OFFSET
0,3
COMMENTS
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/3).
Sum of entries in row n is A000045(n+1) (Fibonacci).
T(n,0) = A053602(n+1) (= number of palindromic compositions of n with parts in {1,2}).
Sum_{k>=0} k*T(n,k) = A275439(n).
LINKS
Krithnaswami Alladi and V. E. Hoggatt, Jr. Compositions with Ones and Twos, Fibonacci Quarterly, 13 (1975), 233-239.
V. E. Hoggatt, Jr., and Marjorie Bicknell, Palindromic compositions, Fibonacci Quart., Vol. 13(4), 1975, pp. 350-356.
FORMULA
G.f.: G(t,z) = (1+z+z^2)/(1-z^2-2tz^3-z^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.
EXAMPLE
Row 4 is [3,2] because the compositions of 4 with parts in {1,2} are 22, 112, 121, 211, and 1111, having asymmetry degrees 0, 1, 0, 1, 0, respectively.
Triangle starts:
1;
1;
2;
1,2;
3,2;
2,6;
5,4,4.
MAPLE
G:=(1+z+z^2)/(1-z^2-2*t*z^3-z^4): Gser:=simplify(series(G, z=0, 25)): for n from 0 to 20 do P[n] := sort(coeff(Gser, z, n)) end do: for n from 0 to 20 do seq(coeff(P[n], t, j), j=0..degree(P[n])) end do; # yields sequence in triangular form
MATHEMATICA
Join[{{1}}, Table[BinCounts[#, {0, 1 + Floor[n/3], 1}] &@ Map[Total, Map[BitXor[Take[# - 1, Ceiling[Length[#]/2]], Reverse@ Take[# - 1, -Ceiling[Length[#]/2]]] &, Flatten[Map[Permutations, DeleteCases[IntegerPartitions@ n, {a_, ___} /; a > 2]], 1]]], {n, 17}]] // Flatten (* Michael De Vlieger, Aug 17 2016 *)
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Emeric Deutsch, Aug 16 2016
STATUS
approved