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A307184
Decimal expansion of the fraction of occupied places on an infinite lattice cover with 4-length segments.
0
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
0,1
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).
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.
FORMULA
4*Integral_{x=0..1} e^(2*(t + t^2/2 + t^3/3)) dx / e^(11/3).
EXAMPLE
0.80389347991537697266629741950321342054687916485770835923972993280709456095...
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]]
PROG
(PARI) intnum(t=0, 1, 4*exp(2*(t + t^2/2 + t^3/3) - 11/3)) \\ Michel Marcus, May 10 2019
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Philipp O. Tsvetkov, Mar 28 2019
STATUS
approved