|
%I #25 Jun 10 2020 17:46:48
%S 1,1,2,4,9,25,70,226,753,2675,9785,37087,143487,566952,2274967,
%T 9257906,38113299,158535204,665364565,2814924441,11993967450,
%U 51433198599,221839745468,961884808879,4190783204515,18339291329225
%N Number of free trees with exactly n nodes with fewer than three neighbors.
%C Generates and uses values from A108521, rooted trees with exactly n generators, a generator being a leaf or node with just one child.
%H Robert A. Russell, <a href="/A335342/b335342.txt">Table of n, a(n) for n = 1..60</a>
%H R. A. Russell, <a href="https://mathoverflow.net/questions/361860/how-many-trees-have-n-nodes-with-fewer-than-three-neighbors">How many trees have n nodes with fewer than three neighbors?</a>, MathOverflow, June 2020.
%F G.f.: A(x) + (x/2-1)*A^2(x) + (x/2)*A(x^2), where A(x) is the g.f. for A108521.
%e 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.
%t 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 *)
%t 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];
%t Table[b[n], {n, 40}]
%t (* a[n] = A108521[n]; d[n] are coefficients of A^2(x) in g.f. *)
%t 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}]
%o (PARI) \\ here S is A108521 as vector.
%o EulerT(v)={Vec(exp(x*Ser(dirmul(v, vector(#v, n, 1/n))))-1, -#v)}
%o S(n)={my(v=[1]); for(n=2, n, v=concat(v, v[#v] + EulerT(concat(v, [0]))[n])); v}
%o 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
%Y Cf. A108521 (rooted trees).
%K nonn,easy
%O 1,3
%A _Robert A. Russell_, Jun 02 2020
|
