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A335344
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Expansion of e.g.f. exp(x^2/(2*(1 - x)^2)).
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7
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1, 0, 1, 6, 39, 300, 2715, 28350, 335265, 4422600, 64298745, 1020477150, 17542820295, 324552648420, 6426708843555, 135573281994150, 3034400481137025, 71801374285040400, 1790523094644709425, 46923435009924823350, 1289032229351717425575
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OFFSET
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0,4
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COMMENTS
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For n>0, a(n) is also the number of ways to split n people into nonempty groups, have each group sit around a circular table, and select 2 people from each table (where two seating arrangements are considered identical if each person has the same left neighbors in both of them). See example below. - Enrique Navarrete, Oct 01 2023
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LINKS
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FORMULA
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a(0) = 1; a(n) = Sum_{k=1..n} binomial(n-1,k-1) * A001286(k) * a(n-k).
D-finite with recurrence a(n) = -(n - 1)*(3*n - 7)*a(n - 2) + 3*(n - 1)*a(n - 1) + (n - 1)*(n - 2)*(n - 3)*a(n - 3). - Robert Israel, Jun 04 2020
a(n) ~ n^(n - 1/6) * exp(1/6 - n^(1/3)/2 + 3*n^(2/3)/2 - n) / sqrt(3). - Vaclav Kotesovec, Jun 11 2020
a(n) = n! * Sum_{k=0..floor(n/2)} binomial(n-1,n-2*k)/(2^k * k!). - Seiichi Manyama, Mar 16 2023
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EXAMPLE
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For n = 5, using one table, there are 4! circular seatings and binomial(5,2) ways to select 2 persons, hence 240 ways. Using two tables, the only way we can select 2 persons from each one is seating 3 persons in one table and 2 in the other, which can be done in 20 ways; then choosing 2 persons from each table can be done in 3 ways, hence giving another 60 ways for a total of 300. - Enrique Navarrete, Oct 01 2023
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MAPLE
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f:= gfun:-rectoproc({a(n) = -(n-1)*(3*n-7)*a(n-2) + 3*(n-1)*a(n-1) + (n - 1)*(n - 2)*(n - 3)*a(n-3), a(0)=1, a(1)=0, a(2)=1}, a(n), remember):
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MATHEMATICA
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nmax = 20; CoefficientList[Series[Exp[x^2/(2 (1 - x)^2)], {x, 0, nmax}], x] Range[0, nmax]!
a[0] = 1; a[n_] := a[n] = (1/2) Sum[Binomial[n - 1, k - 1] (k - 1) k! a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 20}]
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PROG
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(PARI) seq(n)=Vec(serlaplace(exp(x^2/(2*(1 - x)^2) + O(x*x^n)))) \\ Andrew Howroyd, Jun 02 2020
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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