

A291127


Consider the zeros of the polynomial P(m,x) whose coefficients are the divisors of a number m. The sequence lists the numbers m such that P(m,x) contains at least two zeros that are purely imaginary numbers.


2



6, 8, 10, 14, 15, 21, 22, 26, 27, 33, 34, 35, 38, 39, 42, 46, 51, 54, 55, 57, 58, 62, 65, 66, 69, 74, 77, 78, 82, 85, 86, 87, 88, 91, 93, 94, 95, 102, 104, 106, 110, 111, 114, 115, 118, 119, 122, 123, 125, 128, 129, 130, 133, 134, 136, 138, 141, 142, 143, 145
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OFFSET

1,1


COMMENTS

P(m,x) = Sum_{i=1..k} d(i)*x^(i1) where d(1), d(2), ..., d(k) are the k divisors of m.
The number of zeros of the polynomial P(n,x) is given by A032741(n).
We observe that all the zeros of the polynomial are located in the unit circle.
Conjecture: A032741(a(n))= p where p == 3 (mod 4), p prime. Examples:
A032741(m) = 3 for m = 6, 8, 10, 14, 15, 21, 22, 26, 27, 33, ...
A032741(m) = 7 for m = 42, 54, 66, 78, 88, ...
A032741(m) = 11 for m = 156, 204, 228, 276, 294, 342, 348, 372, ...
A032741(m) = 19 for m = 2544, 2832, 2928, 3216, 3408, 3504, 3792, ...
A032741(m) = 23 for m = 24492, 25428, 26052, 26364, 26988, 27924, ...
Except for the two purely imaginary zeros of P(m,x), it seems that the complex zeros are of the form u + u*i where u is a real number.
P(m,x) is the row m polynomial of A027750 with increasing powers of x.
The numbers m = 1 and m = prime obviously do not appear in this sequence. The composite numbers m belonging to irreducible polynomials P(m,x) over the integers given in A292226 also do not appear in this sequence. Moreover, the composite numbers m with factorizable P(m,x) without a factor of the type a*x^2 + b, with positive integers a and b, also do not appear in this sequence; these are the numbers 18, 20, 28, 32, 44, ... (End)
Are there numbers m with more than one pair of purely imaginary solutions?  Wolfdieter Lang, Nov 14 2017
The even and odd parts of P(m,x) are of the form A(x^2) and x*B(x^2) for polynomials A and B with integer coefficients, and pairs of imaginary roots of P(m,x) correspond to negative roots of the gcd of A and B.
Includes the following:
p^k where p is prime and k==3 (mod 4).
p*q^k where k is odd and p, q are prime with either p < q or p > q^k.
p*q*r^k where p, q, r are distinct primes and r > p*q.
(End)


LINKS



EXAMPLE

42 is in the sequence because P(42,x) = 1 + 2x + 3x^2 + 6x^3 + 7x^4 + 14x^5 + 21x^6 + 42x^7 = (1 + 2*x)*(1 + 3*x^2)*(1 + 7*x^4), and the seven zeros are 1/2, +(1/3)*sqrt(3)*i, (1/3)*sqrt(3)*i, r*(1+i), r*(1i), r*(1+i), r*(1i) with r = 7^(3/4)*sqrt(2)/14. The relevant factor for the two purely imaginary zeros is (1 + 3*x^2).  Wolfdieter Lang, Nov 13 2017


MAPLE

F:= proc(n) local x, d, i, A, B, R;
d:= sort(convert(numtheory:divisors(n), list));
A:= add(d[2*i]*x^(i1), i=1..nops(d)/2);
B:= add(d[2*i+1]*x^i, i=0..(nops(d)1)/2);
R:= gcd(A, B);
sturm(sturmseq(R, x), x, infinity, 0) > 0;
end proc:


MATHEMATICA

Position[#, k_ /; k >= 2][[All, 1]] &@ Table[Count[Re /@ Values@ Apply[Join, Solve[Normal@ SeriesData[x, 0, #, 0, Length@ #, 1] == 0, x]], 0] &@ Divisors@ n, {n, 150}] (* Michael De Vlieger, Aug 21 2017 *)


PROG

(PARI) isok(n) = {my(d = divisors(n), p = sum(k=1, #d, x^(k1)*d[k])); #select(x>(real(x)==0), polroots(p)) >= 2; } \\ Michel Marcus, Sep 09 2017


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



