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.

A265290 Decimal expansion of Sum_{n>=1} |phi - c(n)|, where phi is the golden ratio (A001622) and c(n) are the convergents to phi.

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Dec 06 2015

Keywords

Comments

Define the deviance of x > 0 by dev(x) = Sum_{n>=1} |x - c(n,x)|, where c(n,x) = n-th convergent to x. The greatest value of dev(x) occurs when x = golden ratio, so that this constant is the maximal deviance.

Examples

			1.195955786017513596003474800021...
		

Crossrefs

Programs

  • Maple
    x := (3 - sqrt(5))/2:
    evalf(sqrt(5)*add(x^(n^2)*(1 + x^(2*n))/(1 - x^(2*n)), n = 1..16), 100); # Peter Bala, Aug 21 2022
  • Mathematica
    x = GoldenRatio; z = 600; c = Convergents[x, z];
    s1 = Sum[x - c[[2 k - 1]], {k, 1, z/2}]; N[s1, 200]
    s2 = Sum[c[[2 k]] - x, {k, 1, z/2}]; N[s2, 200]
    N[s1 + s2, 200]
    RealDigits[s1, 10, 120][[1]]  (* A265288 *)
    RealDigits[s2, 10, 120][[1]]  (* A265289 *)
    RealDigits[s1 + s2, 10, 120][[1]] (* A265290, dev(x) *)
    d[x_] := If[IntegerQ[1000!*x], Total[Abs[x - Convergents[x]]],
      Total[Abs[x - Convergents[x, 30]]]]
    Plot[{d[x], 1.195}, {x, 0, 1}]

Formula

Equals Sum_{n>=1} 1/(F(2*n-1)*F(2*n)), where F(n) is the n-th Fibonacci number (A000045).
From Amiram Eldar, Oct 05 2020: (Start)
Equals Sum_{k>=1} 1/(phi^k * F(k)).
Equals sqrt(5) * Sum_{k>=1} 1/(phi^(2*k) - (-1)^k) = sqrt(5) * Sum_{k>=1} (-1)^(k+1)/(phi^(2*k) + (-1)^k).
Equals (A290565 + 1/phi)/2. (End)
A rapidly converging series for the constant is sqrt(5)*Sum_{k >= 1} x^(k^2)*(1 + x^(2*k))/(1 - x^(2*k)), where x = (3 - sqrt(5))/2. See A112329. - Peter Bala, Aug 21 2022