%I #13 Aug 17 2016 13:07:27
%S 1,1,1,2,2,2,3,4,4,8,5,16,7,26,4,9,44,12,12,70,32,16,108,76,21,166,
%T 156,8,28,248,308,32,37,368,572,104,49,540,1020,288,65,784,1768,696,
%U 16,86,1132,2976,1568,80,114,1622,4908,3304,304,151,2312,7944,6624,960,200,3280,12652,12768,2640,32
%N Triangle read by rows: T(n,k) is the number of compositions without 2's and having asymmetry degree equal to k (n>=0; 0<=k<=floor(n/4)).
%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/4).
%C Sum of entries in row n is A005251(n+1).
%C T(n,0) = A000931(n+5) (= number of palindromic compositions of n without 2's).
%C Sum_{k >= 0} k*T(n,k) = A275443(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 P. Chinn and S. Heubach, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL6/Heubach/heubach5.html">Integer Sequences Related to Compositions without 2's</a>, J. Integer Seqs., Vol. 6, 2003.
%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^2)/(1-z-z^2+z^4-2tz^4). 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 5 is [3,4] because the compositions of 5 without 2's are 5, 113, 131, 311, 14, 41, and 11111, having asymmetry degrees 0, 1, 0, 1, 1, 1, and 0, respectively.
%e Triangle starts:
%e 1;
%e 1;
%e 1;
%e 2;
%e 2,2;
%e 3,4;
%e 4,8;
%e 5,16.
%p G := (1-z^2)/(1-z-z^2+z^4-2*t*z^4): 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[BinCounts[#, {0, 1 + Floor[n/4], 1}] &@ 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_, ___} /; a == 2]], 1]]], {n, 0, 20}] // Flatten (* _Michael De Vlieger_, Aug 17 2016 *)
%Y Cf. A000931, A005251, A180177, A275443.
%K nonn,tabf
%O 0,4
%A _Emeric Deutsch_, Aug 16 2016