A307154 Decimal expansion of the fraction of occupied places on an infinite lattice cover with 3-length segments.

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

Offset: 0

Author

Philipp O. Tsvetkov, Mar 27 2019

Keywords

• nonn
• cons

Comments

Solution of the discrete parking problem when infinite lattice randomly filled with 3-length segments.

Solution of the discrete parking problem when infinite lattice randomly filled with 2-length segments is equal to 1-1/e^2 (see A219863).

Also, the limit of a(n) = (3 + 2*(n-3)*a(n-3) + (n-1)*(n-3)*a(n-1))/(n*(n-2)); a(0) = 0; a(1) = 0; a(2) = 0 as n tends to infinity.

If the length of the segments that randomly cover infinite lattice tends to infinity, then the fraction of occupied places is equal to Rényi's parking constant (see A050996).

Examples

			0.8236529631773383369006718778116478872139236620539298680914372350071822...
		

Links

• D. G. Radcliffe, Fat men sitting at a bar
• Philipp O. Tsvetkov, Stoichiometry of irreversible ligand binding to a one-dimensional lattice, Scientific Reports, Springer Nature (2020) Vol. 10, Article number: 21308.
• Eric Weisstein's World of Mathematics, Dawson's Integral

Crossrefs

Cf. A050996, A087654, A099288, A219863, A307131, A307132.

Programs

Maple

evalf(3*sqrt(Pi)*(erfi(2)-erfi(1))/(2*exp(4)), 120) # Vaclav Kotesovec, Mar 28 2019

Mathematica

N[-((3 DawsonF[1])/E^3) + 3 DawsonF[2], 200] // RealDigits

PARI

-imag(3*sqrt(Pi)*(erfc(2*I) - erfc(1*I)) / (2*exp(4))) \\ Michel Marcus, May 10 2019

Formula

Equals 3*(Dawson(2) - Dawson(1)/e^3).

Equals 3*sqrt(Pi)*(erfi(2) - erfi(1)) / (2*exp(4)).