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A245963
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Triangle read by rows: T(n,k) is the number of maximal hypercubes Q(p) in the Fibonacci cube Gamma(n) (i.e., Q(p) is an induced subgraph of Gamma(n) that is not a subgraph of a subgraph of Gamma(n) that is isomorphic to the hypercube Q(p+1)).
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3
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1, 0, 1, 0, 2, 0, 1, 1, 0, 0, 3, 0, 0, 3, 1, 0, 0, 1, 4, 0, 0, 0, 6, 1, 0, 0, 0, 4, 5, 0, 0, 0, 1, 10, 1, 0, 0, 0, 0, 10, 6, 0, 0, 0, 0, 5, 15, 1, 0, 0, 0, 0, 1, 20, 7, 0, 0, 0, 0, 0, 15, 21, 1, 0, 0, 0, 0, 0, 6, 35, 8, 0, 0, 0, 0, 0, 1, 35, 28, 1, 0, 0, 0, 0, 0, 0, 21, 56, 9, 0, 0, 0, 0, 0, 0, 7, 70, 36, 1
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OFFSET
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0,5
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COMMENTS
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The nonzero entries in columns 0,1,2,... are rows 0,2,3,... of the Pascal triangle.
Row n contains 1+ceiling(n/2) entries.
Sum of entries in row n = A000931(n+6) (the Padovan sequence).
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LINKS
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FORMULA
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T(n,k) = binomial(k+1,n-2*k+1).
G.f.: (1+t*z*(1+z))/(1-t*(1+z)*z^2).
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EXAMPLE
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Row 3 is 0,1,1. Indeed, the Fibonacci cube Gamma(3) is a square with an additional pendant edge attached to one of its vertices; the pendant edge is a maximal Q(1) and the square is a maximal Q(2).
Triangle starts:
1;
0, 1;
0, 2;
0, 1, 1;
0, 0, 3;
0, 0, 3, 1;
0, 0, 1, 4;
0, 0, 0, 6, 1;
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MAPLE
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T := proc (n, k) options operator, arrow: binomial(1+k, n-2*k+1) end proc: for n from 0 to 20 do seq(T(n, k), k = 0 .. (n+1)*(1/2)) end do; # yields sequence in triangular form
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MATHEMATICA
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Table[Binomial[k + 1, n - 2 k + 1], {n, 0, 17}, {k, 0, Ceiling[n/2]}] // Flatten (* Michael De Vlieger, Jul 16 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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