A275444 - OEIS

%I #10 Aug 17 2016 22:15:42

%S 1,1,2,2,2,3,4,3,10,6,14,4,6,26,12,11,34,36,11,62,68,8,20,82,140,32,

%T 20,144,228,112,37,186,424,264,16,37,316,664,608,80,68,404,1176,1168,

%U 320,68,676,1784,2312,896,32,125,860,3032,4096,2304,192

%N Triangle read by rows: T(n,k) is the number of compositions of n with parts in {1,2,3} 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 A000073(n+2).

%C T(2n,0) = T(2n+1,0) = A001590(n+3).

%C Sum_{k>=0} k*T(n,k) = A275445(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)(1+z^2)/(1-z^2 -z^4 -z^6 -2tz^3*(1+z+z^2 )). 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 [3,4] because the compositions of 4 with parts in {1,2,3} are  13, 31, 22, 211, 121, 112, and 1111, having asymmetry degrees 1, 1, 0, 1, 0, 1, and 0, respectively.

%e Triangle starts:

%e 1;

%e 1;

%e 2;

%e 2,2;

%e 3,10;

%e 6,14,4.

%p G := (1+z)*(1+z^2)/(1-z^2-2*t*z^3-(1+2*t)*z^4-2*t*z^5-z^6): 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 > 3]], 1]]], {n, 0, 17}] // Flatten  (* _Michael De Vlieger_, Aug 17 2016 *)

%Y Cf. A000073, A001590, A275445.

%K nonn,tabf

%O 0,3

%A _Emeric Deutsch_, Aug 17 2016