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Search: a055297 -id:a055297
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Triangle of trees with n nodes and k leaves, 2 <= k <= n.
+10
21
1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 2, 2, 1, 0, 1, 3, 4, 2, 1, 0, 1, 4, 8, 6, 3, 1, 0, 1, 5, 14, 14, 9, 3, 1, 0, 1, 7, 23, 32, 26, 12, 4, 1, 0, 1, 8, 36, 64, 66, 39, 16, 4, 1, 0, 1, 10, 54, 123, 158, 119, 60, 20, 5, 1, 0, 1, 12, 78, 219, 350, 325, 202, 83, 25, 5, 1, 0
OFFSET
2,12
REFERENCES
F. Harary and E. M. Palmer, Graphical Enumeration, Academic Press, NY, 1973, p. 80, Problem 3.9.
FORMULA
G.f.: A(x, y)=(1-x+x*y)*B(x, y)+(1/2)*(B(x^2, y^2)-B(x, y)^2), where B(x, y) is g.f. of A055277.
EXAMPLE
Triangle begins:
n=2: 1
n=3: 1 0
n=4: 1 1 0
n=5: 1 1 1 0
n=6: 1 2 2 1 0
n=7: 1 3 4 2 1 0
n=8: 1 4 8 6 3 1 0
n=9: 1 5 14 14 9 3 1 0
n=10: 1 7 23 32 26 12 4 1 0
n=11: 1 8 36 64 66 39 16 4 1 0
n=12: 1 10 54 123 158 119 60 20 5 1 0
n=13: 1 12 78 219 350 325 202 83 25 5 1 0
PROG
(PARI)
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)}
T(n)={my(u=[y]); for(n=2, n, u=concat([y], EulerMT(u))); my(r=x*Ser(u), v=Vec(r*(1-x+x*y) + (substvec(r, [x, y], [x^2, y^2]) - r^2)/2)); vector(n-1, k, Vecrev(v[1+k]/y^2, k))}
{ my(A=T(10)); for(n=1, #A, print(A[n])) }
CROSSREFS
Row sums give A000055, row sums with weight k give A003228.
The labeled version is A055314.
Central column is A358107.
Left of central column is A359398.
KEYWORD
nonn,tabl
AUTHOR
Christian G. Bower, May 09 2000
STATUS
approved

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