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EXAMPLE
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+---------+
| o o o | a(1) = 2
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| o o |
+---------------------------------------------+
| o o o o o o o o o o o o o o o | a(2) = 7
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| o o o o o o o o o o o o |
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| o o o o o o o |
+---------------------------------------------+
a(3) = 52 while A002658(4) = 56 because there are 56 - 52 = 4 free binary trees admitting height 3 which have two rootings, while the rest have only one rooting. The four trees have degree sequences 32111, 322111, 3222111, 3321111. - Michael Somos, Sep 02 2012
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MATHEMATICA
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bin2[n_] = Binomial[n, 2];
bin3[n_] = Binomial[n, 3];
p[0] = q[0] = 0;
p[1] = q[1] = 1;
q[h1_] := q[h1] = With[{h = h1-1}, q[h] + p[h]];
p[h1_] := p[h1] = With[{h = h1-1}, bin2[1 + p[h]] + p[h] q[h]];
a[h_] := a[h] = bin3[2 + p[h]] + bin2[1 + p[h]] q[h];
b[h_] := b[h] = bin2[1 + p[h]];
e[h_, i_] := e[h, i] = 1 + Sum[d[j, i], {j, h-1}];
d[h_, h_] := 0; d[h_, i_] := p[h] /; i > h;
d[h1_, i1_] := d[h1, i1] = With[{h = h1-1, i = i1-1}, bin2[1 + d[h, i]] + d[h, i] e[h, i]]; d[h_, 1] := d[h, 1] = p[h] - p[h-1];
e[h_, 1] := e[h, 1] = p[h-1];
t1[h_] := Sum[a[h-i] - bin3[2 + d[h-i, i]] - bin2[1 + d[h-i, i]] e[h-i, i], {i, Quotient[h, 2]}];
t2[h_] := Sum[b[h-i+1] - bin2[1 + d[h-i+1, i]], {i, Quotient[h+1, 2]}];
t[h_] := bin2[1 + p[h]] + t1[h] + t2[h];
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