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For any non-negative integer $\sf n$ (and a formal variable $\sf q$),
RR1 : # of partitions with each part $\sf \not\equiv 0,\pm 2 \pmod{5}$ $\sf =$ # of partitions with difference at least $\sf 2$ at distance $\sf 1$
$\sf \prod\limits_{m>0,\,\, m \,\not\equiv \,0, \pm 2 \pmod{5}} \frac{1}{(1-q^m)} = \sum\limits_{n=0}^\infty d_1(n)q^n $
RR2 : # of partitions with each part $\sf \not\equiv 0,\pm 1 \pmod{5}$ $\sf =$ # of partitions with difference at least $\sf 2$ at distance $\sf 1$ and with $\sf 1$ appearing at most $\sf0$ times
$\sf \prod\limits_{m>0,\,\, m \,\not\equiv\, 0,\pm 1 \pmod{5}} \frac{1}{(1-q^m)}=\sum\limits_{n=0}^\infty d_2(n)q^n $
# of partitions of type RR1 $\sf \ge $ # of partitions of type RR2.
(Hint: look at the sum side).
Ehrenpreis's question: Can one deduce this without looking at the sum sides?
RR1 : # of partitions with each part $\sf \not\equiv 0,\pm 2 \pmod{5}$ $\sf =$ # of partitions with difference at least $\sf 2$ at distance $\sf 1$
| \begin{align*} \sf 10 &= \sf 9+1\\ \sf &= \sf 6+4\\ \sf &= \sf 6+1+1+1+1\\ \sf &= \sf 4+4+1+1\\ \sf &=\sf 4+1+1+1+1+1+1\\ \sf &=\sf 1+1+\cdots+1+1\quad\quad \end{align*} | $ \begin{align*} \sf 10 &=\sf 10\\ &= \sf 9+1\\ \sf &= \sf 8 + 2\\ \sf &= \sf 7 + 3 \\ \sf &= \sf 6+4 \\ \sf & = \sf 6+3+1 \quad\quad \end{align*} $ |
RR2 : # of partitions with each part $\sf \not\equiv 0,\pm 1 \pmod{5}$ $\sf =$ # of partitions with difference at least $\sf 2$ at distance $\sf 1$ and with $\sf 1$ appearing at most $\sf0$ times
| \begin{align*} \sf 10 &= \sf 8+2\\ \sf &= \sf 7+3\\ \sf &= \sf 3+3+2+2\\ \sf &=\sf 2+2+2+2+2\quad\quad \end{align*} | $ \begin{align*} \sf 10 &=\sf 10\\ \sf &= \sf 8 + 2\\ \sf &= \sf 7 + 3 \\ \sf &= \sf 6+4 \quad\quad \end{align*} $ |
For any $\sf i=1,\dots,k$,
# of partitions with each part $\sf \not\equiv 0, \pm (k-i+1) \text{ (mod }2k+1)$
Goellnitz-Gordon-Andrews, Capparelli, Nandi, some recent conjectures of M. C. Russell and S. K.
Say $\sf k=4$, hence, there are $\sf 4$ GA identities, the modulus is $\sf 9.$
GA4,1: # of partitions with each part $\sf \not\equiv 0, \pm 4 \text{ (mod } 9)$ $\sf =$ # of partitions with difference at least $\sf 2$ at distance $\sf 3$ and with $\sf 1$ appearing at most $\sf 3$ times.
GA4,2: # of partitions with each part $\sf \not\equiv 0, \pm 3 \text{ (mod } 9)$ $\sf =$ # of partitions with difference at least $\sf 2$ at distance $\sf 3$ and with $\sf 1$ appearing at most $\sf 2$ times.
GA4,3: # of partitions with each part $\sf \not\equiv 0, \pm 2 \text{ (mod } 9)$ $\sf =$ # of partitions with difference at least $\sf 2$ at distance $\sf 3$ and with $\sf 1$ appearing at most $\sf 1$ times.
GA4,4: # of partitions with each part $\sf \not\equiv 0, \pm 1 \text{ (mod } 9)$ $\sf =$ # of partitions with difference at least $\sf 2$ at distance $\sf 3$ and with $\sf 1$ appearing at most $\sf 0$ times.
GG4,1 $\sf \ge$ GG4,2 $\sf \ge$ GG4,3 $\sf \ge$ GG4,4.
Say $\sf k=4$, hence, there are $\sf 4$ AB identities, the modulus is $\sf 8.$
AB4,1: something related to "$\sf \not\equiv 0, \pm 4 \text{ (mod } 8)$" $\sf =$ # of partitions with difference at least $\sf 2$ at distance $\sf 3$ and with $\sf 1$ appearing at most $\sf 3$ times + a parity condition.
AB4,2: # of partitions with each part $\sf \not\equiv 0, \pm 3 \text{ (mod } 8)$ $\sf =$ # of partitions with difference at least $\sf 2$ at distance $\sf 3$ and with $\sf 1$ appearing at most $\sf 2$ times + a parity condition.
AB4,3: # of partitions with each part $\sf \not\equiv 0, \pm 2 \text{ (mod } 8)$ $\sf =$ # of partitions with difference at least $\sf 2$ at distance $\sf 3$ and with $\sf 1$ appearing at most $\sf 1$ times + a parity condition.
AB4,4: # of partitions with each part $\sf \not\equiv 0, \pm 1 \text{ (mod } 8)$ $\sf =$ # of partitions with difference at least $\sf 2$ at distance $\sf 3$ and with $\sf 1$ appearing at most $\sf 0$ times + a parity condition.
AB4,1 $\sf \ge$ AB4,3 and AB4,2 $\sf \ge$ AB4,4.
| Algebra | Level | |
| Rogers-Ramanujan | $\sf A_1^{(1)}$ | $\sf 3$ |
| Gordon-Andrews for modulus $\sf 2k+1$ | $\sf A_1^{(1)}$ | $\sf 2k-1$ |
| Euler | $\sf A_1^{(1)}$ | $\sf 2$ |
| Andrews-Bressoud for modulus $\sf 2k$ | $\sf A_1^{(1)}$ | $\sf 2k-2$ |
| Goellnitz-Gordon (modulus $\sf 8$) | $\sf A_5^{(2)}$ | $\sf 2$ |
| Capparelli | $\sf A_2^{(2)}$ | $\sf 3$ |
| Nandi | $\sf A_2^{(2)}$ | $\sf 4$ |
| M. C. Russell - S. K. mod $\sf 9$ conjectures | $\sf D_4^{(3)}$ | $\sf 3$ |
Motivated by Ehrenpreis's question (RR1 $\sf \ge$ RR2, without looking at sum-sides)
$\sf\begin{align*} G_1 &= \sf \prod\limits_{m>0,\,\, m \,\not\equiv \,0, \pm 2 \pmod{5}} \frac{1}{(1-q^m)} = 1 + q + q^2 + q^3+2q^4+2q^5+3q^6+3q^7+4q^8+5q^9+6q^{10}+\cdots\\ \sf G_2 &= \sf \prod\limits_{m>0,\,\, m \,\not\equiv \,0, \pm 1 \pmod{5}} \frac{1}{(1-q^m)} = 1 +\hphantom{q}+q^2+q^3+\,\,q^4+\,\,q^5\,+2q^6+2q^7+3q^8+4q^9+4q^{10}+\cdots\\ \end{align*}$
Observe: $\sf G_1 \ge G_2.$
$\sf G_1 - G_2 = q+q^4+q^5+q^6+q^7+2q^8+2q^9+\cdots.$
Let $\sf G_3 = (G_1 - G_2)/q = 1+q^3+q^4+q^5+q^6+2q^7+2q^8+\cdots.$
Let $\sf G_4 = (G_2 - G_3)/q^2 = 1+q^4+q^5+q^6+2q^7+2q^8+\cdots$.
Continue:
Let $\sf \bbox{G_i = (G_{i-2} - G_{i-1})/q^{i-2}}.$
Observe that each $\sf G_i = 1 + q^i + \cdots$ : Empirical Hypothesis
Proof of the Empirical Hypothesis not only answers Ehrenpreis's question but also immediately leads to a proof of RR.
$\sf\prod \longrightarrow \sum$
$\sf G_3 = (G_1 - G_2)/q \quad$ corresponds to the exact sequence $\quad\sf 0 \longrightarrow \Lambda_{G_2} \longrightarrow \Lambda_{G_1} \longrightarrow ? \longrightarrow 0. $
$\sf B_{(k-1)(j+1)+i}= \dfrac{B_{(k-1)j+k-i+1}-B_{(k-1)j+k-i+3}}{q^{(j+1)(i-2)}(1+q^{j+1})}$
$\sf B_{(k-1)(j+1)+i} = \dfrac{B_{(k-1)j + k-i+1 } - \mathfrak{B}_{ (k-1)j +k-i+2 } } { q^{ (j+1) (i-1) } } = \dfrac{\mathfrak{B}_{ (k-1)j + k-i+2 } - B_{ (k-1)j+k-i+3 } }{ q^{ (j+1)( i - 2) } } $
"I've caught a cold," the Thing replies, "Out there upon the landing." I turned to look in some surprise, And there, before my very eyes, A little Ghost was standing! He trembled when he caught my eye, And got behind a chair. "How came you here," I said, "and why? I never saw a thing so shy. Come out! Don't shiver there!" Phantasmagoria, Lewis Carroll |