A366185 - OEIS

A366185

Decimal expansion of the real root of the quintic equation x^5 + 3*x^4 + 4*x^3 + x -1 = 0.

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

(list; constant; graph; refs; listen; history; text; internal format)

OFFSET

0,1

COMMENTS

The root appears in the problem of minimizing the area of self-intersection of a folded rectangle. A rectangle with sides a, b (a<b) is folded along the line that passes through the center of the rectangle in order to get the minimum area of crossing intersections: a unique rectangle exists for two solutions with equal area but different shapes - triangle and pentagon.

The unique ratio of sides a/b=T=0.81502370129163... is derived based on the real root of the quintic. If a/b<T ('long' rectangle) the angle to fold is Pi/4. If a/b=1 (square) the angle is Pi/8.

From Mikhail Gaichenkov, May 25 2026: (Start)

According to Oscar Lanzi the neusis construction exists (with no additional tools or curves), and it can be derived using the methodology of Benjamin and Snyder for appropriately constructed quintic and sextic equations with a "conchoid-circle constructions". Now when we form the resolvent sextic a miracle occurs: a rational root u=-1. Using this tool we follow the formulae in the paper and eventually arrive at the following parameters, with the ruler required to pass through (0,0) and the distance between marks rendered as one unit: a=sqrt(2) (horizontal coordinate of one mark, on a line parallel to the vertical axis), (b,c)=(-1/sqrt(2),1/sqrt(2)) (center of the circle that contains the second mark on the line), s=sqrt(2) (radius of this circle).

When this construction is implemented, we find two roots for the mark on the circle (which lies between the ruler's fixed point (0,0) and the mark on the line). One root has the ruler angled Pi/4 from the horizontal axis with the mark on the circle in the first quadrant and one unit from the origin, this corresponds to the auxiliary root u=-1. The second root, with the mark on the circle in the fourth quadrant and closer to the origin (0,0) than the first root. With u having a unit value the distance from this mark on the circle to the origin matches the original quintic root. (End)

LINKS

Table of n, a(n) for n=0..107.

Elliot Benjamin and C. Snyder, On the construction of the regular hendecagon by marked ruler and compass, Mathematical Proceedings of the Cambridge Philosophical Society, Volume 156, Issue 3, May 2014, pp. 409-424.

Mikhail Gaichenkov, QGeometric construction of real root of quintic using marked ruler and compass, Mathoverflow, 2023.

Mikhail Gaichenkov, Folded rectangle

Mikhail Gaichenkov, Quintic equation with integer coefficients, Math Stackexchange, 2023.

Index entries for algebraic numbers, degree 5

EXAMPLE

0.45913372331020753...

MATHEMATICA

First[RealDigits[Root[#^5 + 3*#^4 + 4*#^3 + # - 1 &, 1], 10, 100]] (* Paolo Xausa, Jun 25 2024 *)

PROG

(PARI) polrootsreal(x^5 + 3*x^4 + 4*x^3 + x-1)[1]

CROSSREFS

Sequence in context: A194419 A175380 A263151 * A093088 A234430 A019637