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%I #19 Jan 04 2016 10:43:43
%S 1,0,1,0,2,-1,0,1,0,3,0,1,0,4,-2,0,3,1,0,3,2,-1,0,2,1,0,1,0,5,-2,0,4,
%T 1,0,4,2,4,0,3,-2,0,3,2,0,1,0,6,-2,0,5,1,0,5,2,4,0,4,1,0,4,3,-3,0,4,2,
%U -4,0,3,1,0,3,2,3,0,2,-3,0
%N Sequence that encodes the compliform polynomials associated to the tree of hemitropic sequences.
%C For each integer n >= 1, e_n(x_2, ..., x_n) is a polynomial whose coefficients are integers and has degree 1 in each of the variables, x_2, ..., x_n, (a so-called compliform polynomial). Given the first n terms, 1, c_2, ..., c_n of a hemitropic sequence relative to a subset A of N, (see A265262), one has the following: c_(n+1) = e_n(c_2,...,c_n) if n+1 is not in A, c_(n+1 )= e_n(c_2,...,c_n) + 1 if n + 1 is in A. See Haddad link, formula (8), p. 37. The first few polynomials of the sequence e_n are:
%C e_1 = 1, e_2 = x_2 - 1, e_3 = x_3, e_4 =x_4 - 2x_3 + x_3x_2 - x_2 + 1, e_5 = x_5 - 2x_4 + x_4x_2 + 4x_3 - 2x_3x_2, e_6 =x_6 - 2x_5 + x_5x_2 + 4x_4 + x_4x_3 - 3x_4x_2 - 4x_3 + x_3x_2 + 3x_2 -3, e_7 =x_7 - 2x_6 + x_6x_2 + 4x_5 + x_5x_3 - 3x_5x_2 - 4x_4 - 2x_4x_3 + 4x_4x_2
%C + 4x_3 - x_3x_2 - 4x_2 + 4.
%C Each monomial a.x_ix_j...x_k with i > j > ... > k, is converted into the sequence of integers a, 0, i, j, ..., k, where 0 is used for punctuation. There is no ambiguity. In the display, the monomials a.xixj, ..., xk, are ordered lexicographically in the (reverse) alphabet ..., n, ..., 3, 2. An e_n polynomial is thus converted into an irregular (finite) array:
%C e_1 = 1 --> 1;
%C e_2 = x_2 - 1 --> 1, 0, 2; -1;
%C e_3 = x_3 --> 1, 0, 3;
%C e_4 = x_4 - 2x_3 + x_3x_2 - x_2 + 1 --> 1, 0, 4; -2, 0, 3; 1, 0, 3, 2; -1, 0, 2; 1;
%C e_5 = x_5 - 2x_4 + x_4x_2 + 4x_3 - 2x_3x_2 --> 1, 0, 5; -2, 0, 4; 1, 0, 4, 2; 4, 0, 3; -2, 0, 3, 2;
%C Conversions are one-to-one, bijective. By concatenation of the arrays, the whole sequence of the e_n’s is again an infinite irregular array, with again 0 for punctuation.
%H Labib Haddad, <a href="http://arxiv.org/abs/1507.05849">Some peculiarities of order 2 bases of N and the Erdos-Turan conjecture</a>, arXiv:1507.05849 [math.NT], 2015
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Tur%C3%A1n_conjecture_on_additive_bases">Erdős-Turán conjecture on additive bases</a>
%F An algorithm for the e_n's. For k >+ 1, let P_(k+1) = (x_(k+1) - e_k)^2 - (x_(k+1) - e_k) = x_(k+1)^2 -x_(k+1) -2x_k+1e_k + e_k^2 + e_k: a polynomial in several variables, having degree 2 in the variable x_(k+1).
%F Start with e_1 = 1. Once the polynomials e_1,...,e_(n-1) have been obtained, set E_n =(x_n-e_(n-1))+(x_2-e_1)(x_(n-1)- e_(n-2)) + ... + (x_m - e_(m-1))(x_(n-m+1) - e_(n-m)) with m = floor((n + 1)/2): a polynomial in the variables x_2,...,x_n, not necessarily compliform, whose coefficients are integers, and having degree 1 in x_n.
%F Then, reduce E_n as follows: Let E_(n,n-1) be the remainder in the Euclidean division of E_n by P_(n-1) as polynomials in x_(n-1). Inductively, let E_(n,n-1,...,k) be the remainder in the Euclidean division of E_(n,n-1,k+1) by P_k as polynomials in x_k. This gives e_n = E_(n,n-1,··· ,2), a compliform polynomial. See Haddad link p.32 Corollary.
%Y Cf. A265262.
%K sign,tabf
%O 1,5
%A _Labib Haddad_, Dec 13 2015
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