monotone functions from 2^X to 2^X

[Edwin Clark]

In case anyone else saw this question on sci.math or sci.math.research and
thinks to submit the sequence. I have just submitted it. This formula was
found by Robert Israel.

%I A000001
%S A000001 1, 3, 36, 8000, 796594176, 25039893834551321901, 230156231509903526722108570920314496786496, 478651764962008689839230538296564128023598629748415103570025502338085999191479922367872, 98534292228263589774008553788366851536013089042468221936964638804748172908858256405086673539573392086514266309013017967651602746308015382160831115150912680742123497939800976219242496
%N A000001 The number of monotone functions f: 2^X -> 2^X where 2^X is the powerset of an n-set X. Here f is monotone means that if A is a subset of B then f(A) is a subset of f(B).
%C A000001 Proof of formula by Robert Israel:  If f is monotone, then for each x in X the set G(x) = {A in 2^X: x in f(A)} is an upset, i.e., if A is in G(x) and A \subset B then B is in G(x). Conversely, if for each x in X the set G(x) is an upset, then f is monotone. And the family {G(x): x in X} determines f, since f(A) = {x: A is in G(x)}.  So the cardinality of the set of  monotone set-functions is N(|X|)^|X| where N(|X|) is the cardinality of  the set of upsets G of 2^X, or equivalently monotone Boolean functions.  That is sequence A000372 .
%F A000001 a(n) = A000372[n]^n.
%Y A000001 Cf. A000372
%O A000001 0
%K A000001 ,hard,more,nonn,
%A A000001 W. Edwin Clark (eclark at math.usf.edu), Nov 06 2003
RH 
RA 65.32.45.100
RU 
RI 

A second question in the same posting is answered (by Robert and me
independently) and leads to the following sequenc:

%I A000001
%S A000001 1, 2, 16, 4096, 4294967296, 1208925819614629174706176,
6277101735386680763835789423207666416102355444464034512896
%N A000001 The number of functions f:2^X->2^X where X is an n element
set such that f(A) is a subset of A for all A in 2^X. (2^X denotes the
powerset of X.)
%F A000001 a(n) = 2^sum(i*binomial(n,i),i=0..n) = 2^(2^(n-1)*n)
 
%e A000001 a(1) = 2 since 2^{1} = { {}, {1}} and the functions f
: 2^{1}->2^{1} satisfying f(A) is a subset of A for all A are g and h
where 
g({})={}, g({1})={} and h({}) = {}, h({1})={1}.
%p A000001 a:=n->2^(n*2^(n-1));
%O A000001 0
%K A000001 ,easy,nonn,
%A A000001 W. Edwin Clark (eclark at math.usf.edu), Nov 06 2003
RH 
RA 65.32.45.100
RU