First compute the OGF $T_1(z)$ for rooted unlabeled tree with all
vertices of odd degree except for the root. This has the combinatorial class equation
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\mathcal{T}_1 = \mathcal{Z} + \mathcal{Z}\times
(\textsc{MSET}_{=2}(\mathcal{T}_1)
+\textsc{MSET}_{=4}(\mathcal{T}_1)
+\textsc{MSET}_{=6}(\mathcal{T}_1)
+\cdots)$$
which is
$$\mathcal{T}_1 = \mathcal{Z}\times
(\textsc{MSET}_{=0}(\mathcal{T}_1)
+\textsc{MSET}_{=2}(\mathcal{T}_1)
+\textsc{MSET}_{=4}(\mathcal{T}_1)
+\textsc{MSET}_{=6}(\mathcal{T}_1)
+\cdots)$$
Now the OGF of the cycle index of the symmetric group $S_n$ is
$$\exp
\left(a_1 w + a_2 \frac{w^2}{2} + a_3 \frac{w^3}{3}
+ \cdots\right).$$
It follows that the operator $\textsc{MSET}_{\mathrm{even}}$ applied to
$T_1(z)$ is given by
$$\frac{1}{2}\exp
\left( T_1(z) + \frac{T_1(z^2)}{2} + \frac{T_1(z^3)}{3}
+\cdots\right)
\\+ \frac{1}{2}\exp
\left( -T_1(z) + \frac{T_1(z^2)}{2} - \frac{T_1(z^3)}{3}
+\cdots\right)$$
or
$$\frac{1}{2}\exp
\left(\sum_{q\ge 1} \frac{T_1(z^q)}{q}\right)
+ \frac{1}{2}\exp
\left(\sum_{q\ge 1} (-1)^q \frac{T_1(z^q)}{q}\right).$$
This gives the functional equation
$$T_1(z) =
\frac{z}{2}\exp
\left(\sum_{q\ge 1} \frac{T_1(z^q)}{q}\right)
+ \frac{z}{2}\exp
\left(\sum_{q\ge 1} (-1)^q \frac{T_1(z^q)}{q}\right).$$
To extract coefficients from this start by observing that if
$$A(z) = \sum_{n\ge 1} A_n z^n =
\sum_{q\ge 1} \frac{T(z^q)}{q}
\quad\text{and}\quad
B(z) = \sum_{n\ge 1} B_n z^n =
\sum_{q\ge 1} (-1)^q \frac{T(z^q)}{q}$$
then
$$A_n = \sum_{q|n} \frac{T_{n/q}}{q}
\quad\text{and}\quad
B_n = \sum_{q|n} (-1)^q \frac{T_{n/q}}{q}.$$
The functional equation now becomes
$$T_1(z) = \frac{z}{2} \exp A_1(z) + \frac{z}{2} \exp B_1(z)$$
and coefficient extraction produces
$$[z^n] T_1(z) =
\frac{1}{2}
\sum_{\lambda\vdash n-1}
\frac{1}{l(\lambda)!} {l(\lambda) \choose f}
\prod_{\lambda_i\in\lambda} A_{1,\lambda_i}
+\frac{1}{2}
\sum_{\lambda\vdash n-1}
\frac{1}{l(\lambda)!} {l(\lambda) \choose f}
\prod_{\lambda_i\in\lambda} B_{1,\lambda_i}$$
where the $f_i$ are the multiplicities of the partition
i.e. $\lambda = 1^{f_1} 2^{f_2} 3^{f_3}\cdots$
This recurrence gives a serviceable but not excellent recurrence to
compute $T_{1,n}.$
Now observe that the species $\mathcal{T}_2$ of rooted unlabeled trees
with odd vertex degree throughout is given by
$$\mathcal{T}_2 =
\mathcal{Z} \times \textsc{MSET}_{\mathrm{odd}} (\mathcal{T}_1).$$
This gives the functional equation
$$T_2(z) =
\frac{z}{2}\exp
\left(\sum_{q\ge 1} \frac{T_1(z^q)}{q}\right)
- \frac{z}{2}\exp
\left(\sum_{q\ge 1} (-1)^q \frac{T_1(z^q)}{q}\right).$$
Proceeding as before we extract coefficients to obtain
$$[z^n] T_2(z) =
\frac{1}{2}
\sum_{\lambda\vdash n-1}
\frac{1}{l(\lambda)!} {l(\lambda) \choose f}
\prod_{\lambda_i\in\lambda} A_{1,\lambda_i}
-\frac{1}{2}
\sum_{\lambda\vdash n-1}
\frac{1}{l(\lambda)!} {l(\lambda) \choose f}
\prod_{\lambda_i\in\lambda} B_{1,\lambda_i}$$
Finally use the dissimilarity characteristic theorem for trees as
presented by Harary and Palmer (section 3.2) to obtain for the
generating function $T(z)$ of unrooted unlabeled trees with odd vertex
degree the relation
$$T(z) = T_2(z) - \frac{1}{2}(T_1(z)^2-T_1(z^2)).$$
Since $T_1(z)$ only has non-zero coefficients $T_{1,n}$ at odd values
of $n$ we obtain
$$[z^{2n}] T(z)
= T_{2,2n}
- \frac{1}{2}
\left(\sum_{q=0}^{n-1} T_{1,2q+1} T_{1, 2n-2q-1} -T_{1,n}\right).$$
This gives the sequence (even indices)
$$1, 1, 2, 3, 7, 13, 32, 74, 192, 497, 1379, 3844,
\\ 11111, 32500, 96977, 292600, 894353,\ldots$$
The first two generating functions were verified with Maple's
combstruct package as follows.
with(combstruct);
with(combinat);
with(numtheory);
evens :=
proc(struct)
local rval;
if type(struct, function) then
if op(0, struct) = Prod then
return evens(op(2, struct))
else
rval := add(evens(op(q, struct)),
q=1..nops(struct));
if type(nops(struct), even) then
return 1+rval;
fi;
return rval;
fi;
fi;
return 1;
end;
T1or2 :=
proc(n, what)
option remember;
local Apart, Bpart, A, B, p, cf;
if n=1 then return 1 fi;
A := n -> add(T1(n/q)/q, q in divisors(n));
B := n -> add((-1)^q*T1(n/q)/q, q in divisors(n));
Apart := 0;
Bpart := 0;
p := firstpart(n-1);
while type(p, `list`) do
cf := 1/
mul(q!, q in map(ent->ent[2],
convert(p, `multiset`)));
Apart := Apart + cf*mul(A(q), q in p);
Bpart := Bpart + cf*mul(B(q), q in p);
p := nextpart(p);
od;
if what = T1 then
return (Apart+Bpart)/2;
else
return (Apart-Bpart)/2;
fi;
end;
T1_basic :=
proc(n)
option remember;
local trees, gf;
trees := { T1= Union(Z, Prod(Z, Set(T1, card >=1 ))),
Z=Atom };
gf := add(u^evens(t),
t in allstructs([T1, trees, unlabeled], size=n));
coeff(gf, u, n);
end;
T1 :=
proc(n)
option remember;
if n=1 then return 1 fi;
T1or2(n, T1);
end;
T2_basic :=
proc(n)
option remember;
local trees, gf, t, st;
trees := { T1= Union(Z, Prod(Z, Set(T1, card >=1 ))),
T2= Prod(Z, Set(T1, card >= 1)),
Z=Atom };
gf := 0;
for t in allstructs([T2, trees, unlabeled], size=n) do
st := op(2, t);
if type(nops(st), odd) then
gf := gf + u^evens(st);
fi;
od;
coeff(gf, u, n-1);
end;
T2 :=
proc(n)
option remember;
if n=1 then return 0 fi;
T1or2(n, T2);
end;
T :=
proc(m)
local n;
if type(m, odd) then return 0 fi;
n := m/2;
T2(2*n)
-1/2*(add(T1(2*q+1)*T1(2*n-2*q-1), q=0..n-1)- T1(n));
end;
The last sequence required more work to verify. I used the NAUTY
package to generate the trees and a Perl script to check for odd
degrees as shown below. This confirmed the values from the recurrence
for $n$ up to $20.$ No claims are made regarding proficiency in Perl.
#! /usr/bin/perl -w
#
sub decode_graph {
my ($str) = @_;
sub R {
my (@args) = map {
sprintf "%06b", $_;
} @_;
join '', @args;
}
my (@ents) = map {
ord($_) - 63
} split //, $str;
my $n = shift @ents;
my @adj_data = split //, R(@ents);
my $adj = []; my $pos = 0;
for(my $ind2 = 1; $ind2 < $n; $ind2++){
for(my $ind1 = 0; $ind1 < $ind2; $ind1++){
$adj->[$ind1]->[$ind2] = $adj_data[$pos];
$adj->[$ind2]->[$ind1] = $adj_data[$pos];
$pos++;
}
}
return $adj;
}
MAIN: {
my $mx = shift || 2;
die "out of range for GENG: $mx"
if $mx < 2 || $mx > 32;
for(my $n=2; $n <= $mx; $n+=2){
my $cmd = sprintf "./geng -c %d %d",
$n, $n-1;
my $count = 0;
open GENG, "$cmd 2>/dev/null|";
while(my $tree = <GENG>){
chomp $tree; my $adj = decode_graph $tree;
my $v;
for($v = 0; $v < $n; $v++){
my $deg = 0;
for(my $w = 0; $w < $n; $w++){
my $ent = $adj->[$v]->[$w];
$deg++ if defined($ent) && $ent == 1;
}
last if $deg % 2 == 0;
}
$count++ if $v == $n;
}
close GENG;
print "$count\n";
}
}
This material is inspired by Harary and Palmer,
Graphical Enumeration.
Addendum. For the sake of completeness we treat the labeled
case as well (labeled unrooted trees with odd vertex degree).
The combinatorial class $\mathcal{T}_1$ of labeled rooted trees with
odd vertex degree except at the root has the specification
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\mathcal{T}_1 =
\mathcal{Z}+\mathcal{Z} \times
\left(\textsc{SET}_{=2}(\mathcal{T}_1)
+ \textsc{SET}_{=4}(\mathcal{T}_1)
+ \textsc{SET}_{=6}(\mathcal{T}_1)
+ \cdots\right).$$
This is
$$\mathcal{T}_1 =
\mathcal{Z} \times
\left(\textsc{SET}_{=0}(\mathcal{T}_1)
+ \textsc{SET}_{=2}(\mathcal{T}_1)
+ \textsc{SET}_{=4}(\mathcal{T}_1)
+ \textsc{SET}_{=6}(\mathcal{T}_1)
+ \cdots\right).$$
This gives the functional equation
$$T_1(z) = z \sum_{k\ge 0} \frac{T_1(z)^{2k}}{(2k)!}$$
or
$$T_1(z) =
\frac{z}{2} \exp(T_1(z)) + \frac{z}{2} \exp(-T_1(z)).$$
We can extract coefficients from this using the residue operator.
We have
$$[z^n] T_1(z) = \frac{1}{n} [z^{n-1}] T_1'(z) =
\frac{1}{n} \; \underset{z}{\mathrm{res}} \;
\frac{1}{z^{n}} T_1'(z).$$
Put $T_1(z) = w$ so that
$$z = \frac{2w}{\exp(w)+\exp(-w)}.$$
This gives for the residue
$$\frac{1}{n} \; \underset{w}{\mathrm{res}} \;
\frac{(\exp(w)+\exp(-w))^{n}}{(2w)^{n}}.$$
Now we have
$$(\exp(w)+\exp(-w))^{n}
= \exp(nw)\sum_{q=0}^n {n\choose q} \exp(-2wq).$$
This gives for the residue the value
$$\frac{1}{n\times 2^n} \sum_{p=0}^{n-1} \frac{n^p}{p!}
\sum_{q=0}^n {n\choose q} \frac{(-2q)^{n-1-p}}{(n-1-p)!}.$$
Multiplying by $n!$ because we have an EGF yields
$$T_{1, n} = \frac{1}{2^n} \sum_{p=0}^{n-1} {n-1\choose p} n^p
\sum_{q=0}^n {n\choose q} (-2q)^{n-1-p}
\\ = \frac{1}{2^n} \sum_{q=0}^n {n\choose q} (-2q)^{n-1}
\sum_{p=0}^{n-1} {n-1\choose p} n^p (-2q)^{-p}
\\ = \frac{1}{2^n} \sum_{q=0}^n {n\choose q} (-2q)^{n-1}
\left(1-\frac{n}{2q}\right)^{n-1}
\\ = \frac{1}{2^n}
\sum_{q=0}^n {n\choose q} (n-2q)^{n-1}.$$
This gives at odd indices the sequence
$$1, 3, 65, 3787, 427905, 79549811, 22036379521, 8513206310715,
\\ 4374455745966593, 2885264091484122979, 2376040584184726335681,
\ldots$$
which is OEIS A036778 where we find
confirmation of these computations.
Continuing we find that the combinatorial class $\mathcal{T}_2$ of
rooted labeled trees with odd vertex degree including at the root is
given by
$$\mathcal{T}_2 =
\mathcal{Z} \times \textsc{SET}_{\mathrm{odd}}(\mathcal{T}_1)$$
and has the functional equation
$$T_2(z) = z \sum_{k\ge 0} \frac{T_1(z)^{2k+1}}{(2k+1)!}
= \frac{z}{2} \exp (T_1(z)) - \frac{z}{2} \exp (-T_1(z)).$$
We have
$$T_2'(z) = \frac{1}{z} T_2(z) + T_1(z) T_1'(z).$$
Extracting coefficients we find
$$\frac{T_{2,n}}{(n-1)!} =
\frac{T_{2,n}}{n!} + [z^{n-1}] T_1(z) T_1'(z)$$
which yields with the remaining coefficient extractor
$$T_{2,n} = \frac{n!}{n-1} \; \underset{z}{\mathrm{res}} \;
\frac{1}{z^{n}} T_1(z) T_1'(z).$$
We get from the latter
$$\; \underset{w}{\mathrm{res}} \;
\frac{(\exp(w)+\exp(-w))^{n}}{(2w)^{n}} w
= \frac{1}{2^n} \; \underset{w}{\mathrm{res}} \;
\frac{(\exp(w)+\exp(-w))^{n}}{w^{n-1}}.$$
Similar to before we obtain
$$\frac{n!}{(n-1) \times 2^n} \sum_{p=0}^{n-2} \frac{n^p}{p!}
\sum_{q=0}^n {n\choose q} \frac{(-2q)^{n-2-p}}{(n-2-p)!}
\\ = \frac{n}{2^n}
\sum_{q=0}^n {n\choose q} (n-2q)^{n-2}.$$
The even indices of this give the sequence
$$2, 16, 576, 47104, 6860800, 1562148864, 512260833280, \\
228646878969856, 133296779352342528, 98349146136012390400,\ldots$$
which points us to OEIS A060279
where the calculation is confirmed.
As we are working in the labeled universe the count of unrooted
labeled trees with odd vertex degree throughout is given by
$$\frac{T_{2,n}}{n}$$
which gives the sequence (even indices)
$$1, 4, 96, 5888, 686080, 130179072, 36590059520, 14290429935616,
\\ 7405376630685696, 4917457306800619520,\ldots$$
which is indeed OEIS A007106 as observed by the OP.
