Computing polynomials with cyclic Galois groups in SAGE

Here is some SAGE code to compute a polynomial with degree \(n\) which has cyclic Galois group. The computation is done as in the example https://en.wikipedia.org/wiki/Inverse_Galois_problem#Worked_example:_the_cyclic_group_of_order_three

#!/usr/bin/env spython

import sys
from sage.all import *
from random import sample
from collections import Counter

def t(n,k):
  return binomial(n-floor((k+1)/2),floor(k/2))

def alpha(n,k):
  return (-1)**(k*(k-1)/2)

def pbin(n):
  return sum([alpha(n,k)*t(n,k)*x**(n-k) for k in range(n+1)])


def genCyclicPoly(n):
  x = var('x')
  k = 0
  while True:
    p = k*n+1
    k += 1
    if is_prime(p):
      break
  y = primitive_root(p)
  K = CyclotomicField(p,names='a')
  a = K.gen()
  o = Mod(y**n,p).multiplicative_order()
  s = sum([a**((y**n)**i%p) for i in range(o)])
  pol = s.minpoly('x')
  return (pol,k-1,p,y)

for n in range(1,40+1):
  pol,k,p,y = genCyclicPoly(n)
  print n
  print pol
  print 

For example for \(n = \frac{17-1}{2} = 8 \) we get the following polynomial: \[ x^8 + x^7 - 7x^6 - 6x^5 + 15x^4 + 10x^3 - 10x^2 - 4x + 1 \] Looking at the coefficients of the polynomial and searching for them in OEIS we find the sequence A065941. Consider the following polynomial: \[ p(n,x) = \sum_{k=0}^{n} \binom{n-\left \lfloor \frac{k+1}{2} \right \rfloor}{\left \lfloor \frac{k}{2} \right \rfloor} \cdot (-1)^{\frac{k(k+1)}{2}}\cdot x^{n-k} \] We conjecture that the following is true (let \(p>2\) be prime): \[ Gal(p(n,x)) = C_n \Leftrightarrow n = \frac{p-1}{2} \Leftrightarrow p(n,x) \text{ is irreducible }\] If someone finds a proof or a counterexample to this, please let me know.