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A276064
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Triangle read by rows: T(n,k) is the number of compositions of n with parts in {1,5} and having asymmetry degree equal to k (n>=0; 0<=k<=floor(n/6)).
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1
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1, 1, 1, 1, 1, 2, 1, 2, 2, 2, 1, 4, 2, 4, 2, 6, 3, 8, 3, 8, 4, 4, 12, 4, 4, 10, 12, 6, 16, 12, 5, 16, 24, 8, 24, 28, 6, 26, 40, 8, 10, 36, 52, 8, 8, 40, 60, 32, 13, 56, 84, 32, 11, 58, 96, 80, 17, 84, 136, 88, 15, 80, 160, 160, 16, 23, 120, 220, 192, 16
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OFFSET
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0,6
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COMMENTS
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The asymmetry degree of a finite sequence of numbers is defined to be the number of pairs of symmetrically positioned distinct entries. Example: the asymmetry degree of (2,7,6,4,5,7,3) is 2, counting the pairs (2,3) and (6,5).
Number of entries in row n is 1 + floor(n/6).
Sum of entries in row n is A003520(n).
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REFERENCES
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S. Heubach and T. Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
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LINKS
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FORMULA
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G.f.: G(t,z) = (1+z+z^5)/(1-z^2-2tz^6-z^10). In the more general situation of compositions into a[1]<a[2]<a[3]<..., denoting F(z) = Sum_{j>=1} z^{a[j]}, we have G(t,z) =(1 + F(z))/(1 - F(z^2) - t(F(z)^2 - F(z^2))). In particular, for t=0 we obtain Theorem 1.2 of the Hoggatt et al. reference.
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EXAMPLE
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Row 8 is [1,4] because the compositions of 8 with parts in {1,5} are 5111, 1511, 1151, 1115 and 11111111, having asymmetry degrees 1,1,1,1, and 0, respectively.
Triangle starts:
1;
1;
1;
1;
1;
2;
1, 2;
2, 2.
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MAPLE
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G := (1+z+z^5)/(1-z^2-2*t*z^6-z^10): Gser := simplify(series(G, z = 0, 30)): for n from 0 to 25 do P[n] := sort(coeff(Gser, z, n)) end do: for n from 0 to 25 do seq(coeff(P[n], t, j), j = 0 .. degree(P[n])) end do; # yields sequence in triangular form
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MATHEMATICA
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Table[TakeWhile[BinCounts[#, {0, 1 + Floor[n/4], 1}], # != 0 &] &@ Map[Total, Map[Map[Boole[# >= 1] &, BitXor[Take[# - 1, Ceiling[Length[#]/2]], Reverse@ Take[# - 1, -Ceiling[Length[#]/2]]]] &, Flatten[Map[Permutations, DeleteCases[IntegerPartitions@ n, {___, a_, ___} /; Nor[a == 1, a == 5]]], 1]]], {n, 0, 25}] // Flatten (* Michael De Vlieger, Aug 22 2016 *)
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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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