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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A139340 Decimal expansion of the cube root of the golden ratio. That is, the decimal expansion of ((1+sqrt(5))/2)^(1/3).

Original entry on oeis.org

1, 1, 7, 3, 9, 8, 4, 9, 9, 6, 7, 0, 5, 3, 2, 8, 5, 0, 9, 9, 6, 6, 6, 8, 3, 9, 7, 1, 8, 8, 6, 2, 6, 6, 7, 4, 1, 9, 5, 5, 7, 9, 9, 0, 6, 9, 0, 9, 0, 8, 1, 1, 2, 0, 6, 7, 7, 6, 0, 5, 0, 0, 3, 3, 0, 6, 8, 2, 7, 9, 9, 0, 3, 1, 0, 4, 8, 2, 0, 2, 7, 7, 8, 1, 8, 4, 0, 6, 5, 7, 4, 7, 5, 8, 1, 1, 4, 3, 9, 9, 9, 2, 7, 7, 3
Offset: 1

Views

Author

Mohammad K. Azarian, Apr 14 2008

Keywords

Comments

Larger of the real roots of x^6 - x^3 - 1. - Charles R Greathouse IV, Apr 14 2014

Examples

			1.1739849967053285...
		

Crossrefs

Programs

A270744 (r,1)-greedy sequence, where r(k) = 1/tau^k, where tau = golden ratio.

Original entry on oeis.org

1, 2, 2, 3, 4, 32, 1065, 2038968, 5977146319204, 36314862033946243071181679, 1028280647188781709727717632740627249617427013751977, 958046899855070460620234639622630375078362220775180051610386376308132568342498992099474472596860400289
Offset: 1

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Author

Clark Kimberling, Apr 07 2016

Keywords

Comments

Let x > 0, and let r = (r(k)) be a sequence of positive irrational numbers. Let a(1) be the least positive integer m such that r(1)/m < x, and inductively let a(n) be the least positive integer m such that r(1)/a(1) + ... + r(n-1)/a(n-1) + r(n)/m < x. The sequence (a(n)) is the (r,x)-greedy sequence. We are interested in choices of r and x for which the series r(1)/a(1) + ... + r(n)/a(n) + ... converges to x. (The same algorithm is used to generate sequences listed at A269993.)
Guide to related sequences:
x r(k)
1 1/tau^k A270744
1 k/tau^k A270745
1 2/e^k A270746
1 4/Pi^k A270747
1 2/log(k+1) A270748
1 k/log(k+1) A270749
1 1/(k*log(k+1)) A270750
1 1/(k*tau) A270751
1 1/(k*e) A270752
1 1/(k*sqrt(2)) A270916

Examples

			a(1) = ceiling(r(1)) = ceiling(1/tau) = ceiling(0.618...) = 1;
a(2) = ceiling(r(2)/(1 - r(1)/1)) = 2;
a(3) = ceiling(r(3)/(1 - r(1)/1 - r(2)/2)) = 2.
The first 6 terms of the series r(1)/a(1) + ... + r(n)/a(n) + ... are 0.618..., 0.809..., 0.927..., 0.975..., 0.998..., 0.999967... .
		

Crossrefs

Programs

  • Mathematica
    $MaxExtraPrecision = Infinity; z = 13;
    r[k_] := N[1/GoldenRatio^k, 1000]; f[x_, 0] = x;
    n[x_, k_] := n[x, k] = Ceiling[r[k]/f[x, k - 1]]
    f[x_, k_] := f[x, k] = f[x, k - 1] - r[k]/n[x, k]
     x = 1; Table[n[x, k], {k, 1, z}]
    N[Sum[r[k]/n[x, k], {k, 1, 13}], 200]

Formula

a(n) = ceiling(r(n)/s(n)), where s(n) = 1 - r(1)/a(1) - r(2)/a(2) - ... - r(n-1)/a(n-1).
r(1)/a(1) + ... + r(n)/a(n) + ... = 1.

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.

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.

A290565 Decimal expansion of sum of reciprocal golden rectangle numbers.

Original entry on oeis.org

1, 7, 7, 3, 8, 7, 7, 5, 8, 3, 2, 8, 5, 1, 3, 2, 3, 4, 3, 8, 0, 2, 3, 6, 2, 7, 6, 5, 6, 7, 6, 9, 6, 5, 9, 2, 2, 8, 3, 0, 7, 2, 3, 2, 3, 9, 3, 5, 9, 4, 3, 4, 1, 1, 0, 8, 3, 9, 2, 2, 9, 0, 4, 9, 8, 6, 4, 9, 2, 2, 0, 7, 5, 3, 0, 3, 8, 5, 1, 1, 9, 4, 7, 0, 3, 6, 2, 4, 3, 3, 3, 8, 6, 0, 5, 2, 6, 4, 2, 6, 9, 1
Offset: 1

Views

Author

Bobby Jacobs and Robert G. Wilson v, Aug 06 2017

Keywords

Comments

The constant k in A277266 such that A277266(n) ~ k*n.

Examples

			1/(1*1) + 1/(1*2) + 1/(2*3) + 1/(3*5) + ... = 1 + 1/2 + 1/6 + 1/15 + ... = 1.77387758328513234380...
		

Crossrefs

Programs

  • Mathematica
    RealDigits[ Sum[1/(Fibonacci[k]*Fibonacci[k + 1]), {k, 265}], 10, 111][[1]]
  • PARI
    suminf(n=1, 1/(fibonacci(n)*fibonacci(n+1))) \\ Michel Marcus, Feb 19 2019

Formula

Equals Sum_{n>=1} 1/(Fibonacci(n)*Fibonacci(n+1)).
Equals lim_{n->infinity} A277266(n)/n.
Equals 2 * (Sum_{k>=1} 1/(phi^k * F(k))) - 1/phi = 2 * A265290 - A094214, where phi is the golden ratio (A001622) and F(k) is the k-th Fibonacci number (A000045). - Amiram Eldar, Oct 05 2020
Equals 3/2 + 10*c*Integral_{x=0..infinity} f(x) dx, where c = sqrt(5)/log(phi) = A002163/A002390, phi = (1+sqrt(5))/2 = A001622, and f(x) = sin(x)/((exp(Pi*x/(2*log(phi)))-1)*(7-2*cos(x))*(3+2*cos(x))). - Gleb Koloskov, Sep 12 2021

Extensions

More terms from Alois P. Heinz, Aug 06 2017

A139342 Decimal expansion of e^(-(1+sqrt(5))/2).

Original entry on oeis.org

1, 9, 8, 2, 8, 8, 1, 5, 2, 8, 6, 2, 2, 0, 6, 2, 3, 2, 2, 6, 7, 8, 8, 8, 9, 5, 6, 6, 0, 4, 8, 6, 4, 6, 7, 0, 8, 4, 2, 0, 8, 4, 8, 9, 2, 5, 0, 1, 2, 9, 7, 1, 6, 5, 2, 7, 4, 2, 6, 1, 9, 3, 1, 8, 0, 2, 6, 4, 2, 4, 6, 0, 4, 1, 7, 1, 5, 0, 7, 9, 1, 4, 1, 9, 6, 1, 8, 1, 4, 5, 3, 1, 6, 6, 5, 3, 4, 4, 6, 2, 6, 8, 4, 8
Offset: 0

Views

Author

Mohammad K. Azarian, Apr 14 2008

Keywords

Comments

By the Lindemann-Weierstrass theorem, this constant is transcendental. - Charles R Greathouse IV, May 13 2019

Examples

			0.19828815286220623226788895660486467084208489250129...
		

Crossrefs

Programs

Formula

Equals exp(-A001622).
Equals 1/A139341. - Amiram Eldar, Feb 08 2022

Extensions

Leading zero removed by R. J. Mathar, Feb 05 2009

A242671 Decimal expansion of k2, a Diophantine approximation constant such that the area of the "critical parallelogram" (in this case a square) is 4*k2.

Original entry on oeis.org

7, 2, 3, 6, 0, 6, 7, 9, 7, 7, 4, 9, 9, 7, 8, 9, 6, 9, 6, 4, 0, 9, 1, 7, 3, 6, 6, 8, 7, 3, 1, 2, 7, 6, 2, 3, 5, 4, 4, 0, 6, 1, 8, 3, 5, 9, 6, 1, 1, 5, 2, 5, 7, 2, 4, 2, 7, 0, 8, 9, 7, 2, 4, 5, 4, 1, 0, 5, 2, 0, 9, 2, 5, 6, 3, 7, 8, 0, 4, 8, 9, 9, 4, 1, 4, 4, 1, 4, 4, 0, 8, 3, 7, 8, 7, 8, 2, 2, 7, 4
Offset: 0

Views

Author

Jean-François Alcover, May 20 2014

Keywords

Comments

Quoting Steven Finch: "The slopes of the 'critical parallelogram' are (1+sqrt(5))/2 [phi] and (1-sqrt(5))/2 [-1/phi]."
Essentially the same as A229780, A134972, A134945, A098317 and A002163. - R. J. Mathar, May 23 2014
Let W_n be the collection of all binary words of length n that do not contain two consecutive 0's. Let r_n be the ratio of the total number of 1's in W_n divided by the total number of letters in W_n. Then lim_{n->oo} r_n = 0.723606... Equivalently, lim_{n->oo} A004798(n)/(n*A000045(n+2)) = 0.723606... - Geoffrey Critzer, Feb 04 2022
The limiting frequency of the digit 0 in the base phi representation of real numbers in the range [0,1], where phi is the golden ratio (A001622) (Rényi, 1957). - Amiram Eldar, Mar 18 2025

Examples

			k2 = 0.723606797749978969640917366873127623544...
		

References

  • Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Section 2.23, p. 176.

Crossrefs

Programs

  • Mathematica
    RealDigits[(1+1/Sqrt[5])/2, 10, 100] // First
  • PARI
    (1 + 1/sqrt(5))/2 \\ Stefano Spezia, Dec 07 2024

Formula

Equals (1 + 1/sqrt(5))/2.
Equals 1/A094874. - Michel Marcus, Dec 01 2018
From Amiram Eldar, Feb 11 2022: (Start)
Equals phi/sqrt(5), where phi is the golden ratio (A001622).
Equals lim_{k->oo} Fibonacci(k+1)/Lucas(k). (End)
From Amiram Eldar, Nov 28 2024: (Start)
Equals A344212/2 = A296184/5 = A300074^2 = sqrt(A229780).
Equals Product_{k>=1} (1 - 1/A081007(k)). (End)
Equals 1 - A244847. - Amiram Eldar, Mar 18 2025

A337669 Decimal expansion of Product_{n>=3} (1-1/Fibonacci(n)).

Original entry on oeis.org

1, 8, 9, 7, 8, 9, 1, 4, 3, 6, 1, 7, 8, 6, 6, 0, 3, 6, 3, 4, 9, 4, 8, 2, 5, 1, 4, 2, 9, 0, 3, 4, 0, 5, 3, 1, 2, 7, 2, 9, 8, 1, 4, 9, 1, 3, 1, 9, 2, 8, 8, 7, 5, 2, 2, 9, 0, 6, 8, 9, 0, 7, 0, 7, 1, 0, 8, 2, 0, 4, 6, 9, 4, 5, 7, 0, 3, 7, 0, 4, 4, 5, 6, 6, 7, 9, 7
Offset: 0

Views

Author

Michel Marcus, Sep 15 2020

Keywords

Examples

			0.18978914361786603634948251429...
		

Crossrefs

Programs

  • Mathematica
    With[{b = 1/GoldenRatio}, RealDigits[(Sqrt[5]/6)*b^(-5/4) * EllipticTheta[2, 0, b] * EllipticTheta[3, 0, b] * EllipticTheta[4, 0, b]/EllipticTheta[4, 0, b^4], 10, 100][[1]]] (* Amiram Eldar, May 27 2021 *)
  • PARI
    prodinf(n=3, 1-1/fibonacci(n))

Formula

Equals (sqrt(5)/6) * b^(-5/4) * theta_2(b) * theta_3(b) * theta_4(b)/theta_4(b^4), where theta_i are the Jacobi theta functions and b = 1/phi = A094214 (Duverney and Tachiya, 2021). - Amiram Eldar, May 27 2021
Equals (sqrt(5) * phi^(5/4) / 3) * eta(tau_0)^3 * eta(4*tau_0) / eta(2*tau_0)^2, where phi is the golden ratio (A001622), tau_0 = i*log(phi)/Pi, and i = sqrt(-1) (Duverney et al., 2022). - Amiram Eldar, Mar 26 2024

Extensions

More terms from Jinyuan Wang, Sep 19 2020

A084242 Least k, 1 <= k <= n, such that the number of elements of the continued fraction for n/k is maximum.

Original entry on oeis.org

1, 1, 2, 3, 3, 4, 4, 5, 5, 6, 7, 7, 8, 9, 11, 9, 10, 11, 11, 11, 13, 13, 14, 13, 14, 15, 17, 17, 18, 19, 18, 23, 19, 21, 22, 22, 23, 22, 25, 29, 23, 26, 25, 27, 26, 27, 29, 31, 30, 29, 28, 33, 33, 31, 34, 41, 32, 36, 33, 37, 33, 35, 37, 39, 47, 37, 41, 42, 40, 41, 41, 53, 45, 43
Offset: 1

Views

Author

Benoit Cloitre, Jun 21 2003

Keywords

Comments

Also, for n > 1, the smallest number k such that the Euclidean algorithm for (n,k) requires the maximum number of steps, A034883(n). - T. D. Noe, Mar 24 2011

Crossrefs

Programs

  • PARI
    a(n)=if(n<0,0,s=1; while(abs(vecmax(vector(n,i,length(contfrac(n/i))))-length(contfrac(n/s)))>0,s++); s)

Formula

For k > 1, a(F(k)) = F(k-1) where F(k) denotes the k-th Fibonacci number.
Probably, lim_{n->oo} (1/n)*Sum_{k=1..n} a(k) = 1/phi = A094214.

A337668 Decimal expansion of Product_{n>=1} (1+1/Fibonacci(n)).

Original entry on oeis.org

1, 3, 1, 5, 0, 9, 6, 6, 6, 5, 7, 7, 8, 4, 1, 8, 4, 3, 6, 7, 6, 1, 2, 4, 3, 3, 3, 7, 0, 6, 2, 6, 6, 5, 8, 9, 3, 2, 1, 9, 0, 1, 9, 9, 5, 4, 3, 1, 6, 4, 2, 8, 4, 7, 0, 1, 3, 5, 4, 1, 0, 0, 7, 4, 7, 1, 5, 7, 6, 9, 8, 0, 4, 6, 0, 2, 7, 6, 1, 1, 8, 0, 9, 1, 1, 8, 2, 4
Offset: 2

Views

Author

Michel Marcus, Sep 15 2020

Keywords

Examples

			13.15096665778418436761243337...
		

Crossrefs

Programs

  • Mathematica
    With[{b = 1/GoldenRatio}, RealDigits[2*b^(-5/4)*EllipticTheta[2, 0, b]/EllipticTheta[4, 0, b^4], 10, 100][[1]]] (* Amiram Eldar, May 27 2021 *)
  • PARI
    prodinf(n=1, 1+1/fibonacci(n))

Formula

Equals 2 * b^(-5/4) * theta_2(b)/theta_4(b^4), where theta_i are the Jacobi theta functions and b = 1/phi = A094214 (Duverney and Tachiya, 2021). - Amiram Eldar, May 27 2021
Equals 4 * phi^(5/4) * eta(4*tau_0) / eta(tau_0), where phi is the golden ratio (A001622), tau_0 = i*log(phi)/Pi, and i = sqrt(-1) (Duverney et al., 2022). - Amiram Eldar, Mar 26 2024

Extensions

More terms from Jinyuan Wang, Sep 19 2020
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