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A362596
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Number of parking functions of size n avoiding the patterns 213 and 321.
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5
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1, 1, 3, 13, 60, 275, 1238, 5480, 23922, 103267, 441798, 1876366, 7921488, 33275758, 139194812, 580180598, 2410827422, 9990993443, 41308185542, 170439003998, 701953309592, 2886284314298, 11850433719572, 48591008205608, 199002198798980, 814117064956430
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OFFSET
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0,3
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LINKS
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FORMULA
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For n>=1, a(n) = (n^2 - 3*n + 4)/4*A000108(n) + 4^(n - 1)/2.
G.f.: 1+((9*x^2 - 10*x + 2)*sqrt(1 - 4*x) - 23*x^2 + 14*x - 2)/(2*(1 - 4*x)^(3/2)*x).
D-finite with recurrence 2*(n+1)*a(n) +2*(-15*n+1)*a(n-1) +(167*n-193)*a(n-2) +2*(-204*n+467)*a(n-3) +184*(2*n-7)*a(n-4)=0. - R. J. Mathar, Jan 11 2024
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EXAMPLE
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For n=3 the a(3)=13 parking functions, given in block notation, are {1},{2},{3}; {1,2},{},{3}; {1,2},{3},{}; {1},{2,3},{}; {1,2,3},{},{}; {1},{3},{2}; {1,3},{},{2}; {1,3},{2},{}; {2},{3},{1}; {2,3},{},{1}; {2,3},{1},{}; {3},{1},{2}; {3},{1,2},{}.
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PROG
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(PARI) a(n)=if(n==0, 1, (n^2 - 3*n + 4)*binomial(2*n, n)/(4*(n+1)) + 4^n/8) \\ Andrew Howroyd, Apr 27 2023
(Python)
from math import comb
def A362596(n): return ((n*(n-3)+4)*comb(n<<1, n)//(n+1)>>2)+(1<<(n<<1)-3) if n>1 else 1 # Chai Wah Wu, Apr 27 2023
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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