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A255908
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Triangle read by rows: T(n,L) = number of rho-labeled graphs with n edges whose labeling is bipartite with boundary value L.
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0
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2, 4, 8, 8, 32, 48, 16, 128, 288, 384, 32, 512, 1728, 3072, 3840, 64, 2048, 10368, 24576, 38400, 46080, 128, 8192, 62208, 196608, 384000, 552960, 645120, 256, 32768, 373248, 1572864, 3840000, 6635520, 9031680, 10321920, 512, 131072, 2239488, 12582912, 38400000, 79626240, 126443520, 165150720, 185794560, 1024, 524288, 13436928, 100663296, 384000000, 955514880, 1770209280, 2642411520, 3344302080, 3715891200
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OFFSET
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1,1
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COMMENTS
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A graph with n edges is rho-labeled if there exists a one-to-one mapping from its vertex set to {0,1,...,2n} such that every edge receives as label the absolute difference of its end-vertices and the edge labels are x_1, x_2, ..., x_n where x_i = i or x_i = 2n + 1 - i. A rho-labeling of a bipartite graph is said to be bipartite when the labels of one stable set are smaller than the labels on the other stable set. The largest of the smaller vertex labels is its boundary value.
The columns:
T(n, 0) = 2^n,
T(n, 1) = 2^(2n-1),
T(n, 2) = 2^(n+1)*3^(n-2),
T(n, 3) = 3*2^(3n-5),
T(n, 4) = 3*2^(n+3)*5^(n-4),
T(n, 5) = 5*2^(2n-2)*3^(n-4), etc.
The diagonals:
the main, T(n, n-1) = 2^n*n*(n-1!) = 2*A002866,
the second diagonal, T(n, n-2) = 2^n*(n-1)^2*(n-2)! = 4*A014479,
the third diagonal, T(n, n-3) = 2^n*(n-2)^3*(n-3)!,
the k_th diagonal, T(n, n-k) = 2^n*(n-k)^k*(n-k)!, etc.
... (End)
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LINKS
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FORMULA
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For n>=1, 0<=L<=n-1, T(n,L)=(2^n)*(L!)*(L+1)^(n-L).
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EXAMPLE
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For n=5 and L=1, T(5,1)=(2^5)*(1!)*(1+1)^(5-1)=512.
For n=9 and L=3, T(9,3)=12582912.
Triangle, T, begins:
-----------------------------------------------------------------------------
n\L | 0 1 2 3 4 5 6
----|------------------------------------------------------------------------
1 | 2;
2 | 4, 8;
3 | 8, 32, 48;
4 | 16, 128, 288, 384;
5 | 32, 512, 1728, 3072, 3840;
6 | 64, 2048, 10368, 24576, 38400, 46080;
7 | 128, 8192, 62208, 196608, 384000, 552960, 645120;
8 | 256, 32768, 373248, 1572864, 3840000, 6635520, 9031680, ...
...
For n=2 and L=1, T(2,1)=8, because: the bipartite graph <{v1,v2,v3},{x1=v1v2,x2=v1v3}> has rho-labelings (2,1,3),(2,1,4) with L=1 on the stable set {x2} and rho-labelings (1,2,0),(0,4,1) with L=1 on the stable set {x1,x3}; the bipartite graph <{v1,v2,v3,v4},{x1=v1v2,x2=v3v4}> has rho-labeling (0,4,1,3),(1,2,0,3) with L=1 on the stable set {v1,v3} and rho-labeling (4,0,3,1),(2,1,3,0) with L=1 on the stable set {v2,v4}. - Danny Rorabaugh, Apr 03 2015
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MATHEMATICA
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t[n_, l_] := 2^n*l!(l+1)^(n-l); Table[ t[n, l], {n, 8}, {l, 0, n-1}] // Flatten (* Robert G. Wilson v, Jul 05 2015 *)
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PROG
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(Magma) [2^n*Factorial(l)*(l+1)^(n-l): l in [0..n-1], n in [1..10]]; // Bruno Berselli, Aug 05 2015
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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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