A306668 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.

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

Offset: 0

Author

Alois P. Heinz, Mar 04 2019

Keywords

• sign

Comments

The total number of binary bracketings of 0^0^...^0 with n 0's is A000108(n-1) for n > 0.

Examples

			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.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1670
• Eric Weisstein's World of Mathematics, Binary Bracketing
• Wikipedia, Zero to the power of zero

Crossrefs

Cf. A000079, A000108, A055113, A111160, A211192.

Programs

Maple

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);

Mathematica

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)];
a /@ Range[0, 29] (* Jean-François Alcover, Apr 02 2021, after 2nd Maple program *)

Formula

a(n) = A111160(n-1) - A055113(n) for n > 0.

a(n) is odd <=> n in { A000079 }.