A120981
Triangle read by rows: T(n,k) is the number of ternary trees with n edges and having k vertices of outdegree 1 (n >= 0, k >= 0).
Original entry on oeis.org
1, 0, 3, 3, 0, 9, 1, 27, 0, 27, 18, 12, 162, 0, 81, 15, 270, 90, 810, 0, 243, 138, 270, 2430, 540, 3645, 0, 729, 189, 2898, 2835, 17010, 2835, 15309, 0, 2187, 1218, 4536, 34776, 22680, 102060, 13608, 61236, 0, 6561, 2280, 32886, 61236, 312984, 153090, 551124
Offset: 0
T(2,0)=3 because we have (Q,L,M), (Q,L,R) and (Q,M,R), where Q denotes the root and L (M,R) denotes a left (middle, right) child of Q.
Triangle starts:
1;
0, 3;
3, 0, 9;
1, 27, 0, 27;
18, 12, 162, 0, 81;
15, 270, 90, 810, 0, 243;
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T:=proc(n,k) if k<=n then (1/(n+1))*binomial(n+1,k)*sum(3^(3*j-n+2*k)*binomial(n+1-k,j)*binomial(j,n-k-2*j),j=0..n+1-k) else 0 fi end: for n from 0 to 10 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form
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T[n_, k_] := (1/(n+1))*Binomial[n+1, k]*Sum[3^(2k - n + 3j)*Binomial[n + 1 - k, j]*Binomial[j, n - k - 2j], {j, 0, n - k + 1}];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 02 2018 *)
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T(n,k) = binomial(n+1, k)*sum(j=0, n+1-k, 3^(2*k-n+3*j)*binomial(n+1-k, j)*binomial(j, n-k-2*j))/(n+1); \\ Andrew Howroyd, Nov 06 2017
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from sympy import binomial
def T(n, k): return binomial(n + 1, k)*sum([3**(2*k - n + 3*j)*binomial(n + 1 - k, j)*binomial(j, n - k - 2*j) for j in range(n + 2 - k)])//(n + 1)
for n in range(21): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Nov 07 2017
A120985
Number of ternary trees with n edges and having no vertices of degree 2. A ternary tree is a rooted tree in which each vertex has at most three children and each child of a vertex is designated as its left or middle or right child.
Original entry on oeis.org
1, 3, 9, 28, 93, 333, 1272, 5085, 20925, 87735, 372879, 1602450, 6953824, 30438138, 134255403, 596154495, 2662813341, 11955684591, 53927330037, 244250703252, 1110401393067, 5065143385647, 23176155530394, 106344639962973
Offset: 0
a(1)=3 because we have (Q,L), (Q,M) and (Q,R), where Q denotes the root and L (M,R) denotes a left (middle, right) child of Q.
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a:=n->sum(3^(n-3*j)*binomial(n+1,2*j+1)*binomial(n-2*j,j),j=0..n/2)/(n+1): seq(a(n),n=0..27);
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Table[1/(n+1)*Sum[3^(n-3*j)*Binomial[n+1,2*j+1]*Binomial[n-2*j,j],{j,0,n/2}],{n,0,20}] (* Vaclav Kotesovec, Oct 19 2012 *)
A120429
Triangle read by rows: T(n,k) is the number of ternary trees with n edges and having k leaves (i.e., vertices of degree 0; n>=0, k>=1). A ternary tree is a rooted tree in which each vertex has at most three children and each child of a vertex is designated as its left or middle or right child.
Original entry on oeis.org
1, 3, 9, 3, 27, 27, 1, 81, 162, 30, 243, 810, 360, 15, 729, 3645, 2970, 405, 3, 2187, 15309, 19845, 5670, 252, 6561, 61236, 115668, 56700, 6426, 84, 19683, 236196, 612360, 459270, 98658, 4536, 12, 59049, 885735, 3018060, 3214890, 1122660
Offset: 0
T(2,2)=3 because we have (Q,L,M), (Q,L,R) and (Q,M,R), where Q denotes the root and L (M,R) denotes a left (middle, right) child of Q.
Triangle starts:
1;
3;
9, 3;
27, 27, 1;
81, 162, 30;
243, 810, 360, 15;
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T:=proc(n,k) if k<=n+1-ceil(n/3) then (1/(n+1))*binomial(n+1,k)*sum(3^(n+j-2*k+2)*binomial(n+1-k,j)*binomial(j,k-1-j),j=0..n+1-k) else 0 fi end: 1; for n from 1 to 11 do seq(T(n,k),k=1..n+1-ceil(n/3)) od; # yields sequence in triangular form
A120983
Triangle read by rows: T(n,k) is the number of ternary trees with n edges and having k vertices of outdegree 3 (n >= 0, k >= 0).
Original entry on oeis.org
1, 3, 12, 54, 1, 261, 12, 1323, 105, 6939, 810, 3, 37341, 5859, 63, 205011, 40824, 840, 1143801, 277830, 9072, 12, 6466230, 1861380, 86670, 360, 36960300, 12335895, 764478, 6435, 213243435, 81120204, 6377778, 89100, 55, 1240219269, 530408736
Offset: 0
T(3,1)=1 because we have (Q,L,M,R), where Q denotes the root and L (M,R) denotes a left (middle, right) child of Q.
Triangle starts:
1;
3;
12;
54, 1;
261, 12;
1323, 105;
6939, 810, 3;
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T:=(n,k)->(1/(n+1))*binomial(n+1,k)*sum(3^j*binomial(n+1-k,j)*binomial(j,n-3*k-j),j=0..n+1-k): for n from 0 to 14 do seq(T(n,k),k=0..floor(n/3)) od; # yields sequence in triangular form
Showing 1-4 of 4 results.
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