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A306668
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Difference between numbers of binary bracketings of 0^0^...^0 with n 0's giving the result 1 and those giving the result 0, with conventions that 0^0=1^0=1^1=1, 0^1=0.
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4
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0, -1, 1, 0, 3, 4, 20, 50, 189, 588, 2100, 7116, 25344, 89298, 321178, 1156298, 4206059, 15356796, 56424836, 208137800, 771229684, 2867771004, 10700980956, 40050890172, 150328400292, 565699287186, 2133889856550, 8067040670100, 30559571239890, 115986196679730
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OFFSET
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0,5
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COMMENTS
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The total number of binary bracketings of 0^0^...^0 with n 0's is A000108(n-1) for n > 0.
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LINKS
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FORMULA
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EXAMPLE
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There are A000108(3) = 5 binary bracketings of 0^0^0^0: ((0^0)^0)^0, (0^0)^(0^0), (0^(0^0))^0, 0^((0^0)^0), 0^(0^(0^0)). Only 0^((0^0)^0) evaluates to 0: 0^((0^0)^0) = 0^(1^0) = 0^1 = 0. The four other bracketings evaluate to 1. Thus a(4) = 4-1 = 3.
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MAPLE
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b:= proc(n) option remember; `if`(n<2, [n, 0], add(((f, g)-> [f[1]*g[2],
f[1]*g[1] +f[2]*g[1] +f[2]*g[2]])(b(i), b(n-i)), i=1..n-1))
end:
a:= n-> (v-> v[2]-v[1])(b(n)):
seq(a(n), n=0..29);
# second Maple program:
a:= proc(n) option remember; `if`(n<2, -n, ((35*n^3-147*n^2+220*n-120)*
a(n-1)+18*(n-2)*(5*n-6)*(2*n-5)*a(n-2))/((2*(5*n-11))*(2*n-1)*n))
end:
seq(a(n), n=0..29);
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MATHEMATICA
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a[n_] := a[n] = If[n<2, -n, ((35n^3 - 147n^2 + 220n - 120) a[n-1] + 18(n-2) (5n - 6)(2n - 5) a[n-2])/((2(5n - 11))(2n - 1)n)];
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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