Support-character carry recursion has only simple extra standards

Put $z_a(x)=x^a+x^{-a}$ for $a>0$, and use $z_0=2$ in identities involving it. Summing over the lower digit signs gives $F_M(x)={\chi }_{n_r{s}_r-1}(x){\prod }_{j<r,n_j>0}{z}_{n_j{s}_j}(x)\mathrm{.}$ Indeed the positive highest exponential times the sign product minus its negative counterpart, divided by $x-x^{-1}$, is exactly the sum of the indicated ${\chi }_m$. The lower signed sums cannot make the exponent negative, since their absolute sum is at most $s_r-1$, while the highest contribution is at least $s_r$. No zero digit contributes a factor $2$, since the support is a set.

In positive characteristic $p$, define $C_A(y)$ by the same formula as step 1.1 using ordinary base-$p$ digits of the positive integer $A$, and put $C_0=0$. If $A=hp+d$ with $h\ge 1$ and $0\le d<p$, this formula gives $C_A(y)={\chi }_{p-1}(y)C_h(y^p)$ times $z_d(y)$ when $d>0$, and times $1$ when $d=0$. When $1\le A<p$, it gives $C_A={\chi }_{A-1}$. For $A\ge 2$, write $z_1{C}_{A-1}=C_A+{\sum }_{0<B<A}{\gamma }_{A,B}{C}_B\mathrm{.}$ The coefficients are defined recursively as follows. If $A-1<p$, the only extra is $B=A-2$ with coefficient $1$ when $B>0$. Otherwise write $A-1=hp+d$. If $d=0$, there are no extras. If $1\le d\le p-2$, the only extra is $B=A-2$, with coefficient $2$ for $d=1$ and $1$ otherwise. If $d=p-1$, take the extra $B=A-2$ with coefficient $2$ for $p=2$ and $1$ otherwise, and add the extras $B=p{B}^{\mathrm{\prime }}$ with coefficients ${\gamma }_{h+1,B^{\mathrm{\prime }}}$. This recursion decreases $A$ and proves the displayed identity: use $z_1{\chi }_j={\chi }_{j+1}+{\chi }_{j-1}$ at the base, and $z_1{z}_d=z_{d+1}+z_{d-1}$ at the other stages, remembering $z_0=2$. All coefficients are nonnegative, and every extra has $B\equiv A(\mathrm{m}\mathrm{o}\mathrm{d}2)$. In characteristic zero define $C_A={\chi }_{A-1}$ and use just the base recurrence with extra $B=A-2$.

In positive characteristic every extra of step 2.1 satisfies $\frac{A-B}{2}<p^{{\nu }_p(B)}(p-b),b\equiv B\mathrm{/}{p}^{{\nu }_p(B)}(\mathrm{m}\mathrm{o}\mathrm{d}p),1\le b<p\mathrm{.}$ For an unscaled extra $B=A-2$, the left side is $1$. In the base case $1\le B\le p-2$, the right side is at least $2$. In the noncarry case, either $d=1$ and $p\mid B$, making the right side at least $p\ge 2$, or $B<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>p=d-1\le p-3$, making it at least $3$. In the carry case $p>2$ gives $B<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>p=p-2$ and right side $2$; when $p=2$, $B$ is positive and even, so the right side is again at least $2$. A scaled extra $A=p(h+1),B=p{B}^{\mathrm{\prime }}$ multiplies both sides of the inductive inequality for $(h+1,B^{\mathrm{\prime }})$ by $p$. This proves the assertion by induction.

Suppose $\mathrm{\ell }<\mathrm{\infty }$. Put $S_{\mathrm{\ell }}(x)={\chi }_{\mathrm{\ell }-1}(x)$. For $M=h\mathrm{\ell }+d$ with $h\ge 1$ and $0\le d<\mathrm{\ell }$, step 1.1 factors $F_M=S_{\mathrm{\ell }}(x)C_h(x^{\mathrm{\ell }})$ times $z_d(x)$ if $d>0$ and times $1$ otherwise. If $n<\mathrm{\ell }$, then $z_1{F}_n=F_{n+1}+F_{n-1}$, omitting $F_0$, including $n=\mathrm{\ell }-1$. If $n\ge \mathrm{\ell }$, write $n=h\mathrm{\ell }+d$. For $d=0$ one gets just $F_{n+1}$. For $1\le d\le \mathrm{\ell }-2$ one gets $F_{n+1}$ and the extra $F_{n-1}$ with coefficient $2$ if $d=1$ and $1$ otherwise. For $d=\mathrm{\ell }-1$, put $A=h+1$: step 2.1 gives $F_{n+1}$, the extras $F_{B\mathrm{\ell }}$ with coefficients ${\gamma }_{A,B}$, and $F_{n-1}$ with coefficient $2$ if $\mathrm{\ell }=2$ and $1$ otherwise. These are the asserted $\beta$; if two indices coincide their coefficients add. Immediate extras have $r=n-2$ and $\mathrm{\ell }\nmid n$, so [F2] makes $W_{n,r}$ simple. The scaled extras have $n+1=A\mathrm{\ell },r+1=B\mathrm{\ell }$ and satisfy the interval of [F2] by step 3.1, so are also simple; in characteristic zero use the critical clause of [F2]. If $\mathrm{\ell }=\mathrm{\infty }$, the recurrence $z_1{\chi }_{n-1}={\chi }_n+{\chi }_{n-2}$ gives the claim and [F2] gives simplicity of every extra.

The Laurent polynomials ${\chi }_m$ are linearly independent over $\mathbb{Z}$: the largest exponent of ${\chi }_m$ is $m$ with coefficient $1$, so a finite nonzero combination has nonzero coefficient at its largest index. Comparing their coefficients in step 4.1, using $z_1{\chi }_0={\chi }_1$ and the out-of-range convention, proves the last identity of the Statement. Finally $\delta =0$ gives $[2{]}_{\delta }=0$ and $[1{]}_{\delta }=1$, hence $\mathrm{\ell }=2$. When $n>0$ is even one has $n\ge \mathrm{\ell }$ and $d=0$ in step 4.1, so its only term is $F_{n+1}$ and ${\beta }_{n,0}=0$.