A307184 Decimal expansion of the fraction of occupied places on an infinite lattice cover with 4-length segments.

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

Offset: 0

Author

Philipp O. Tsvetkov, Mar 28 2019

Keywords

• nonn
• cons

Comments

The solution of the discrete parking problem when infinite lattice randomly filled with L-length segments at L=4.

At L=3 it is equal to 3*(Dawson(2) - Dawson(1)/e^3) (see A307154).

At L=2 it is equal to 1-1/e^2 (see A219863).

The general solution of the discrete parking problem when infinite lattice randomly filled with L-length segments is equal to L*e(-2H(L-1))*Integral_{x=0..1} e^(2*(t + t^2/2 + t^3/3 + ... + t^(L-1)/(L-1))) dx, where H(L) is harmonic number.

Also, the limit of the following recurrence as n tends to infinity: a(n) = (4 + 2(n-4)*a(n-4) + (n-1)*(n-4)*a(n-1))/(n*(n-3)); a(0) = 0; a(1) = 0; a(2) = 0; a(3) = 0.

If L tends to infinity, then the fraction of occupied places is equal to Rényi's parking constant (see A050996).

Examples

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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.

Crossrefs

Cf. A219863, A050996, A307132, A307131, A307154.

Programs

Maple

evalf(Integrate(4*exp(2*(t + t^2/2 + t^3/3) - 11/3), t= 0..1), 120); # Vaclav Kotesovec, Mar 28 2019

Mathematica

RealDigits[ N[(4*Integrate[E^(2*(t + t^2/2 + t^3/3)), {t, 0, 1}])/E^(11/3), 200]][[1]]

PARI

intnum(t=0, 1, 4*exp(2*(t + t^2/2 + t^3/3) - 11/3)) \\ Michel Marcus, May 10 2019

Formula

4*Integral_{x=0..1} e^(2*(t + t^2/2 + t^3/3)) dx / e^(11/3).