A318601 Triangle read by rows: T(n,k) is the number of hypertrees on n unlabeled nodes with k edges, (0 <= k < n).
1, 0, 1, 0, 1, 1, 0, 1, 1, 2, 0, 1, 2, 3, 3, 0, 1, 2, 6, 7, 6, 0, 1, 3, 9, 17, 18, 11, 0, 1, 3, 13, 30, 51, 44, 23, 0, 1, 4, 17, 53, 109, 148, 117, 47, 0, 1, 4, 23, 79, 213, 372, 443, 299, 106, 0, 1, 5, 28, 119, 370, 827, 1276, 1306, 793, 235
Offset: 1
Examples
Triangle begins: 1; 0, 1; 0, 1, 1; 0, 1, 1, 2; 0, 1, 2, 3, 3; 0, 1, 2, 6, 7, 6; 0, 1, 3, 9, 17, 18, 11; 0, 1, 3, 13, 30, 51, 44, 23; 0, 1, 4, 17, 53, 109, 148, 117, 47; 0, 1, 4, 23, 79, 213, 372, 443, 299, 106; ... Case n=4: There are 4 possible graphs which are the complete graph on 4 nodes which has 1 block, a triangle joined to a single vertex which has 2 blocks and two trees which have 3 blocks. Row 4 is then 0, 1, 1, 2. o---o o---o o---o o--o--o | X | / \ | | o---o o---o o---o o . Case n=5, k=3: The following illustrates how the dissymmetry thereom for each unlabeled hypertree gives 1 = vertex rooted + edge (block) rooted - directed edge (vertex of block) rooted. o---o / \ 1 = 3 + 2 - 4 o---o---o . o o / \ / 1 = 3 + 2 - 4 o---o---o . o o / \ / \ 1 = 4 + 3 - 6 o---o o .
Links
- Andrew Howroyd, Table of n, a(n) for n = 1..1275
Crossrefs
Programs
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PARI
\\ here b(n) is A318602 as vector of polynomials. EulerMT(u)={my(n=#u, p=x*Ser(u), vars=variables(p)); Vec(exp( sum(i=1, n, substvec(p + O(x*x^(n\i)), vars, apply(v->v^i,vars))/i ))-1)} b(n)={my(v=[1]); for(i=2, n, v=concat([1], EulerMT(y*EulerMT(v)))); v} G(n)={my(u=b(n)); apply(p->Vecrev(p), Vec(y*Ser(EulerMT(u))*(1-x*Ser(u)) + (1 - y)*Ser(u)))} { my(T=G(10)); for(n=1, #T, print(T[n])) }
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