A299930 Prime numbers represented by a cyclotomic binary form f(x, y) with x and y odd prime numbers and x > y.

19, 37, 79, 97, 109, 127, 139, 163, 223, 229, 277, 283, 313, 349, 397, 421, 433, 439, 457, 607, 643, 691, 727, 733, 739, 877, 937, 997, 1063, 1093, 1327, 1423, 1459, 1489, 1567, 1579, 1597, 1627, 1657, 1699, 1753, 1777, 1801, 1987, 1999, 2017, 2089, 2113, 2203

Offset: 1

Author

Peter Luschny, Feb 25 2018

Keywords

• nonn

Comments

A cyclotomic binary form over Z is a homogeneous polynomial in two variables which has the form f(x, y) = y^EulerPhi(k)*CyclotomicPolynomial(k, x/y) where k is some integer >= 3. An integer n is represented by f if f(x,y) = n has an integer solution.

We say a prime number p decomposes into x and y if x and y are odd prime numbers and there exists a cyclotomic binary form f such that p = f(x,y). The transitive closure of this relation can be displayed as a binary tree, the cbf-tree of p. A cbf-tree is squarefree if all its leafs are distinct. Examples are:

.

33751 23833 310567

/ \ / \ / \

131 79 163 19 359 283

/ \ / \ / \ / \

7 3 11 3 5 3 19 13

/ \

5 3

.

The leaves of these trees are in A299956. Related to the question whether the root of a cbf-tree can be reconstructed from its leafs is A299733.

Links

• Étienne Fouvry, Claude Levesque, Michel Waldschmidt, Representation of integers by cyclotomic binary forms, arXiv:1712.09019 [math.NT], 2017.

Crossrefs

Cf. A299956 (complement), A293654, A296095, A299214, A299498, A299733, A299928, A299929, A299956, A299964.

Programs

Julia

using Nemo
function isA299930(n)
    !isprime(ZZ(n)) && return false
    R, z = PolynomialRing(ZZ, "z")
    K = Int(floor(5.383*log(n)^1.161)) # Bounds from
    M = Int(floor(2*sqrt(n/3)))  # Fouvry & Levesque & Waldschmidt
    N = QQ(n)
    P(u) = (p for p in u:M if isprime(ZZ(p)))
    for k in 3:K
        e = Int(eulerphi(ZZ(k)))
        c = cyclotomic(k, z)
        for y in P(3), x in P(y+2)
            N == y^e*subst(c, QQ(x, y)) && return true
    end end
    return false
end
A299930list(upto) = [n for n in 1:upto if isA299930(n)]
println(A299930list(2203))