &@ Reverse@ Position[r, #] &,Abs@ Differences@ NestWhileList[Function[k, k - SelectFirst[Reverse@ r, # < k &]], n + 1, # > 1 &]], {n, 373}]] (* Michael De Vlieger, Jun 09 2017 *)
  • Python
    from sympy import fibonacci
    def a(n):
        x=0
        while n>0:
            k=0
            while fibonacci(k)<=n: k+=1
            x+=10**(k - 3)
            n-=fibonacci(k - 1)
        return x
    def ok(n): return str(a(n))[-3:]=="101"
    print([n for n in range(4, 501) if ok(n)]) # Indranil Ghosh, Jun 08 2017
    
  • Python
    from math import isqrt
    def A134860(n): return 3*(n+isqrt(5*n**2)>>1)+(n<<1)-1 # Chai Wah Wu, Aug 10 2022

Formula

a(n) = A(A(B(n))), n>=1, with A=A000201, the lower Wythoff sequence and B=A001950, the upper Wythoff sequence.

Extensions

This is the result of merging two sequences which were really the same. - N. J. A. Sloane, Jun 10 2017

A219641 a(n) = n minus (number of 1's in Zeckendorf expansion of n).

Original entry on oeis.org

0, 0, 1, 2, 2, 4, 4, 5, 7, 7, 8, 9, 9, 12, 12, 13, 14, 14, 16, 16, 17, 20, 20, 21, 22, 22, 24, 24, 25, 27, 27, 28, 29, 29, 33, 33, 34, 35, 35, 37, 37, 38, 40, 40, 41, 42, 42, 45, 45, 46, 47, 47, 49, 49, 50, 54, 54, 55, 56, 56, 58, 58, 59, 61, 61, 62, 63, 63, 66
Offset: 0

Views

Author

Antti Karttunen, Nov 24 2012

Keywords

Comments

See A014417 for the Fibonacci number system representation, also known as Zeckendorf expansion.

Crossrefs

Cf. A007895, A014417. A022342 gives the positions of records, resulting the same sequence with duplicates removed: A219640. A035336 gives the positions of values that occur only once: A219639. Cf. also A219637, A219642. Analogous sequence for binary system: A011371, for factorial number system: A219651.

Programs

  • Mathematica
    zeck = DigitCount[Select[Range[0, 500], BitAnd[#, 2*#] == 0&], 2, 1];
    Range[0, Length[zeck]-1] - zeck (* Jean-François Alcover, Jan 25 2018 *)
  • Python
    from sympy import fibonacci
    def a(n):
        k=0
        x=0
        while n>0:
            k=0
            while fibonacci(k)<=n: k+=1
            x+=10**(k - 3)
            n-=fibonacci(k - 1)
        return str(x).count("1")
    print([n - a(n) for n in range(101)]) # Indranil Ghosh, Jun 09 2017
  • Scheme
    (define (A219641 n) (- n (A007895 n)))
    

Formula

a(n) = n - A007895(n).

A134861 Wythoff BAA numbers.

Original entry on oeis.org

2, 10, 15, 23, 31, 36, 44, 49, 57, 65, 70, 78, 86, 91, 99, 104, 112, 120, 125, 133, 138, 146, 154, 159, 167, 175, 180, 188, 193, 201, 209, 214, 222, 230, 235, 243, 248, 256, 264, 269, 277, 282, 290, 298, 303, 311, 319, 324, 332, 337, 345, 353, 358, 366, 371
Offset: 1

Views

Author

Clark Kimberling, Nov 14 2007

Keywords

Comments

The lower and upper Wythoff sequences, A and B, satisfy the complementary equation BAA = A+2B-3.
Also numbers with suffix string 0010, when written in Zeckendorf representation (with leading zero's for the first term). - A.H.M. Smeets, Mar 20 2024
The asymptotic density of this sequence is 1/phi^4 = A094214^4 = 0.145898... . - Amiram Eldar, Mar 24 2025

Crossrefs

Let A = A000201, B = A001950. Then AA = A003622, AB = A003623, BA = A035336, BB = A101864. The eight triples AAA, AAB, ..., BBB are A134859, A134860, A035337, A134862, A134861, A134863, A035338, A134864, resp.

Programs

  • Mathematica
    A[n_] := Floor[n * GoldenRatio]; B[n_] := Floor[n * GoldenRatio^2]; a[n_] := B[A[A[n]]]; Array[a, 100] (* Amiram Eldar, Mar 24 2025 *)
  • Python
    from sympy import floor
    from mpmath import phi
    def A(n): return floor(n*phi)
    def B(n): return floor(n*phi**2)
    def a(n): return B(A(A(n))) # Indranil Ghosh, Jun 10 2017
    
  • Python
    from math import isqrt
    def A134861(n): return 3*((n+isqrt(5*n**2)>>1)-1)+(n<<1) # Chai Wah Wu, Aug 10 2022

Formula

a(n) = B(A(A(n))), n>=1, with A=A000201, the lower Wythoff sequence and B=A001950, the upper Wythoff sequence.

A134862 Wythoff ABB numbers.

Original entry on oeis.org

8, 21, 29, 42, 55, 63, 76, 84, 97, 110, 118, 131, 144, 152, 165, 173, 186, 199, 207, 220, 228, 241, 254, 262, 275, 288, 296, 309, 317, 330, 343, 351, 364, 377, 385, 398, 406, 419, 432, 440, 453, 461, 474, 487, 495, 508, 521, 529, 542, 550, 563, 576, 584, 597
Offset: 1

Views

Author

Clark Kimberling, Nov 14 2007

Keywords

Comments

The lower and upper Wythoff sequences, A and B, satisfy the complementary equation ABB = 2A+3B.
The asymptotic density of this sequence is 1/phi^5 = phi^5 - 11 = A244593 - 4 = 0.0901699... . - Amiram Eldar, Mar 24 2025

Crossrefs

Let A = A000201, B = A001950. Then AA = A003622, AB = A003623, BA = A035336, BB = A101864. The eight triples AAA, AAB, ..., BBB are A134859, A134860, A035337, A134862, A134861, A134863, A035338, A134864, resp.

Programs

  • Mathematica
    A[n_] := Floor[n * GoldenRatio]; B[n_] := Floor[n * GoldenRatio^2]; a[n_] := A[B[B[n]]]; Array[a, 100] (* Amiram Eldar, Mar 24 2025 *)
  • Python
    from sympy import floor
    from mpmath import phi
    def A(n): return floor(n*phi)
    def B(n): return floor(n*phi**2)
    def a(n): return A(B(B(n))) # Indranil Ghosh, Jun 10 2017
    
  • Python
    from math import isqrt
    def A134862(n): return 5*(n+isqrt(5*n**2)>>1)+3*n # Chai Wah Wu, Aug 10 2022

Formula

a(n) = A(B(B(n))), n>=1, with A=A000201, the lower Wythoff sequence and B=A001950, the upper Wythoff sequence.

A134864 Wythoff BBB numbers.

Original entry on oeis.org

13, 34, 47, 68, 89, 102, 123, 136, 157, 178, 191, 212, 233, 246, 267, 280, 301, 322, 335, 356, 369, 390, 411, 424, 445, 466, 479, 500, 513, 534, 555, 568, 589, 610, 623, 644, 657, 678, 699, 712, 733, 746, 767, 788, 801, 822, 843, 856, 877, 890, 911, 932, 945
Offset: 1

Views

Author

Clark Kimberling, Nov 14 2007

Keywords

Comments

The lower and upper Wythoff sequences, A and B, satisfy the complementary equation BBB = 3A+5B.
The asymptotic density of this sequence is 1/phi^6 = A094214^6 = 0.05572809... . - Amiram Eldar, Mar 24 2025

Crossrefs

Let A = A000201, B = A001950. Then AA = A003622, AB = A003623, BA = A035336, BB = A101864. The eight triples AAA, AAB, ..., BBB are A134859, A134860, A035337, A134862, A134861, A134863, A035338, A134864, resp.

Programs

  • Maple
    a:=n->floor(n*((1+sqrt(5))/2)^2): [a(a(a(n)))$n=1..55]; # Muniru A Asiru, Nov 24 2018
  • Mathematica
    Nest[Quotient[#(3+Sqrt@5),2]&,#,3]&/@Range@100 (* Federico Provvedi, Nov 24 2018 *)
    b[n_]:=Floor[n GoldenRatio^2]; a[n_]:=b[b[b[n]]]; Array[a, 60] (* Vincenzo Librandi, Nov 24 2018 *)
  • Python
    from sympy import floor
    from mpmath import phi
    def B(n): return floor(n*phi**2)
    def a(n): return B(B(B(n))) # Indranil Ghosh, Jun 10 2017
    
  • Python
    from math import isqrt
    def A134864(n): return (m:=5*n)+(((n+isqrt(n*m))&-2)<<2) # Chai Wah Wu, Aug 10 2022

Formula

a(n) = B(B(B(n))), n>=1, with B=A001950, the upper Wythoff sequence.
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