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A152876
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Number of permutations of {1,2,...,n} having no consecutive triples of the form (odd, even, odd) or (even, odd, even).
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2
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1, 1, 2, 4, 16, 60, 288, 1584, 10368, 74880, 604800, 5356800, 51840000, 544320000, 6147187200, 74579097600, 962415820800, 13241346048000, 192255565824000, 2957575348224000, 47721518530560000, 811595019755520000, 14407079038894080000, 268390402745794560000
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OFFSET
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0,3
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COMMENTS
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LINKS
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FORMULA
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a(2n) = (n!)^2*Sum_{j=0..floor(n/2)} (binomial(n-j, j))^2*(2*n^2 - 6*j*n + 6*j^2)/(n-j)^2;
a(2n+1) = n!*(n+1)!*Sum_{j=1..floor(n/2)} binomial(n+1-j, j)*binomial(n-j, j)*(2*n^2 + 2*n - 6*j*n - 3*j + 6*j^2)/((n+1-j)*(n-j)).
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EXAMPLE
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a(3) = 4 because we have 132, 213, 231 and 312.
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MAPLE
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ae := proc (n) options operator, arrow: 2*(sum((n^2-3*n*j+3*j^2)*binomial(n-j, j)^2/(n-j)^2, j = 0 .. floor((1/2)*n))) end proc: ao := proc (n) options operator, arrow: sum(binomial(n+1-k, k)*binomial(n-k, k)*(2*n^2+2*n-6*k*n-3*k+6*k^2)/((n+1-k)*(n-k)), k = 0 .. floor((1/2)*n)) end proc: a := proc (n) if `mod`(n, 2) = 0 then factorial((1/2)*n)^2*ae((1/2)*n) else factorial((1/2)*n-1/2)*factorial((1/2)*n+1/2)*ao((1/2)*n-1/2) end if end proc: 1, 1, seq(a(n), n = 2 .. 22);
# second Maple program:
b:= proc(o, u, t) option remember; `if`(u+o=0, 1,
`if`(t=4, 0, o*b(o-1, u, `if`(t=3, 5, 2)))+
`if`(t=5, 0, u*b(o, u-1, `if`(t=2, 4, 3))))
end:
a:= n-> b(ceil(n/2), floor(n/2), 1):
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MATHEMATICA
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b[o_, u_, t_] := b[o, u, t] = If[u+o == 0, 1, If[t==4, 0, o*b[o-1, u, If[t==3, 5, 2]]] + If[t==5, 0, u*b[o, u-1, If[t==2, 4, 3]]]]; a[n_] := b[Ceiling[n/2], Floor[n/2], 1]; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Jun 29 2015, after Alois P. Heinz *)
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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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