A275440 - OEIS

%I #17 Jan 05 2025 19:51:40

%S 1,1,1,2,1,2,3,2,2,6,5,8,3,14,4,8,22,4,5,30,20,13,52,24,8,60,68,8,21,

%T 112,92,8,13,116,192,56,34,228,284,64,21,218,484,248,16,55,446,768,

%U 312,16,34,402,1132,872,144,89,848,1900,1184,160,55,730

%N Triangle read by rows: T(n,k) is the number of compositions of n into odd parts, 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 A000045(n) (Fibonacci).

%C T(n,0) = A053602(n) (= number of palindromic compositions of n into odd parts).

%C Sum(k*T(n,k), k>=0) = A275441(n).

%D S. Heubach and T. Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.

%H K. Alladi and V. E. Hoggatt, Jr. <a href="https://web.archive.org/web/2024*/https://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 M. Bicknell, <a href="https://web.archive.org/web/2024*/https://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^2)/(1-2z^2-2tz^4+z^6). 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 [3,2] because the compositions of 5 into odd parts are 5, 113, 131, 311, and 11111, having asymmetry degrees 0, 1, 0, 1, 0, respectively.

%e Triangle starts:

%e 1;

%e 1;

%e 1;

%e 2;

%e 1,2;

%e 3,2;

%e 2,6;

%e 5,8; .

%p G := (1-z^4)*(1+z-z^2)/(1-2*z^2-2*t*z^4+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)/2, Ceiling[Length[#]/2]], Reverse@ Take[(# - 1)/2, -Ceiling[Length[#]/2]]]] &, Flatten[Map[Permutations, DeleteCases[IntegerPartitions@ n, {___, a_, ___} /; EvenQ@ a]], 1]]], {n, 0, 20}] // Flatten (* _Michael De Vlieger_, Aug 17 2016 *)

%Y Cf. A000045, A053602, A275441.

%K nonn,tabf

%O 0,4

%A _Emeric Deutsch_, Aug 16 2016