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A307154
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Decimal expansion of the fraction of occupied places on an infinite lattice cover with 3-length segments.
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1
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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
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OFFSET
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0,1
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COMMENTS
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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).
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LINKS
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FORMULA
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Equals 3*(Dawson(2) - Dawson(1)/e^3).
Equals 3*sqrt(Pi)*(erfi(2) - erfi(1)) / (2*exp(4)).
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EXAMPLE
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0.8236529631773383369006718778116478872139236620539298680914372350071822...
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MAPLE
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evalf(3*sqrt(Pi)*(erfi(2)-erfi(1))/(2*exp(4)), 120) # Vaclav Kotesovec, Mar 28 2019
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MATHEMATICA
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N[-((3 DawsonF[1])/E^3) + 3 DawsonF[2], 200] // RealDigits
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PROG
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(PARI) -imag(3*sqrt(Pi)*(erfc(2*I) - erfc(1*I)) / (2*exp(4))) \\ Michel Marcus, May 10 2019
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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