A275445 Sum of the asymmetry degrees of all compositions of n with parts in {1,2,3}.
0, 0, 0, 2, 4, 10, 22, 50, 106, 222, 458, 936, 1890, 3788, 7540, 14924, 29388, 57620, 112540, 219062, 425112, 822726, 1588314, 3059470, 5881254, 11284514, 21614774, 41336300, 78936358, 150533496, 286708744, 545428024, 1036468344, 1967555208, 3731449176, 7070218506, 13384916364, 25319020898, 47857031870, 90391975562, 170614347714
Offset: 0
Examples
a(4) = 4 because the compositions of 4 with parts in {1,2,3} are 13, 31, 22, 211, 121, 112, and 1111 and the sum of their asymmetry degrees is 1 + 1 + 0 + 1 + 0 + 1 + 0 = 4.
References
- S. Heubach and T. Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- 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.
- Index entries for linear recurrences with constant coefficients, signature (1,2,2,0,-4,-6,-6,-3,-1).
Crossrefs
Cf. A275444.
Programs
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Maple
g := 2*z^3*(1+z+z^2)/((1+z)*(1+z^2)*(1-z-z^2-z^3)^2): gser := series(g, z = 0, 45): seq(coeff(gser, z, n), n = 0 .. 40);
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Mathematica
Table[Total@ Map[Total, Map[Map[Boole[# >= 1] &, BitXor[Take[# - 1, Ceiling[Length[#]/2]], ReverseTake[# - 1, -Ceiling[Length[#]/2]]]] &, Flatten[Map[Permutations, DeleteCases[IntegerPartitions@ n, {a_, _} /; a > 3]], 1]]], {n, 0, 24}] // Flatten (* Michael De Vlieger, Aug 17 2016 *) -
PARI
concat(vector(3), Vec(2*x^3*(1+x+x^2)/((1+x)*(1+x^2)*(1-x-x^2-x^3)^2) + O(x^50))) \\ Colin Barker, Aug 28 2016
Formula
G.f. g(z) = 2*z^3*(1+z+z^2)/((1+z)*(1+z^2)*(1-z-z^2-z^3)^2). In the more general situation of compositions into a[1]=1} z^(a[j]), we have g(z) = (F(z)^2 - F(z^2))/((1+F(z))*(1-F(z))^2).
a(n) = Sum_{k>=0} k*A275444(n,k).
Comments