A371943 Number of permutations that end with a consecutive pattern 123, and avoid consecutive patterns 123 and 213 elsewhere.
0, 0, 0, 1, 2, 10, 28, 116, 388, 1588, 5960, 25168, 102856, 453608, 1985008, 9163360, 42486128, 205065136, 1000056928, 5035366208, 25689681760, 134588839648, 715328668736, 3889568161408, 21463055829568, 120839175460160, 690344333849728, 4015753752384256
Offset: 0
Keywords
Examples
For n=0, 1, 2, there are no permutations ending with 123. Hence, a(0)=a(1)=a(2)=0. For n=3, a(3)=1 since 123 is the only permutation that ends with 123. For n=4, a(4)=2 with qualified permutations 3124, 4123. For n=5, a(5)=10 with qualified permutations 14235, 15234, 24135, 25134, 34125, 35124, 43125, 45123, 53124, 54123.
Links
- Sergi Elizalde and Yixin Lin, Penney's game for permutations, arXiv:2404.06585 [math.CO], 2024.
Crossrefs
Cf. A059480.
Programs
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Maple
a:= proc(n) option remember; `if`(n<3, 0, `if`(n=3, 1, 2*a(n-1)+2*(n-2)*(a(n-2)-a(n-3))-(n-2)*(n-3)*a(n-4))) end: seq(a(n), n=0..30); # Alois P. Heinz, Apr 13 2024 -
Mathematica
a[n_]:=a[n]=If[n<3, 0, If[n==3, 1, 2*a[n-1]+2*(n-2)*(a[n-2]-a[n-3])-(n-2)*(n-3)*a[n-4]]]; Table[a[n], {n,0,27}] (* Stefano Spezia, Apr 13 2024 *) RecurrenceTable[{a[n] == 2*a[n-1] + 2*(n-2)*(a[n-2] - a[n-3]) - (n-2)*(n-3)*a[n-4], a[0] == 0, a[1] == 0, a[2] == 0, a[3] == 1}, a, {n, 0, 30}] (* Vaclav Kotesovec, Apr 14 2024 *) nmax = 30; FullSimplify[CoefficientList[Series[-1 + E^(x*(2 + x)/2) * (1 + x) + E^((1 + x)^2/2) * Sqrt[Pi/2] * (2 + x)*(Erf[1/Sqrt[2]] - Erf[(1 + x)/Sqrt[2]]), {x, 0, nmax}], x] * Range[0, nmax]!] (* Vaclav Kotesovec, Apr 14 2024 *) -
Python
def aList(len): b = [0, 0, 0, 1, 4] a = [0, 0, 0, 1, 2] for i in range(4, len): b.append(b[i] + i * b[i - 1]) a.append(a[i] + i * a[i - 1] + b[i]) return a print(aList(27))
Formula
a(0)=a(1)=a(2)=0, a(3)=1, a(4)=2; a(n) = a(n-1)+(n-1)*a(n-2)+b(n-1), where b(n) = b(n-1)+(n-1)*b(n-2) is the same sequence as A059480, up to the first initial terms. Here, our b(n) has initial terms 0, 0, 0, 1, 4.
From Vaclav Kotesovec, Apr 14 2024: (Start)
a(n) ~ c * n^((n+1)/2) * exp(sqrt(n) - n/2), where c = exp(-1/4) / sqrt(2) - exp(1/4) * sqrt(Pi) * erfc(1/sqrt(2)) / 2 = 0.189615662815288097469466802437...
E.g.f.: -1 + exp(x*(2 + x)/2) * (1 + x) + exp((1 + x)^2/2) * sqrt(Pi/2) * (2 + x) * (erf(1/sqrt(2)) - erf((1 + x)/sqrt(2))). (End)
Comments