A374883 Decimal expansion of phi*(2*phi + 1) (i.e., (7 + 3*sqrt(5))/2), where phi is the golden ratio.

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

Offset: 1

Author

Marco Ripà, Jul 22 2024

Keywords

• nonn
• cons
• easy

Comments

The author conjectures that this is the minimum volume of an axis-aligned bounding box which includes the shortest minimum-link circuit joining all the vertices of the cube {0,1}^3 (i.e., the closed polygonal chains consisting of exactly 6 edges visiting all the points of the set {(0,0,0),(0,0,1),(0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1)}).

In detail, such a circuit of 6 links is given by (1/2,1+phi,1/2)-((1-phi)/2,0,(1+phi)/2)-((phi+1)/2,0, (1-phi)/2)-(1/2,1+phi,1/2)-((phi+1)/2,0,(phi+1)/2)-((1-phi)/2,0,(1-phi)/2(1/2,1+phi,1/2), where phi := (1+sqrt(5))/2 (see A001622).

Then, phi*(2*phi + 1) = phi^2*(phi + 1) since phi - 1 = 1/phi.

Examples

			6.8541019662496845446137605030969...
		

References

• Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 138-139.

Links

• Roberto Rinaldi and Marco Ripà, Optimal cycles enclosing all the nodes of a k-dimensional hypercube, arXiv:2212.11216 [math.CO], 2022.
• Marco Ripà, General uncrossing covering paths inside the Axis-Aligned Bounding Box, Journal of Fundamental Mathematics and Applications, Volume 4, 2021, Number 2, Pages 154-166.

Crossrefs

Cf. A001622, A090550, A104457, A134973, A205769, A225227, A261547, A363755, A373537, A374149, A374260.

Programs

Mathematica

RealDigits[3*GoldenRatio + 2, 10, 120][[1]] (* Amiram Eldar, Jul 23 2024 *)

Formula

Equals (7 + 3*sqrt(5))/2.

Equals phi^2*(phi + 1), where phi = (1 + sqrt(5))/2.

Equals A104457^2 = 2*A205769. - Hugo Pfoertner, Jul 22 2024

Equals A090550 + 1 = A134973 + 5. - Amiram Eldar, Jul 23 2024

Equals phi^4. - Stefano Spezia, May 28 2025