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I'm exploring "Singular Binary Bracelets": size p=2q+1 with p,q prime, q unit beads, resultant amplitude squared is integer. I will try to explain why I think that 'Algebraic Number fields' is needed, and show which questions I'm stuck with.
Define z as principal p-th root of 1: $z= exp(2 \pi i /p)$.
Consider the $f(p,k)= \sum_{i=1}^p b(i) z^i$ with $b(i)= 0|1$ and $\sum_{ i=1}^p b(i)=k$;
For prime $p$, let $p = 2q+1$ with $q$ an odd prime, limiting the values of $p$ to 7,11,23,47,59,83,... alias A005385 : "safe primes".

The above describes a reversible necklace of $p$ beads, $k$ of them being white ( thus $1$), the rest black (thus $0$), where $f(p,k)$ represents the vector sum of all unit vectors pointing to the white beads.

With the restrictions on $p$, all values of $f(p,k)$ are distinct for a given $p$, apart from rotations, reflections and black-white interchange (bracelets instead of necklaces, and $k$ ranging from $0$ to $p-1$. So all distinct bracelets have distinct vector sums.
Let $g(p,k)$ be the count of distinct values of $f(p,k)$, and hence distinct bracelets, then $g(p,k)$ is given by A052307. Remark that by black-white interchange, $f(p,k) = -f(p,p-k)$, so we need only look at the range $k= 0\,..\, (p-1)/2$.

It is clear that $g(p,0)=g(p,1)=1$, but, for $k>1$, $g(p,k) \pmod q =0$ except $g(p,q) \pmod q = 1$. Why?
This shows that within the $g(p,k)$ bracelets, one stands apart as a 'singular bracelet'.

It seems that all values of $|f(p,k)|^2$ for $k>1$ are the roots of a degree-q polynomial, with each of the q roots of this polynomial corresponding to a distinct value of $|f(p,k)|^2$. This 'groups' the values of $f(p,k)$ into polynomial families of size $q$. Why? What do the bracelets $f(p,k)$ within a family have in common?

For the special case of $f(p,q)$, exactly one bracelet is 'singular', corresponding to a degree 1 polynomial. This bracelet is the only one such that $|f(p,q)|^2$, alias the length squared of the sum vector, has an integer value equal to $(q+1)/2$ or equivalently $(p+1)/4$. Why?

The 'singular bracelets' $b(i)$ seem at first sight mostly random, but still show some remarkable properties when compared in their 'standard configuration'. They can always be rotated/reflected such that 

   i=  0    1    2    3  ..   q-1   q  q+1   ..  p-3    p-2   p-1
b(i)=  0    0  b(2) b(3) .. b(q-1)  s  1-s  .. 1-b(3) 1-b(2)   1

with $s=1$ if $q \pmod 4=3$ and $s=0$ if $q \pmod 4=1$

Properties of 'singular bracelets' (conjectured): 

1./ b(0)=0 and b(i) = 1-b(p-i)
2./ b(i)=0 if i is square or a triple of a square,
3./ b(i)=1-s for i= 2k^2 or 6k^2, k>0 
4./ b(i)=0 for i=(q+1)/2
5./ b(q)=s and b(q+1)= -s
6./ b(i)=b(j) if both i and j are of the form (1 or 3)*(1 or a square)*(1 or a given prime r to an odd power)

note that for $r=1$ and $r=2$ these are the cases 2./ and 3./ above.

These conjectures hold for the 20 'singular bracelets' of size $p=7$ up to $p=719$. (The count of different bracelets $f(719,359)$ is approx. $10^{212}$, and only 1 of those is 'singular').

With some luck, I stumbled upon a Mathematica implementation to produce the 'singular bracelets' in standard configuration:

a = ToNumberField[1/2 + I Sqrt[p]/2, Exp[2 Pi I/p]]; b = Append[1 - Last[a], 1]

I'm looking for an Algebraic Number Fields introductory text at amateur level so that I can eventually get to grips with the mess above. Suggestions ?

28 sept 2018:
$b(i)=0$ iff $i=(1+6k)^2 \pmod p$ for $k$ in $0 .. p-1$

Wouter M.
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    I would perhaps recommend an introductory text to number theory in general. There are several good recommendations here on this site, e.g., here. – Dietrich Burde Aug 30 '18 at 18:42
  • If I got it right, the numbers $|f(p,k)|^2$ are all algebraic integers in the real subfield $$L=\Bbb{Q}(z)\cap \Bbb{R}=\Bbb{Q}(\cos(2\pi/p))$$ of the cyclotomic field. We have, indeed, $[L:\Bbb{Q}]=(p-1)/2$ implying that the minimal polynomial of $|f(p,k)|^2$ is of degree that is a factor of $(p-1)/2=q$. Confirming your observation. – Jyrki Lahtonen Aug 30 '18 at 18:43
  • And the sets of values of $f(p,k)$ "grouped" together by the fact that their squared moduli share the same minimal polynomial, are at least (possibly there is more to it) the Galois conjugates. Galois conjugates mean here simply that if $a$ is an integer, coprime to $p$, then $f(p,k)=\sum_ib(i)z^i$ and its "$a$-twist" $f^{[a]}(p,k)=\sum_ib(i)z^{ai}$ share the same minimal polynomial (hence so do their squared moduli). The $a$-twist comes from the sequence $b(a^{-1}i)_{i=1}^p$ with $a^{-1}i$ calculated modulo $p$. – Jyrki Lahtonen Aug 30 '18 at 18:50
  • @JyrkiLahtonen , if I understand your remark correctly, then the "$\alpha$-twist" is equivalent to rotating the bracelet over $\alpha$ positions, right? – Wouter M. Aug 30 '18 at 20:16
  • No. Say with $p=11, a=3$, the $3$-twist of $(b(1),b(2),\ldots,b(11))$ would be $$(b(4),b(8),b(1),b(5),b(9),b(2),b(6),b(10),b(3),b(7),b(11)).$$ The indices in the twisted sequence are incremented by four because $3^{-1}=4\pmod{11}$. – Jyrki Lahtonen Aug 30 '18 at 20:36
  • @JyrkiLahtonen, thank you, got it. The "family groups" are then just 'twisted' bracelets (I picture them as 'streched' by $a$ as in 1->1+a, 2->2+2a, ..). – Wouter M. Aug 30 '18 at 21:10
  • Yes, that¨s a goodw way of thinking about it. – Jyrki Lahtonen Aug 30 '18 at 21:12
  • Am I correct to think that $f$ is a function of $b$? – Peter Taylor Aug 30 '18 at 21:48
  • @PeterTaylor: see line 5 in the Question: $b(i)$ are the binary $0|1$ coefficients of $z^i$ ($b$ stands for 'binary' : the beads of the bracelet). – Wouter M. Aug 31 '18 at 10:22
  • Line 5 is what's confusing me. If $f$ is just a function of $n$ and $k$ then I'm not sure where $b$ comes from. – Peter Taylor Aug 31 '18 at 10:24
  • @PeterTaylor: f(n,k) stands for a set of polynomials of order n with exactly k of the coefficients equal to $1$ and the rest equal to $0$. They are counted by $g(n,k)$. – Wouter M. Aug 31 '18 at 11:04
  • @JyrkiLahtonen: your 'Galois twisters' are actually the answer to the whole question! My 'Singular Bracelets' are invariant under that $a$-stretching. The $f(n,k)$ changes, but they are all rotated/reflected versions of each other. If you post it as an answer, then I can applaud you for it. – Wouter M. Aug 31 '18 at 11:08
  • I will do my best, but a bit later. – Jyrki Lahtonen Aug 31 '18 at 11:26
  • Aha. So $f(p,k)= \left{ \sum_{i \in s} z^s \mid s \subseteq {1,\ldots,p} \wedge |s|=k \right}$ and "values of $|f(p,k)|^2$" means "values of $|x|^2$ for $x \in f(p,k)$". – Peter Taylor Aug 31 '18 at 11:41
  • @PeterTaylor: no, an example to clarify: $f(7,3)=b(1)z+b(2)z^2+b(3)z^3+b(4)z^4+b(5)z^5+b(6)z^6+b(7)z^7$ with only $3$ of the $b(i)$ equal to $1$ and the rest zero. $f(7,3)$ represents the set of all $g(7,3) = 4$ polynomials of this type up to rotations and reflections (hence: 'bracelets'). They are $\text{(1,1,1,0,0,0,0),(0,0,0,1,0,1,1),(0,0,1,0,0,1,1),(0,0,1,0,1,0,1)}$ – Wouter M. Aug 31 '18 at 13:50

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