It seems difficult to get $a_n=b_n$ without doing all the work needed for $a_n=c_n.$ By making the setting sufficiently abstract it seems possible to at least avoid making this blatant.
The idea is to replace each integer $n$ by a generalized integer $I_n$, replace $a_n=\Phi_n(1)$ by some appropriate $A_n$, define a generalized factorial $[I_n]!=I_1I_2\cdots I_n$ and then a generalized binomial coefficient $${n \brack m}=\frac{[I_n]!}{[I_m]![I_{n-m}]!},$$ and deduce that $\gcd\left( {n \brack 1},{n \brack 2},\cdots,{n \brack {n-1}}\right)=A_n$.
The familiar and obvious choice is $I_n=1+X+\cdots+X^{n-1}.$ Then the cyclotomic polynomials $A_n=\Phi_n(X)$ are defined by $$I_n=\prod_{d\mid n}A_d.$$ This seems a good start since setting $X=1$ might yield the desired result. I will , however, make the setting more abstract simply because it seems too hard to ignore the immediate observation that $A_p=1+X+X^2+\cdots+X^{p-1}$ which is then the base case of an easy induction proof that $A_{p^k}=1+X^{q}+X^{2q}+\cdots+X^{(p-1)q}$ for $q=p^{k-1}$ and $k \ge 1$. At this stage $a_n=c_n$ (when $X=1$) is almost immediate.
So to start, simply say that we have a commutative ring $\mathcal{R}$ and within it two sequences of elements $I_1,I_2,I_3,\cdots$ and $A_1,A_2,A_3\cdots$ with the property $$I_n=\prod_{d\mid n}A_d.$$ There are at least two approaches. One is to start with the $A_i$ and use that equation as the definition of the $I_n.$ It seems more satisfying to start with the $I_n$, and stipulate the (somewhat weak) condition
$A_d \mid A_n$ when $d\mid n$ $(*)$
Based just on $(*)$ we have that there are elements $A_1,A_2,\cdots$ uniquely defined by $I_n=\prod_{d\mid n}A_d.$ (Although we will not need it, it then follows that $C_n=\prod_{d\mid n}A_n^{\mu(n/d)}$ where $\mu$ is the classic Mobius function.)
Examples include
- $I_n=n$ in $\mathbb{N}$
- The Fibonacci numbers in $\mathbb{N}$
- $I_n=u^n-v^n$ in $\mathbb{Z}[u,v]$
- $I_n=\frac{u^n-v^n}{u-v}=\Sigma_{k=0}^{n-1}u^{n-k-1}v^{k}$ in $\mathbb{Z}[u,v]$
Asides: We will always be able, if desired, to assume $A_1=I_1=1$ by replacing $I_n$ with $\frac{I_n}{I_1}$ as in 4. Example 1 is the case $u=v=1$ of 4 while example 2 is the case $u,v=(1\pm \sqrt{5})/2$ with the convenient feature $uv=-1$. The obvious choice above was example 4 with $u=X,v=1.$
In any case we can define a generalized factorial $$[I_n]!=I_1I_2I_3\cdots I_n$$ and it follows that $$[I_n]!=\prod_{k=1}^nA_k^{\lfloor n/k\rfloor}$$
We can also define a generalized binomial coefficient by
$${n \brack m}=\frac{[I_n]!}{[I_m]![I_{n-m}]!}=\prod_{k \le n}A_k^{e_{n,m,k}}$$
where the exponent $$e_{n,m,k}=\lfloor n/k \rfloor-\lfloor m/k \rfloor-\lfloor (n-m)/k \rfloor$$
Note that $0 \le e_{n,m,k} \le 1$ (not that we need this fact).
It follows that
- $A_n$ divides all of ${n \brack m}$ for $1 \le m \le n-1$ while $A_1$ divides none of them (in the sense that $e_{n,m,1}=0$)
- ${n \brack 1}=\prod A_d$ where $d$ ranges over the divisors of $n$ with $d \gt 1$
- If $d \mid n$ then $A_d$ does not divide $n \brack d$ in the sense that $e_{n,d,d}=0$
As a specific example, ${6 \brack 1}={6 \brack 5}=A_2A_3A_6$, ${6 \brack 2}={6 \brack 4}=A_3A_5A_6$ and ${6 \brack 3}=A_2A_4A_5A_6.$
We would like to be able to say that $\gcd\left({6 \brack 1},{6 \brack 2},{6 \brack 3}\right)=A_6.$ However this requires a stronger condition than $(*).$ After all, we could start with arbitrary $A_d$ (including $A_3=A_5$) and use $I_n=\prod_{d\mid n}A_d$ to define $I_n$.
Everything said so far only requires $\mathcal{R}$ to be a multiplicative semigroup. In a ring $\mathcal{R}$ I will take as the definition of $\gcd(U,V)=W$ ($W$ is a gcd of $U$ and $V$) to be
$W \mid U$ and $W \mid V$ and $US+VT=W$ for some $S,T \in \mathcal{R}$.
Note that we do not assume that every pair $U,V$ have a $\gcd$. In $\mathbb{Z}[X]$ We do not have a $\gcd$ for $U=\Phi_4=X^2+1$ and $V=\Phi_2=X+1$. We can get $US+VT=2$ but not $1$.However $\gcd(X^3+X^2+X+1,X+1)=X+1.$
A stronger property which holds for example 1, and example 2 is
$$\gcd(I_m,I_n)=I_{\gcd(m,n)} \tag{**}$$ in the sense above.
I don't think it holds for example 4, but it does when $v=1$. The fact that it hold in the case of example 2 is perhaps due to the extra relation $uv=-1$.
I believe that with this definition we can deduce that $\gcd(A_m,A_n)=1$ when $\gcd(m,n)=1$. That together with the divisibility observations above should prove the claim below.
CLAIM In a ring $\mathcal{R},$ given elements $A_i$ with $\gcd(A_m,A_n)=1$ when $\gcd(m,n)=1$, define $I_n$ and ${n \brack m}$ as above. Then $\gcd(I_m,I_n)=I_{\gcd(m,n)} $ and $\gcd\left( {n \brack 1},{n \brack 2},\cdots,{n \brack {n-1}}\right)=A_n$. Furthermore, given only that the $I_m$ are elements such that $\gcd(I_m,I_n)=I_{\gcd(m,n)} ,$ the $A_i$ are well defined, $\gcd(A_m,A_n)=1$ when $\gcd(m,n)=1$, and everything follows.
Actually finding the cofactors might be impractical.
The fact that they exist follows (in certain of the examples above) from
- $1n-1m=n-m$
- $F_{m+1}F_n-F_{n+1}F_M=(-1)^nF_{n-m}$
- $1\frac{X^n-1}{X-1}-X^{n-m}\frac{X^m-1}{X-1}=\frac{X^{n-m}-1}{X-1}$
but even in the first case we don't have a simple way to explicitly find $s,t$ with $ns+mt=\gcd(n,m)$, even in the case that the right-hand side is $1.$ Furthermore, the claim depends on iterated application of facts such as:
Given (in some ring) that $\gcd(U,V)=\gcd(U,V')=1$ in the sense that there are $S,T,S',T'$ with $US+TV=1$ and $U'S'+T'V'=1$ It follows that $\gcd(U,VV')=1$ since $$U{\LARGE(}USS'+ST'V'+S'TV{\LARGE)}+VV'(TT')=1.$$
It is something I have asked about before, although not very clearly.