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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

User: Labib Haddad

Labib Haddad's wiki page.

Labib Haddad has authored 2 sequences.

A265674 Sequence that encodes the compliform polynomials associated to the tree of hemitropic sequences.

Original entry on oeis.org

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, 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, -4, 0, 3, 1, 0, 3, 2, 3, 0, 2, -3, 0
Offset: 1

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Author

Labib Haddad, Dec 13 2015

Keywords

Comments

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:
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
+ 4x_3 - x_3x_2 - 4x_2 + 4.
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:
e_1 = 1 --> 1;
e_2 = x_2 - 1 --> 1, 0, 2; -1;
e_3 = x_3 --> 1, 0, 3;
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;
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;
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.

Links

Crossrefs

Cf. A265262.

Formula

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).
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.
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.

A265262 The tree of hemitropic sequences read by rows, arising from an Erdős-Turán conjecture.

Original entry on oeis.org

1, 1, 2, 0, 1, 1, 2, 0, 1, 1, 2, 1, 2, 2, 3, 0, 1, 1, 2, 0, 1, 1, 2, 0, 1, 1, 2, 1, 2, 2, 3, 0, 1, 1, 2, 0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 0, 1, 1, 2, 0, 1, 1, 2, 0, 1, 1, 2, 0, 1, 1, 2, 0, 1, 1, 2, 0, 1, 1, 2
Offset: 0

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Author

Labib Haddad and Michel Marcus, Dec 06 2015

Keywords

Comments

Let A be a subset of the set N of nonnegative integers. Let pA(n) be the number of pairs (x, y) of elements of A such that n = x + y and x <= y. The sequence pA = [pA(0), pA(1), ... , pA(n), ... ] is called the profile of A. A Sidon set is a subset A whose profile has only 0's and 1's.
An [order 2 additive] basis of N is a subset A whose profile has no 0’s. Erdős and Turán conjectured that the profile of a basis is always unbounded (see the Erdős and Turán link). The conjecture is, up till now, still undecided.
The tree below displays the infinite sequences [1, pA(2), . . . ], associated to the profiles pA = [1, 1, pA(2), . . . ] of all the subsets A of N to which 0 and 1 belong. Those are the so-called hemitropic sequences. The Erdős-Turán conjecture would not hold if (and only if) the tree contained an infinite bounded branch with no 0's.
The length of the n-th row is 2^n. The right leaf of a node is equal to the left leaf + 1.

Examples

			First few levels of the tree:
                       1;
           1,                      2;
     0,          1,          1,         2;
  0,    1,    1,    2,    1,    2,    2,   3;
0, 1, 1, 2, 0, 1, 1, 2, 0, 1, 1, 2, 1, 2, 2, 3;
...
First few rows of the irregular array:
1;
1, 2;
0, 1, 1, 2;
0, 1, 1, 2, 1, 2, 2, 3;
0, 1, 1, 2, 0, 1, 1, 2, 0, 1, 1, 2, 1, 2, 2, 3;
...

Links

Crossrefs

Cf. A004137, A066062, A217793.

Programs

  • Maple
    with(ListTools):
v:=n->Reverse( convert(n,base,2)):
m:=n->nops(v(n)):
c:=n-> v(n)[m(n)] + sum(v(n)[k]*v(n)[m(n)-k],k=1..floor(m(n)/2)):
d:= h->[seq(c(n),n=2^h..2^(h+1)-1)]: # the h-th row
f:= p->[seq(c(n),n=1..p)]: # the first p terms
  • PARI
    f(t,n,va) = 1+ sum(k=1, n+1, va[k]*t^k);
row(n) = {if (n==0, vc = [1], vc = []; for (ni = 2^n, 2^(n+1)-1, b = binary(ni); ft = f(t, n, b); pt = (f(t, n, b)^2 + f(t^2, n, b))/2; vc = concat(vc, polcoeff(pt, n+1)););); vc;}
tabf(nn) = for (n=0, nn, vrow = row(n); for (k=1, #vrow, print1(vrow[k], ", ")); print());

Formula

For each k>=0, let u(k)=1 if k is in A, u(k)=0 is k is not in A. Then pA(n) = Sum_{k=0..floor(n/2)} u(k)*u(n-k). See formula (5) on p. 8 and p. 28 in Haddad link.

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