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A278136
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Triangle read by rows: T(n,k) is the maximum number of disjoint subgraphs of the Fibonacci cube Gamma(n) that are isomorphic to the hypercube of dimension k.
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2
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1, 2, 1, 3, 1, 5, 2, 1, 8, 4, 1, 13, 6, 2, 1, 21, 10, 5, 1, 34, 17, 7, 2, 1, 55, 27, 12, 6, 1, 89, 44, 22, 8, 2, 1, 144, 72, 34, 14, 7, 1, 233, 116, 56, 28, 9, 2, 1, 377, 188, 94, 42, 16, 8, 1, 610, 305, 150, 70, 35, 10, 2, 1, 987, 493, 244, 122, 51, 18, 9, 1
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OFFSET
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0,2
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COMMENTS
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Number of entries in row n is 1 + ceiling(n/2).
T(n,0) = F(n+2) = A000045(n+2) (Fibonacci); number of vertices of Gamma(n).
Sum of entries in row n is A278137(n).
T(n,1) = floor(F(n+2)/2) (see Lemma 2.1 in the Gravier et al. paper).
The generating function of column k is x^{2k-1}/((1-x^3)^k*(1-x-x^2)) (k>=0) (see Corollary 2.5 in the Gravier et al. paper).
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LINKS
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FORMULA
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T(n,k) = Sum_{i=k-1..floor((n+k-2)/3)} binomial(i,k-1)*F(n+k-3i-1), where F(j) = A000045(j) (Fibonacci); (see Corollary 2.4 in the Gravier et al. paper).
T(n,k) = T(n-2,k-1) + T(n-3,k) (n>=3, k>=1) (see Theorem 2.2 in the Gravier et al. paper).
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EXAMPLE
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Row 3 is 5,2,1. Indeed, the Fibonacci cube Gamma(3) has 5 vertices A, B, C, D, E and edges AB, BC, CD, DA, DE and so it has at most 2 disjoint edges and it has one square.
Triangle starts:
1;
2, 1;
3, 1;
5, 2, 1;
8, 4, 1;
13, 6, 2, 1;
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MAPLE
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with(combinat): F := proc (k) options operator, arrow: fibonacci(k) end proc; T := proc (n, k) options operator, arrow: sum(binomial(i, k-1)*F(n+k-3*i-1), i = k-1 .. floor((1/3)*n+(1/3)*k-2/3)) end proc: for n from 0 to 20 do seq(T(n, k), k = 0 .. ceil((1/2)*n)) end do; # yields sequence in triangular form
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MATHEMATICA
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Flatten[Table[Sum[Binomial[i, k-1] Fibonacci[n+k-3i-1], {i, k-1, Floor[(n+k-2)/3]}], {n, 0, 14}, {k, 0, Ceiling[n/2]}]] (* Indranil Ghosh, Mar 05 2017 *)
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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