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 1-3 of 3 results.

A079586 Decimal expansion of Sum_{k>=1} 1/F(k) where F(k) is the k-th Fibonacci number A000045(k).

Original entry on oeis.org

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

Views

Author

Benoit Cloitre, Jan 26 2003

Keywords

Comments

André-Jeannin proved that this constant is irrational.
This constant does not belong to the quadratic number field Q(sqrt(5)) (Bundschuh and Väänänen, 1994). - Amiram Eldar, Oct 30 2020

Examples

			3.35988566624317755317201130291892717968890513373...
		

References

  • Daniel Duverney, Number Theory, World Scientific, 2010, 5.22, pp.75-76.
  • Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 358.

Crossrefs

Programs

  • Maple
    Digits := 120: c := Pi/2 + I*arccsch(2):
    Jeannin := n -> sqrt(5/4)*add(I^(1-j)/sin(j*c), j = 1..n):
    evalf(Jeannin(1000)); # Peter Luschny, Nov 15 2023
  • Mathematica
    digits = 105; Sqrt[5]*NSum[(-1)^n/(GoldenRatio^(2*n + 1) - (-1)^n), {n, 0, Infinity}, WorkingPrecision -> digits, NSumTerms -> digits] // RealDigits[#, 10, digits] & // First (* Jean-François Alcover, Apr 09 2013 *)
    First@RealDigits[Sqrt[5]/4 ((Log[5] + 2 QPolyGamma[1, 1/GoldenRatio^4] - 4 QPolyGamma[1, 1/GoldenRatio^2])/(2 Log[GoldenRatio]) + EllipticTheta[2, 0, 1/GoldenRatio^2]^2), 10, 105] (* Vladimir Reshetnikov, Nov 18 2015 *)
  • PARI
    /* Fast computation without splitting into even and odd indices, see the Arndt reference */
    lambert2(x, a, S)=
    {
    /* Return G(x,a) = Sum_{n>=1} a*x^n/(1-a*x^n) (generalized Lambert series)
       computed as Sum_{n=1..S} x^(n^2)*a^n*( 1/(1-x^n) + a*x^n/(1-a*x^n) )
       As series in x correct up to order S^2.
       We also have G(x,a) = Sum_{n>=1} a^n*x^n/(1-x^n) */
        return( sum(n=1,S, x^(n^2)*a^n*( 1/(1-x^n) + a*x^n/(1-a*x^n) ) ) );
    }
    inv_fib_sum(p=1, q=1, S)=
    {
    /* Return Sum_{n>=1} 1/f(n) where f(0)=0, f(1)=1, f(n) = p*f(n-1) + q*f(n-1)
       computed using generalized Lambert series.
       Must have p^2+4*q > 0 */
        my(al,be);
        \\ Note: the q here is -q in the Horadam paper.
        \\ The following numerical examples are for p=q=1:
        al=1/2*(p+sqrt(p^2+4*q));  \\ == +1.6180339887498...
        be=1/2*(p-sqrt(p^2+4*q));  \\ == -0.6180339887498...
        return( (al-be)*( 1/(al-1) + lambert2(be/al, 1/al, S) ) ); \\ == 3.3598856...
    }
    default(realprecision,100);
    S = 1000; /* (be/al)^S == -0.381966^S == -1.05856*10^418 << 10^-100 */
    inv_fib_sum(1,1,S) /* 3.3598856... */ /* Joerg Arndt, Jan 30 2011 */
    
  • PARI
    suminf(k=1, 1/(fibonacci(k))) \\ Michel Marcus, Feb 19 2019
    
  • Sage
    m=120; numerical_approx(sum(1/fibonacci(k) for k in (1..10*m)), digits=m) # G. C. Greubel, Feb 20 2019

Formula

Alternating series representation: 3 + Sum_{k >= 1} (-1)^(k+1)/(F(k)*F(k+1)*F(k+2)). - Peter Bala, Nov 30 2013
From Amiram Eldar, Oct 04 2020: (Start)
Equals sqrt(5) * Sum_{k>=0} (1/(phi^(2*k+1) - 1) - 2*phi^(2*k+1)/(phi^(4*(2*k+1)) - 1)), where phi is the golden ratio (A001622) (Greig, 1977).
Equals sqrt(5) * Sum_{k>=0} (-1)^k/(phi^(2*k+1) - (-1)^k) (Griffin, 1992).
Equals A153386 + A153387. (End)
From Gleb Koloskov, Sep 14 2021: (Start)
Equals 1 + c1*(c2 + 32*Integral_{x=0..infinity} f(x) dx),
where c1 = sqrt(5)/(8*log(phi)) = A002163/(8*A002390),
c2 = 2*arctan(2)+log(5) = 2*A105199+A016628,
phi = (1+sqrt(5))/2 = A001622,
f(x) = sin(x)*(4+cos(2*x))/((exp(Pi*x/log(phi))-1)*(2*cos(2*x)+3)*(7-2*cos(2*x))) (End)
From Amiram Eldar, Jan 27 2022: (Start)
Equals 3 + 2 * Sum_{k>=1} 1/(F(2*k-1)*F(2*k+1)*F(2*k+2)) (Bruckman, 1987).
Equals 2 + Sum_{k>=1} 1/A350901(k) (André-Jeannin, Problem H-450, 1991).
Equals lim_{n->oo} A350903(n)/(A350904(n)*A350902(n)) (André-Jeannin, 1991). (End)
Equals sqrt(5/4)*Sum_{j>=1} i^(1-j)/sin(j*c) where c = Pi/2 + i*arccsch(2). - Peter Luschny, Nov 15 2023
Equals lim_{n->oo} A203006(n)/A003266(n) (Z.W. M. Trzaska, 1996). - Raul Prisacariu, Sep 04 2024

A350903 Numerators of the sequence of fractions defined by u(n) = ((5*F(n)*F(n-1)*F(2*n-1)*u(n-1) + F(n-1)*L(n)*u(n-2))/(L(n-1)*F(n))), with u(0) = 0 and u(1) = 1, where F(n) = A000045(n) and L(n) = A000032(n).

Original entry on oeis.org

0, 1, 10, 84, 8225, 999146, 161691205, 4081394133187, 801267937794945, 451272063930179690869, 955797228958312695758495, 12869303093903467063139191673469, 141131682569461636438244407470674215, 5214528077594695050414454970728001934806021
Offset: 0

Views

Author

Amiram Eldar, Jan 21 2022

Keywords

Comments

See A350902 for details.

Examples

			The sequence of fractions begins with 0, 1, 10, 84, 8225/3, 999146/5, 161691205/4, 4081394133187/195, 801267937794945/28, 451272063930179690869/4420, ...
		

Crossrefs

Cf. A000032, A000045, A079586, A350902, A350904 (denominators).

Programs

  • Mathematica
    With[{F = Fibonacci, L = LucasL}, u[0] = 0; u[1] = 1; u[n_] := u[n] = (5*F[n]*F[n - 1]*F[2*n - 1]*u[n - 1] + F[n - 1]*L[n]*u[n - 2])/(L[n - 1]*F[n]); Numerator @ Array[u, 15, 0]]

A350904 Denominators of the sequence of fractions defined by u(n) = ((5*F(n)*F(n-1)*F(2*n-1)*u(n-1) + F(n-1)*L(n)*u(n-2))/(L(n-1)*F(n))), with u(0) = 0 and u(1) = 1, where F(n) = A000045(n) and L(n) = A000032(n).

Original entry on oeis.org

1, 1, 1, 1, 3, 5, 4, 195, 28, 4420, 1001, 550732, 94248, 20757737, 150585864, 596098336680, 84878386593, 17090110926980520, 1216260982575912, 13296541287045886485, 484071647034823848, 3418959485072391296664264, 19630886922468003512297
Offset: 0

Views

Author

Amiram Eldar, Jan 21 2022

Keywords

Comments

See A350902 for details.

Crossrefs

Cf. A000032, A000045, A079586, A350902, A350903 (numerators).

Programs

  • Mathematica
    With[{F = Fibonacci, L = LucasL}, u[0] = 0; u[1] = 1; u[n_] := u[n] = (5*F[n]*F[n - 1]*F[2*n - 1]*u[n - 1] + F[n - 1]*L[n]*u[n - 2])/(L[n - 1]*F[n]); Denominator @ Array[u, 25, 0]]
Showing 1-3 of 3 results.