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A344216
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a(n) = n!*((n+1)/2 + 2*Sum_{k=2..n-1}(n-k)/(k+1)).
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3
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1, 3, 16, 104, 768, 6336, 57888, 581472, 6379200, 75977280, 977045760, 13499930880, 199537067520, 3142504512000, 52546707763200, 929908914278400, 17366044153651200, 341336836618444800, 7044417438363648000
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OFFSET
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1,2
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COMMENTS
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Conjecture: a(n) is the number of linear intervals in the weak order on the symmetric group S_n. An interval is linear if it is isomorphic to a total order. The conjecture has been checked up to a(7) = 57888.
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LINKS
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FORMULA
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a(n) = (1/2) * n! * (4 * (n + 1) * H(n) - 9*n + 3), where H(n) are the harmonic numbers H(n) = A001008(n)/A002805(n).
a(n) = n! * [x^n] (3 - 8*x - 4*log(1 - x))/(2*(x - 1)^2).
a(n) = ((2*n^2 - 5*n - 1)*a(n-1) - (n^3 - 3*n^2 + 2*n)*a(n-2))/(n - 3) for n >= 4. (End)
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EXAMPLE
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For S_3, among the 17 intervals in the hexagon-shaped lattice, only the full lattice is not linear.
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MAPLE
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a := n -> (1/2)*n!*(4*(n + 1)*harmonic(n) - 9*n + 3):
# Or:
egf := (3 - 8*x - 4*ln(1 - x))/(2*(x - 1)^2):
ser := series(egf, x, 24): a := n -> n!*coeff(ser, x, n):
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MATHEMATICA
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Join[{1}, RecurrenceTable[{(n - 3) a[n] == (2 n^2 - 5 n - 1) a[n - 1] - (n^3 - 3 n^2 + 2 n) a[n - 2], a[2] == 3, a[3] == 16}, a, {n, 2, 19}]] (* Peter Luschny, May 13 2021 *)
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PROG
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(Sage)
def a(n):
return factorial(n)*((n+1)/2+2*sum((n-k)/(k+1) for k in range(2, n)))
(PARI) a(n) = n!*((n+1)/2+2*sum(k=2, n-1, (n-k)/(k+1))); \\ Michel Marcus, May 13 2021
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CROSSREFS
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Cf. A007767 for all intervals in the weak order on S_n.
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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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