OFFSET
1,3
COMMENTS
Generates and uses values from A108521, rooted trees with exactly n generators, a generator being a leaf or node with just one child.
LINKS
Robert A. Russell, Table of n, a(n) for n = 1..60
R. A. Russell, How many trees have n nodes with fewer than three neighbors?, MathOverflow, June 2020.
FORMULA
G.f.: A(x) + (x/2-1)*A^2(x) + (x/2)*A(x^2), where A(x) is the g.f. for A108521.
EXAMPLE
For n=4, we have 1) a node with four neighbors, 2) two adjacent nodes with three neighbors each, 3) two adjacent nodes with two neighbors each, and 4) two adjacent nodes, one having two neighbors and the other three neighbors.
MATHEMATICA
a[1] = 1; a[n_] := a[n] = 1+a[n-1] + Total[Product[Binomial[a[i]-1+Count[#, i], Count[#, i]], {i, DeleteCases[DeleteDuplicates[#], 1]}] & /@ IntegerPartitions[n, {2, n-1}]]; (* A108521 *)
b[1] = 1; b[n_] := b[n] = If[n > 2, 1, 0] + If[EvenQ[n], a[n/2] (a[n/2] + 1)/2, a[(n-1)/2] (a[(n-1)/2]+1)/2] + If[n > 3, Total[If[Max[#] <= If[EvenQ[n], n/2-1, (n-1)/2], Product[Binomial[a[i] - 1 + Count[#, i], Count[#, i]], {i, DeleteCases[DeleteDuplicates[#], 1]}], 0] & /@ IntegerPartitions[n, {3, n-1}]], 0];
Table[b[n], {n, 40}]
(* a[n] = A108521[n]; d[n] are coefficients of A^2(x) in g.f. *)
a[0] = 0; a[1] = 1; a[n_] := a[n] = a[n-1] + (DivisorSum[n, a[#] # &, #<n &] + Sum[c[k] b[n-k], {k, n-1}])/n; b[n_] := b[n] = (c[n] + Sum[c[k] b[n-k], {k, n-1}])/n; c[n_] := c[n] = DivisorSum[n, a[#] # &]; d[n_] := d[n] = Sum[2 a[k] a[n-k], {k, Floor[(n-1)/2]}] + If[EvenQ[n], a[n/2]^2, 0]; Table[a[n] - d[n] + (d[n-1] + If[OddQ[n], a[(n-1)/2], 0])/2, {n, 40}]
PROG
(PARI) \\ here S is A108521 as vector.
EulerT(v)={Vec(exp(x*Ser(dirmul(v, vector(#v, n, 1/n))))-1, -#v)}
S(n)={my(v=[1]); for(n=2, n, v=concat(v, v[#v] + EulerT(concat(v, [0]))[n])); v}
seq(n)={my(p=x*Ser(S(n))); Vec(p + (x/2-1)*p^2 + (x/2)*subst(p, x, x^2))} \\ Andrew Howroyd, Jun 06 2020
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Robert A. Russell, Jun 02 2020
STATUS
approved