A367752
Number of shapes of labeled rooted hypertrees with n vertices.
Original entry on oeis.org
1, 1, 4, 29, 256, 3007, 42932, 721121, 13982563, 306967231, 7527903208, 203977383469, 6051630040496, 195111205542541, 6792697846367791, 253966747582533681, 10149075292428481965, 431705938073882999275, 19474660918369182445456, 928660364396786865580881
Offset: 1
For n = 3 the a(3) = 4 solutions are:
- the corolla with a black root which have 3 white children,
- and the 3 possible labeling of the hypertree with a white root which have 2 white children connected to it via a hyperedge.
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my(x='x+O('x^30)); Vec(serlaplace(serreverse(log(1+x)*exp(-exp(x)+x+1)))) \\ Michel Marcus, Nov 30 2023
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R.=PowerSeriesRing(QQ);(ln(1+t)*exp(-exp(t)+t+1)).reverse().egf_to_ogf().list()[1:]
A367753
Number of shapes of labeled forest of rooted hypertrees with n vertices.
Original entry on oeis.org
1, 2, 8, 55, 507, 5969, 85605, 1445420, 28110615, 618760615, 15207174501, 412790636977, 12265337498000, 395962288858946, 13800754780797740, 516494067220932259, 20658199248901273576, 879406216174705907137, 39698270229941320201019, 1894212537494300993244732
Offset: 1
-
my(x='x+O('x^30)); Vec(serlaplace(serreverse((1+x)*log(1+log(1+x))*exp(-x)))) \\ Michel Marcus, Nov 30 2023
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R.=PowerSeriesRing(QQ,30);(1+t)*(ln(1+ln(1+t))*exp(-t)).reverse().egf_to_ogf().list()[1:]
A370948
Triangle read by rows: T(n,k) is the number of labeled forests of rooted Greg hypertrees with n white vertices and weight k, 0 <= k < n.
Original entry on oeis.org
1, 3, 1, 22, 15, 1, 262, 271, 53, 1, 4336, 6020, 2085, 165, 1, 91984, 160336, 81310, 13040, 487, 1, 2381408, 4996572, 3364011, 851690, 73024, 1407, 1, 72800928, 178613156, 150499951, 53119521, 7696794, 383649, 4041, 1
Offset: 1
Triangle T(n,k) begins:
n\k 0 1 2 3 4 ...
1 1;
2 3, 1;
3 22, 15, 1;
4 262, 271, 53, 1;
5 4336, 6020, 2085, 165, 1;
...
Series reversion as e.g.f. is related to
A092271.
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T(n)={my(x='x+O('x^(n+1))); [Vecrev(p) | p <- Vec(serlaplace(serreverse( (log(1+y*x)/y - exp(x) + x + 1)*exp(-x) )))]}
{ my(A=T(8)); for(i=1, #A, print(A[i])) } \\ Andrew Howroyd, Mar 06 2024
Showing 1-3 of 3 results.
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