A211192 Consider all distinct functions f representable as x -> x^x^...^x with n x's and parentheses inserted in all possible ways; sequence gives difference between numbers of f with f(0)=1 and numbers of f with f(0)=0, with conventions that 0^0=1^0=1^1=1, 0^1=0.
0, -1, 1, 0, 2, 1, 8, 10, 39, 72, 225, 506, 1434, 3550, 9767, 25391, 69293, 185061, 505843, 1372744, 3769842, 10339104, 28546539, 78890525, 218945822, 608657861, 1697106780, 4740593393, 13272626627, 37224982494, 104599603493, 294384019508, 829836855332
Offset: 0
Keywords
Examples
There are A000081(4) = 4 functions f representable as x -> x^x^...^x with 4 x's and parentheses inserted in all possible ways: ((x^x)^x)^x, (x^x)^(x^x) == (x^(x^x))^x, x^((x^x)^x), x^(x^(x^x)). Only x^((x^x)^x) evaluates to 0 at x=0: 0^((0^0)^0) = 0^(1^0) = 0^1 = 0. Three functions evaluate to 1 at x=0: ((0^0)^0)^0 = (1^0)^0 = 1^0 = 1, (0^0)^(0^0) = 1^1 = 1, 0^(0^(0^0)) = 0^(0^1) = 0^0 = 1. Thus a(4) = 3-1 = 2. a(8) = A222380(8) - A222379(8) = 77 - 38 = 39.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1000
- Alois P. Heinz, Plot of A000081(8) = 115 = 77 + 38 functions with 8 x's
- Wikipedia, Zero to the power of zero
Programs
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Maple
g:= proc(n, i) option remember; `if`(n=0, [0, 1], `if`(i<1, 0, (v->[v[1]- v[2], v[2]])(add(((l, h)-> [binomial(l[2]+l[1]+j-1, j)*(h[1]+h[2]), binomial(l[1]+j-1, j)*h[2]])(g(i-1$2), g(n-i*j, i-1)), j=0..n/i)))) end: a:= n-> (f-> f[1]-f[2])(g(n-1$2)): seq(a(n), n=0..40); -
Mathematica
g[n_, i_] := g[n, i] = If[n==0, {0, 1}, If[i<1, {0, 0}, ({#[[1]]-#[[2]], #[[2]]}&)[Sum[Function[{l, h}, {(h[[1]]+h[[2]])*Binomial[j+l[[1]]+l[[2]] -1, j], h[[2]]*Binomial[j+l[[1]]-1, j]}][g[i-1, i-1]], g[n-i*j, i-1]]], {j, 0, Quotient[n, i]}]]; a[n_] := (#[[1]]-#[[2]]&)[g[n-1, n-1]]; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Feb 22 2017, translated from Maple *)
Formula
From Alois P. Heinz, Mar 01 2019: (Start)
a(n) is even <=> n in { A258592 }.
a(n) is odd <=> n in { A263831 }. (End)
Comments