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A297897
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Triangular array read by row: T(m,n) = number of ways to obtain a single sphere by gluing the (labeled) sides of a (2m+1)-gon and a (2n+1)-gon, m >= n >= 0.
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1
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1, 3, 15, 10, 60, 260, 35, 231, 1050, 4375, 126, 882, 4140, 17640, 72324, 462, 3366, 16170, 70070, 291060, 1183644, 1716, 12870, 62920, 276276, 1159704, 4756752, 19253520, 6435, 49335, 244530, 1085175, 4594590, 18981270, 77297220, 311949495, 24310, 189618, 950300
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OFFSET
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0,2
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COMMENTS
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Number of ways to obtain a sphere by gluing the sides of a single 2n-gon equals Catalan number A000108(n).
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LINKS
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FORMULA
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T(m,n) = (2*m*n+m+n+1)/(m+n+1) * binomial(2*m+1,m) * binomial(2*n+1,n).
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EXAMPLE
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Array starts:
m=0: 1
m=1: 3, 15
m=2: 10, 60, 260
m=3: 35, 231, 1050, 4375
m=4: 126, 882, 4140, 17640, 72324
m=5: 462, 3366, 16170, 70070, 291060, 1183644
m=6: 1716, 12870, 62920, 276276, 1159704, 4756752, 19253520
m=7: 6435, 49335, 244530, 1085175, 4594590, 18981270, 77297220, 311949495
...
For m=n=1, let P and Q be triangles. They can be glued into a sphere in two manners: (1) by gluing each side of P to a side of Q, which can be done in 2*3=6 ways, where factor 2 stands for choosing orientation of gluing and factor 3 accounts for matchings of the edges across P and Q to glue with respect to the chosen orientation; or (2) by first gluing a pair of edges of P (chosen in 3 ways) together and gluing a pair of edges of Q (chosen in 3 ways) together, and then gluing the remaining single edges of P and Q, which overall can be done in 3*3=9 ways. Hence, T(1,1) = 6 + 9 = 15.
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MATHEMATICA
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T[n_, k_] := With[{m=n+k+1}, 4^m (2*n*k+m) (n+1/2)! (k+1/2)!/(Pi m (n+1)! (k+1)!)];
Table[T[n, k], {n, 0, 7}, {k, 0, n}] // Flatten (* Peter Luschny, Feb 28 2018 *)
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PROG
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(PARI) { A297897(m, n) = (2*m*n+m+n+1) * binomial(2*m+1, m) * binomial(2*n+1, n) / (m+n+1); }
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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