cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A307131 Numerator of the expected fraction of occupied places on n-length lattice randomly filled with 2-length segments.

Original entry on oeis.org

1, 2, 5, 4, 37, 52, 349, 338, 11873, 14554, 157567, 466498, 11994551, 41582906, 618626159, 614191052, 7545655031, 92853583996, 1755370057489, 8737266957604, 365468962351379, 2002633668589496, 45904893141293831
Offset: 1

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Author

Philipp O. Tsvetkov, Mar 26 2019

Keywords

Comments

The limit of expected fraction of occupied places on n-length lattice randomly filled with 2-length segments at n tends to infinity is equal to 1-1/e^2 (see A219863).

Examples

			0, 1, 2/3, 5/6, 4/5, 37/45, 52/63, 349/420, 338/405, 11873/14175, ...
		

Crossrefs

Cf. A219863, A231580, A307132 (denominators).

Programs

  • Mathematica
    RecurrenceTable[{f[n] == (2 + 2 (n - 2) f[n - 2] + (n - 1) (n - 2) f[n - 1])/(n (n - 1)),f[0] == 0, f[1] == 0}, f, {n, 2, 100}] // Numerator

Formula

Numerator of f(n), where f(0)=0; f(1)=0 and f(n) = (2 + 2(n-2)f(n-2) + (n-1)(n-2)f(n-1))/(n(n-1)) for n>1.

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

Original entry on oeis.org

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

Views

Author

Philipp O. Tsvetkov, Mar 27 2019

Keywords

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...
		

Crossrefs

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)).

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

Original entry on oeis.org

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

Views

Author

Philipp O. Tsvetkov, Mar 28 2019

Keywords

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

			0.80389347991537697266629741950321342054687916485770835923972993280709456095...
		

Crossrefs

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).
Showing 1-3 of 3 results.