Let $a(n)$ be A002326 (i.e., multiplicative order of $2 \operatorname{mod} 2n+1$).
Let $b(n)$ be A179382 (i.e., the smallest period of pseudo-arithmetic progression with initial term $1$ and difference $2n-1$).
Let $c(n)$ be A136042 (i.e., base-$2$ MR-expansion of $\frac{1}{29}$).
Here base-$2$ MR-expansion of $\frac{1}{x}$ is the integer sequence $\{s(1),s(2),s(3),\ldots\}$ where $s(i)$ is the smallest exponent $d$ such that $2^{d}x(i)>1$ and where $x(i+1)=2^{d} x(i)-1$, with initialization $x(1)=x$.
- Let
$$ \ell(n)=\left\lfloor\log_2 n\right\rfloor $$
Let $\operatorname{tr}(n)$ be A007814 i.e. number of trailing zeros in the binary expansion of $n$.
Let $d(n, m)$ be an integer sequence such that we define $A:=\ell((2m+1)n)-\ell(2m+1)+1$ and $B:=\left\lfloor\frac{2^{A+\ell(2m+1)}}{2m+1}\right\rfloor-n+1$ and $d(n,m)=2^{A-1}$ if $n=B$, $d(n,m)=2^{A}-d(B,m)$ otherwise.
I conjecture that base-$2$ MR-expansion of $\frac{1}{2n+1}$ is periodic with period length $b(n+1)$.
I also conjecture that $d(n,m)$ is an arising sequence of numbers $k$ such that values of $\operatorname{tr}(d(n,m))$ belong to numbers $q$ congruent to a specific set of integers modulo $a(m)$.
To get this set, take first $b(m+1)$ values of the base-$2$ MR-expansion of $\frac{1}{2n+1}$, change $s(1)$ to $0$ and then consecutively apply $s(i+1):=s(i)+s(i+1)$.
Here is the PARI/GP program to check it numerically:
a(n) = znorder(Mod(2, 2*n+1))
oddres(n) = n>>valuation(n, 2)
b(n) = my(d=2*n-1, k=1, t=1); while((t=oddres(t+d))>1, k++); k
b1(n) = my(A1 = logint(n, 2) + 2, A2 = A1, B = 2^A2/(2*n+1) - 1, C = 0); until(A1 == A2, A2 = 0; until(2^A2*B>1, A2++); B = 2^A2*B - 1; C++); C
d(n, m) = m = 2*m+1; my(L = logint(m, 2), A = logint(m*n, 2) - L + 1, B = 2^(A+L)\m - n + 1, C = 0, D = 0); until(B == n, D += (-1)^C*2^A; n = B; A = logint(m*n, 2) - L + 1; B = 2^(A+L)\m - n + 1; C++); D + (-1)^C*2^(A-1)
s(n) = my(A = b1(n), B = 1/(2*n+1), v1); v1 = vector(A, i, 0); for(i=1, A, my(C = 0); until(2^C*B>1, C++); B = 2^C*B - 1; v1[i] = C); v1[1] = 0; for(i=2, A, v1[i] += v1[i-1]); v1
test(n) = b(n+1) == b1(n)
my(z=1, m = 1, v1); v1 = s(m); for(k=1, #v1, while(!(valuation(d(z, m), 2)==v1[k]), z++); print(z); z++;);
Is there a way to prove it?
