A127454 Decimal expansion of transcendental solution to round pegs in square holes problem.

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

Offset: 1

Author

Jonathan Vos Post, Jan 13 2007

Keywords

• cons
• nonn

Comments

This value "must be determined numerically. As a result, a round peg fits better into a square hole than a square peg fits into a round hole only for integer dimensions n < 9."

Examples

			8.13794104609137237652983898405322337009672530976244376958353099224630941205660...
		

Links

• Robert G. Wilson v, Table of n, a(n) for n = 1..1001
• David Singmaster, On Round Pegs in Square Holes and Square Pegs in Round Holes, Math. Mag. 37, 335-339, 1964.
• Eric Weisstein's World of Mathematics, Peg.

Crossrefs

Cf. A194940.

Programs

Mathematica

RealDigits[ FindRoot[ Pi^x*x^(x/2) == 2^(2 x) Gamma[1 + x/2]^2, {x, 8}, WorkingPrecision -> 121][[1, 2]], 10, 111][[1]] (* Robert G. Wilson v, Jul 03 2014 *)

Formula

Where the real number ratio crosses 1 in (Pi^n)(n^(n/2))/(2^(2n))(Gamma(1+n/2))^2. n such that (Pi^n)(n^(n/2)) = (2^(2n))(Gamma(1+n/2))^2.

Extensions

More terms from Eric W. Weisstein, Jan 15 2007