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.

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

Offset: 1

Author

Michel Lagneau, Aug 18 2017

Keywords

• nonn

Comments

P(m,x) = Sum_{i=1..k} d(i)*x^(i-1) 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.

From Wolfdieter Lang, Nov 07 2017: (Start)

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

From Robert Israel, Nov 14 2017: (Start)

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)

Examples

			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*(1-i), r*(-1+i), r*(-1-i) 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
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A027750, A032741, A292226.

Programs

Maple

F:= proc(n) local x,d,i,A,B,R;
  d:= sort(convert(numtheory:-divisors(n),list));
  A:= add(d[2*i]*x^(i-1),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:
select(F, [$1..1000]); # Robert Israel, Nov 14 2017

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 *)

PARI

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