A265288 Decimal expansion of Sum_{n >= 1} (phi - c(2*n-1)), where phi is the golden ratio (A001622), and c(n) is the n-th convergent to the continued fraction expansion of phi.

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

Offset: 0

Author

Clark Kimberling, Dec 06 2015

Keywords

• nonn
• cons

Comments

Define the lower deviance of x > 0 by dL(x) = Sum_{n>=1} (x - c(2*n-1,x)), where c(k,x) = k-th convergent to x. The greatest lower deviance occurs when x = golden ratio, so that this constant is the absolute maximal lower deviance.

Guide to related constants (as sequences):

x Sum{x-c(2*n-1)} Sum{c(2*n)-x} Sum|c(2*n)-c(2*n-1)|

(1+sqrt(5))/2 A265288 A265289 A265290

sqrt(2) A265291 A265292 A265293

sqrt(3) A265294 A265295 A265296

sqrt(5) A265297 A265298 A265299

sqrt(6) A265300 A265301 A265302

sqrt(8) A265303 A265304 A265305

e A265306 A265307 A265308

Examples

			0.75720437504600733864782526067377483...
The convergents to x are c(1) = 1, c(2) = 2, c(3) = 3/2, c(4) = 5/3, ..., so that
A265288 = (x - 1) + (x - 3/2) + (x - 8/5) + ... ;
A265289 = (2 - x) + (5/3 - x) + (13/8 - x ) + ... ;
A265290 = (2 - 1) + (5/3 - 3/2) + (13/8 - 8/5) + ...
		

Crossrefs

Cf. A000045, A001227, A001622, A153386, A265289, A265290.

Programs

Maple

x := -(3 - sqrt(5))/2:
evalf(sqrt(5)*add(x^(n*(n+1)/2)/(x^n - 1), n = 1..24), 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 *)

Formula

Equals Sum_{k >= 1} 1/(phi^(2*k-1) * F(2*k-1)), where F(k) is the k-th Fibonacci number (A000045). - Amiram Eldar, Oct 05 2020

From Peter Bala, Aug 19 2022: (Start)

The constant equals Sum_{k >= 1} (-1)^(k+1)/F(2*k). The constant also equals (3/5)*Sum_{k >= 1} (-1)^(k+1)/(F(2*k)*F(2*k+2)*F(2*k+4)) + 11/15.

A rapidly converging series for the constant is sqrt(5) * Sum_{k >= 1} x^(k*(k+1)/2)/ (x^k - 1) at x = phi - 2 = -(3 - sqrt(5))/2. (End)