A381684 Decimal expansion of the isoperimetric quotient of a truncated tetrahedron.

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

Offset: 0

Author

Paolo Xausa, Mar 04 2025

Keywords

• nonn
• cons
• easy

Comments

Polya (1954) defines the isoperimetric quotient of a solid as 36*Pi*V^2/S^3, where V and S are the volume and surface area of the solid, respectively.

The isoperimetric quotient of a sphere is 1.

Examples

			0.4662292826432950646084875599089894958106273300491...
		

References

• George Polya, Mathematics and Plausible Reasoning, Vol. 1: Induction and Analogy in Mathematics, Princeton University Press, Princeton, New Jersey, 1954. See pp. 188-189, exercise 43.

Links

• Paolo Xausa, Table of n, a(n) for n = 0..10000
• P. K. Aravind, How Spherical Are the Archimedean Solids and Their Duals?, The College Mathematics Journal, Vol. 42, Issue 2, 2011, pp. 98-107.
• Index entries for transcendental numbers.

Crossrefs

Cf. A377274 (surface area), A377275 (volume).

Cf. A000796, A002194.

Programs

Mathematica

First[RealDigits[529*Pi/(2058*Sqrt[3]), 10, 100]]

PARI

529*Pi/2058/sqrt(3) \\ Charles R Greathouse IV, Aug 19 2025

Formula

Equals 36*Pi*A377275^2/(A377274^3).

Equals 529*Pi/(2058*sqrt(3)) = 529*A000796/(2058*A002194).