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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.
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%I #30 Nov 17 2017 03:40:54

%S 6,8,10,14,15,21,22,26,27,33,34,35,38,39,42,46,51,54,55,57,58,62,65,

%T 66,69,74,77,78,82,85,86,87,88,91,93,94,95,102,104,106,110,111,114,

%U 115,118,119,122,123,125,128,129,130,133,134,136,138,141,142,143,145

%N 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.

%C 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.

%C The number of zeros of the polynomial P(n,x) is given by A032741(n).

%C We observe that all the zeros of the polynomial are located in the unit circle.

%C Conjecture: A032741(a(n))= p where p == 3 (mod 4), p prime. Examples:

%C A032741(m) = 3 for m = 6, 8, 10, 14, 15, 21, 22, 26, 27, 33, ...

%C A032741(m) = 7 for m = 42, 54, 66, 78, 88, ...

%C A032741(m) = 11 for m = 156, 204, 228, 276, 294, 342, 348, 372, ...

%C A032741(m) = 19 for m = 2544, 2832, 2928, 3216, 3408, 3504, 3792, ...

%C A032741(m) = 23 for m = 24492, 25428, 26052, 26364, 26988, 27924, ...

%C 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.

%C From _Wolfdieter Lang_, Nov 07 2017: (Start)

%C P(m,x) is the row m polynomial of A027750 with increasing powers of x.

%C 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)

%C Are there numbers m with more than one pair of purely imaginary solutions? - _Wolfdieter Lang_, Nov 14 2017

%C From _Robert Israel_, Nov 14 2017: (Start)

%C 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.

%C Includes the following:

%C p^k where p is prime and k==3 (mod 4).

%C p*q^k where k is odd and p, q are prime with either p < q or p > q^k.

%C p*q*r^k where p, q, r are distinct primes and r > p*q.

%C (End)

%H Robert Israel, <a href="/A291127/b291127.txt">Table of n, a(n) for n = 1..10000</a>

%e 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

%p F:= proc(n) local x,d,i,A,B,R;

%p d:= sort(convert(numtheory:-divisors(n),list));

%p A:= add(d[2*i]*x^(i-1),i=1..nops(d)/2);

%p B:= add(d[2*i+1]*x^i,i=0..(nops(d)-1)/2);

%p R:= gcd(A,B);

%p sturm(sturmseq(R,x),x,-infinity,0) > 0;

%p end proc:

%p select(F, [$1..1000]); # _Robert Israel_, Nov 14 2017

%t 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 *)

%o (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

%Y Cf. A027750, A032741, A292226.

%K nonn

%O 1,1

%A _Michel Lagneau_, Aug 18 2017