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A277919
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Triangle read by rows: CL(n,k) is the number of labeled subgraphs with k edges of the n-cycle C_n.
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3
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1, 1, 1, 3, 2, 1, 7, 6, 3, 1, 15, 16, 10, 4, 1, 31, 40, 30, 15, 5, 1, 63, 96, 84, 50, 21, 6, 1, 127, 224, 224, 154, 77, 28, 7, 1, 255, 512, 576, 448, 258, 112, 36, 8, 1, 511, 1152, 1440, 1248, 810, 405, 156, 45, 9, 1, 1023, 2560, 3520, 3360, 2420, 1362, 605, 210, 55, 10, 1
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OFFSET
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0,4
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LINKS
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FORMULA
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The identity CL(n,k) = 2^(n-2*k) * CL(n,n-k) can be proved combinatorially.
G.f.: (1 - 2*x + 2*x^2)/((1-x)*(1 - y*x - 2*x + y*x^2)). - Andrew Howroyd, Sep 27 2019
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EXAMPLE
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For row 3 of the triangle below: there are 7 labeled subgraphs of the triangle C_3 with 0 edges, 6 with 1 edge, 3 with 2 edges, and 1 with 3 edges (C_3 itself).
Triangle begins:
1;
1, 1;
3, 2, 1;
7, 6, 3, 1;
15, 16, 10, 4, 1;
31, 40, 30, 15, 5, 1;
63, 96, 84, 50, 21, 6, 1;
127, 224, 224, 154, 77, 28, 7, 1;
255, 512, 576, 448, 258, 112, 36, 8, 1;
511, 1152, 1440, 1248, 810, 405, 156, 45, 9, 1;
1023, 2560, 3520, 3360, 2420, 1362, 605, 210, 55, 10, 1;
...
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PROG
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(PARI) T(n)={[Vecrev(p) | p<-Vec((1 - 2*x + 2*x^2)/((1-x)*(1 - y*x - 2*x + y*x^2)) + O(x*x^n))]}
{ my(A=T(12)); for(n=1, #A, print(A[n])) } \\ Andrew Howroyd, Sep 27 2019
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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