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A216041
Number of redundant function representations of x^x^...^x with n x's and parentheses inserted in all possible ways.
2
0, 0, 0, 1, 5, 22, 84, 314, 1144, 4143, 14954, 54020, 195526, 709927, 2586629, 9459464, 34722823, 127923631, 472950024, 1754436962, 6528898588, 24369211839, 91214280785, 342315888666, 1287836972679, 4856186764942, 18351269337823, 69488543849735
OFFSET
1,5
COMMENTS
A000081(n) distinct functions are representable as x -> x^x^...^x with n x's and parentheses inserted in all possible ways. The number of valid parenthesizations is A000108(n-1). So the number of redundant representations is A000108(n-1) - A000081(n).
For n>=6 we have a(n) > A000081(n), so the number of redundant function representations is larger than the number of essential representations.
LINKS
FORMULA
a(n) = A000108(n-1) - A000081(n).
EXAMPLE
a(4) = 1: there are A000108(3) = 5 valid parenthesizations of x^x^x^x, namely x^(x^(x^x)), x^((x^x)^x), (x^(x^x))^x, (x^x)^(x^x), ((x^x)^x)^x, but only A000081(4) = 4 distinct functions. (x^(x^x))^x and (x^x)^(x^x) represent the same function x -> x^(x^x*x), so 1 representation is redundant.
MAPLE
with(numtheory):
b:= proc(n) option remember; `if`(n<=1, n,
(add(add(d*b(d), d=divisors(j))*b(n-j), j=1..n-1))/(n-1))
end:
C:= n-> binomial(2*n, n)/(n+1):
a:= n-> C(n-1) -b(n):
seq(a(n), n=1..40);
MATHEMATICA
b[n_] := b[n] = If[n <= 1, n, Sum[DivisorSum[j, #*b[#]&]*b[n-j], {j, 1, n-1}]/(n-1)];
c[n_] := Binomial[2*n, n]/(n+1);
a[n_] := c[n-1] - b[n];
Table[a[n], {n, 1, 40}] (* Jean-François Alcover, Mar 24 2017, translated from Maple *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Alois P. Heinz, Aug 30 2012
STATUS
approved