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%I #13 Aug 29 2016 02:56:20
%S 1,1,2,1,2,4,2,2,8,7,8,4,3,26,4,13,24,24,6,66,28,8,23,62,104,8,10,158,
%T 120,64,42,148,352,80,16,19,350,416,344,16,75,334,1052,448,160,33,756,
%U 1252,1440,208,32,136,726,2860,1936,1024,32,61,1578,3448,5176,1440,384,244,1534,7312,7056,5072,512,64
%N Triangle read by rows: T(n,k) is the number of compositions of n with parts in {1,2,4,6,8,10,...} and having asymmetry degree equal to k (n>=0; 0<=k<=floor(n/3)).
%C 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).
%C Number of entries in row n is 1 + floor(n/3).
%C Sum of entries in row n is A028495(n).
%C T(n,0) = A276055(n)
%C Sum_{k>=0} k*T(n,k) = A276054(n).
%D S. Heubach and T. Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
%H Krithnaswami Alladi and V. E. Hoggatt, Jr. <a href="http://www.fq.math.ca/Scanned/13-3/alladi1.pdf">Compositions with Ones and Twos</a>, Fibonacci Quarterly, 13 (1975), 233-239.
%H V. E. Hoggatt, Jr., and Marjorie Bicknell, <a href="http://www.fq.math.ca/Scanned/13-4/hoggatt1.pdf">Palindromic compositions</a>, Fibonacci Quart., Vol. 13(4), 1975, pp. 350-356.
%F G.f.: G(t,z) = (1-z^4)(1+z -z^3)/(1-2z^2-2tz^3-z^4+(3-2t)z^6+2tz^7-z^8). In the more general situation of compositions into a[1]<a[2]<a[3]<..., denoting F(z) = Sum(z^{a[j]},j>=1}, 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.
%e Row 4 is [4,2] because the compositions of 4 with parts in {1,2,4,6,8,...} are 4, 22, 211, 121, 112, and 1111, having asymmetry degrees 0, 0, 1, 0, 1, and 0, respectively.
%e Triangle starts:
%e 1;
%e 1;
%e 2;
%e 1,2;
%e 4,2;
%e 2,8;
%e 7,8,4.
%p G := (1-z^4)*(1+z-z^3)/(1-2*z^2-2*t*z^3-z^4+(3-2*t)*z^6+2*t*z^7-z^8): 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
%t Table[TakeWhile[BinCounts[#, {0, 1 + Floor[n/3], 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, EvenQ@ a]]], 1]]], {n, 0, 18}] // Flatten (* _Michael De Vlieger_, Aug 28 2016 *)
%Y Cf. A028485, A276054, A276055.
%K nonn,tabf
%O 0,3
%A _Emeric Deutsch_, Aug 17 2016
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