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%I #19 Nov 11 2014 04:55:14
%S 1,2,4,16,48,192,704,3072,12032,52736,219136,985088,4218880,19144704,
%T 84066304,387088384,1725497344,7989886976,36128948224,168658206720,
%U 770103574528,3611291549696,16636941697024,78453223194624,363787840389120,1721209150504960
%N Number of rooted binary trees with n leaves and each internal vertex colored in one of two colors.
%C For n>1, a(n) is the number of bipolar networks one can build from n identical impedances by combining smaller networks either in series or in parallel.
%C Also for n>1, given two symmetric binary operations f(x,y) and g(x,y), such as two different means of x and y, one can use them (and just them) to form up to a(n) distinct expressions with n arguments x1,x2,...,x5.
%H Stanislav Sykora, <a href="/A248748/b248748.txt">Table of n, a(n) for n = 1..1000</a>
%F a(n) = A000992(n)*2^(n-1).
%e a(5)=48 because there are three binary trees with 5 leaves, namely, (1,((1,1),(1,1))); (1,(1,(1,(1,1)))); (1,((1,1),(1,(1,1))); and each of their four (5-1) internal vertices can be colored in 2 ways, giving rise to 3*2^4 = 48 possibilities. The "coloring" can be indicated by means of two different kinds of parentheses, for example (1,[(1,1),[1,1]]).
%e It also implies that 5 identical impedances can be wired together in 48 ways, iterating only simple series/parallel bondings.
%e Also, given two different means f(x,y) and g(x,y) of two numbers (e.g., an arithmetic and a geometric one), these can be combined to form 48 distinct means of 5 arguments x1,x2,x3,x4,x5. One such mean, for example, is f(x1,g(f(x2,x3),g(x4,x5))), corresponding to (1,[(1,1),[1,1]]).
%o (PARI) v=vector(1000); v[1]=1; \\ Use any desired size
%o for(n=2,#v, v[n]=sum(k=1,n\2,v[k]*v[n-k])); \\ v = A000992
%o for(n=1,#v, v[n]*=2^(n-1)); v \\ Final multiplication and result display
%Y Cf. A000992.
%K nonn
%O 1,2
%A _Stanislav Sykora_, Oct 13 2014
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