The two congruences for $DSOME(n)$

For a nonnegative integer $n$, let $DSOME(n)$ be the sum of the odd parts minus the sum of the even parts, taken over all partitions of $n$ into distinct parts. We prove Baruah and Gogoi's Conjecture 4.1: for every $n\geq0$,

$$DSOME(50n+21)\equiv0\pmod8, \qquad DSOME(100n+71)\equiv0\pmod{16}.$$

The first congruence was proved by Bardhan and Saikia in Theorem 4.11 and Remark 4.12. We prove the second. We will also use their quadratic-representation involution, explicitly given below. The additional step is a coefficientwise theta-series calculation modulo $16$; two parity decompositions then finish the argument.

1. A weighted pentagonal identity

All series in the proof are formal power series with integer coefficients. Put

$$f=\prod_{r\geq1}(1-q^r),\qquad X=\sum_{m\geq1}(-1)^m q^{m^2},\qquad X_d=X(q^d),\qquad D=q\frac{d}{dq}.$$

Thus $X_1=X$. Write $\phi=1+2X$. The generating-function identity in Baruah and Gogoi, Theorem 1.1, together with $\phi=\prod_{r\geq1}(1-q^r)^2/\prod_{r\geq1}(1-q^{2r})$, gives

$$\mathcal F(q):=\sum_{n\geq0}DSOME(n)q^n =\frac{f}{8}\left(\phi^{-1}-\phi^3\right). \tag{1}$$

Set $P_j=j(3j-1)/2$ for $j\in\mathbb Z$. We use the classical identities recorded as (2.2) and (4.54) in Bardhan and Saikia:

$$f=\sum_{j\in\mathbb Z}(-1)^j q^{P_j}, \qquad f\phi^2=\sum_{j\in\mathbb Z}(1-6j)q^{P_j}. \tag{2}$$

Define $U=X(1+X)$ and $Y=fU$. Since $\phi^2=1+4U$, (2) implies

$$Y=\sum_{j\in\mathbb Z}(-1)^j v_j q^{P_j}, \qquad v_{2k}=-3k,\quad v_{2k+1}=3k+1. \tag{3}$$

A direct calculation, valid for every integer $j$, gives

$$3P_j=2v_j^2+v_j.$$

Consequently $v_j\equiv P_j\pmod4$. Substituting this into $3(P_j-v_j)=2v_j(v_j-1)$ modulo $8$ gives $v_j\equiv2P_j^2-P_j\pmod8$. Therefore

$$Y\equiv(2D^2-D)f\pmod8, \qquad Y\equiv Df\pmod4. \tag{4}$$

In particular, both $f$ and $Y$ have zero coefficient at every index not of the form $P_j$.

2. The coefficient calculation modulo $16$

Fix an integer $N\geq0$ such that $N\equiv3\pmod4$ and $N\neq P_j$ for every $j\in\mathbb Z$. For a series $H$, write $\langle H\rangle=[q^N]H$, and put

$$a_d=\langle fX_d\rangle \qquad(d=1,2,4,8,16).$$

We claim that

$$DSOME(N)\equiv -2Na_1-12a_4-8(a_2+a_8+a_{16}) \pmod{16}. \tag{5}$$

Here are the details. Factoring the numerator of (1) gives the exact identity

$$\mathcal F =-Y\left(1+\frac{2X^2}{1+2X}\right).$$

Since $X$ has zero constant term, expansion of the inverse is valid formally. With

$$Z=X^2-2X^3+4X^4,$$

we obtain $\mathcal F\equiv-Y-2YZ\pmod{16}$. Our assumption on $N$ gives $\langle Y\rangle=0$, so

$$DSOME(N)\equiv-2\langle YZ\rangle\pmod{16}. \tag{6}$$

For any series $H$,

$$\begin{aligned} \langle(Df)H\rangle &=N\langle fH\rangle-\langle fDH\rangle,\\ \langle(D^2f)H\rangle &=N^2\langle fH\rangle -2N\langle fDH\rangle+\langle fD^2H\rangle. \end{aligned}$$

Using (4), then $N\equiv3\pmod4$, yields

$$\langle YZ\rangle\equiv (2-N)\langle fZ\rangle +5\langle fDZ\rangle +2\langle fD^2Z\rangle \pmod8. \tag{7}$$

The following congruences follow coefficientwise by separating odd and even $m$ in the definition of $X$:

$$\begin{aligned} DX&\equiv X+X_4-2X_{16}\pmod4,\\ D^2X&\equiv X+X_4\pmod2,\\ (DX)^2&\equiv X_2+X_8\pmod2. \end{aligned} \tag{8}$$

For the last line, square the preceding parity expression and use $X_d^2\equiv X_{2d}\pmod2$.

Differentiating $Z$ gives

$$\begin{aligned} DZ&\equiv2U\,DX\pmod8,\\ 2D^2Z&\equiv4(DX)^2+4U\,D^2X\pmod8. \end{aligned}$$

Substitution into (7), followed by (8), gives

$$\begin{aligned} \langle YZ\rangle\equiv \big\langle f\big(&-NX^2+4X^4+6UX_4\\ &-4UX_{16}+4X_2+4X_8\big)\big\rangle \pmod8. \end{aligned} \tag{9}$$

Indeed, before using $N\equiv3\pmod4$, the coefficient of $X^3$ in this simplification is $2N+2$, which vanishes modulo $8$.

To simplify (9), use $\langle fX^2\rangle=-a_1$, which follows from $\langle Y\rangle=0$, and $X^4\equiv X_4\pmod2$. Also (4) and $DX_d\equiv0\pmod d$ give

$$\langle YX_4\rangle\equiv Na_4\pmod4, \qquad \langle YX_{16}\rangle\equiv Na_{16}\pmod2.$$

It follows that

$$\begin{aligned} \langle YZ\rangle &\equiv Na_1+(4+6N)a_4 -4Na_{16}+4a_2+4a_8\\ &\equiv Na_1+6a_4+4(a_2+a_8+a_{16}) \pmod8. \end{aligned}$$

Together with (6), this proves (5).

3. The progression $100n+71$

Now let $N=100n+71$ with $n\geq0$, and put

$$M=24N+1=5(480n+341).$$

Thus the exponent of $5$ in $M$ is exactly one. In particular $M$ is not a square, so $N$ is not pentagonal, since $24P_j+1=(6j-1)^2$. Also $N\equiv3\pmod4$, and (5) applies.

By (2), the sums $a_d$ have the explicit form

$$a_d= \sum_{\substack{j\in\mathbb Z,\ m\geq1\\ P_j+dm^2=N}}(-1)^{j+m}. \tag{10}$$

For $d=2$, a summand would give $M=(6j-1)^2+48m^2$. Reduction modulo $5$ forces both $6j-1$ and $m$ to be divisible by $5$, because $2$ is a quadratic nonresidue modulo $5$. This contradicts the exponent of $5$ in $M$. The same argument for $d=8$ uses the nonresidue $3$. Hence

$$a_2=a_8=0. \tag{11}$$

Consider the finite set

$$\mathcal R= \{(j,m)\in\mathbb Z\times\mathbb Z_{>0}: P_j+m^2=N\}.$$

We use the involution from the proof of Bardhan and Saikia's Theorem 4.11. For $(j,m)\in\mathcal R$, put $u=6j-1$. Then $u^2+24m^2=M$. Neither $u$ nor $m$ is divisible by $5$, and there is a unique $s\in\{1,-1\}$ such that $5\mid u+sm$. Define

$$u' =\frac{24sm-u}{5},\qquad m'=\frac{|su+m|}{5},\qquad j'=\frac{u'+1}{6}. \tag{12}$$

Both $u'$ and $m'$ are integers, $u'\equiv-1\pmod6$, and $m'>0$: equality $su+m=0$ would make $M=25m^2$. Moreover,

$$(u')^2+24(m')^2=M,$$

so $(j',m')\in\mathcal R$. To check the inverse explicitly, let $\varepsilon$ be the sign of $su+m$ and put $s'=s\varepsilon$. Then

$$s'm'=\frac{u+sm}{5},\qquad u'+s'm'=5sm.$$

Thus $s'$ is the unique sign selected for $(u',m')$, and applying (12) again returns $u$ and $m$.

Furthermore,

$$m'\equiv m+1\pmod2,\qquad j'=\frac{4sm-j+1}{5}\equiv j+1\pmod2.$$

The involution therefore exchanges even and odd $m$ and preserves the sign $(-1)^{j+m}$.

For $d=2,4,8$, set

$$E_d=\sum_{\substack{(j,m)\in\mathcal R\\ d\mid m}}(-1)^j.$$

Pairing by (12) gives $a_1=2E_2$. In addition, simply separating even and odd values of $m/2$ and $m/4$ gives

$$E_2+a_4=2E_4, \qquad E_4+a_{16}=2E_8. \tag{13}$$

Using $N\equiv3\pmod4$ and (13), we obtain

$$\begin{aligned} NE_2+3a_4+2a_{16} &\equiv3(E_2+a_4)+2a_{16}\\ &=6E_4+2a_{16}\\ &\equiv2(E_4+a_{16}) =4E_8\equiv0\pmod4. \end{aligned}$$

Finally, (5), (11) and $a_1=2E_2$ imply

$$DSOME(N)\equiv -4(NE_2+3a_4+2a_{16}) \equiv0\pmod{16}.$$

This proves the second congruence for every $n\geq0$. Together with the previously proved modulo-$8$ congruence, it establishes all of Conjecture 4.1.