The sequence ${\displaystyle c_{n}=\sum _{i=0}^{n}a_{i}b_{n-i}}$ is called the discrete convolution or the Cauchy product of the sequences a_n and b_n. For integers ${\displaystyle e}$ and ${\displaystyle f}$ define the convolution sum ${\displaystyle S_{e,f}(n)=\sum _{i=1}^{n-1}\sigma _{e}(i)\sigma _{f}(n-i)}$. Note that ${\displaystyle S_{e,f}(n)=S_{f,e}(n)}$

For odd integers ${\displaystyle e<=f,\;\;e+f=2,4,6,8,12}$, the sum ${\displaystyle S_{e,f}(n)}$ can be evaluated in terms of ${\displaystyle \sigma _{e+f+1},\sigma _{e+f-1},\cdots ,\sigma _{3},\sigma ,}$. Namely:

${\displaystyle \sum _{i=1}^{n-1}\sigma (i)\sigma (n-i)={\frac {5}{12}}\sigma _{3}(n)+\left({\frac {1}{12}}-{\frac {1}{2}}n\right)\sigma (n).}$

${\displaystyle \sum _{i=1}^{n-1}\sigma (i)\sigma _{3}(n-i)={\frac {7}{80}}\sigma _{5}(n)+\left({\frac {1}{24}}-{\frac {1}{8}}n\right)\sigma _{3}(n)-{\frac {1}{240}}\sigma (n).}$

${\displaystyle \sum _{i=1}^{n-1}\sigma (i)\sigma _{5}(n-i)={\frac {5}{126}}\sigma _{7}(n)+\left({\frac {1}{24}}-{\frac {1}{12}}n\right)\sigma _{5}(n)+{\frac {1}{504}}\sigma (n).}$

${\displaystyle \sum _{i=1}^{n-1}\sigma _{3}(i)\sigma _{3}(n-i)={\frac {1}{120}}\sigma _{7}(n)-{\frac {1}{120}}\sigma _{3}(n).}$

${\displaystyle \sum _{i=1}^{n-1}\sigma (i)\sigma _{7}(n-i)={\frac {11}{480}}\sigma _{9}(n)+\left({\frac {1}{24}}-{\frac {1}{16}}n\right)\sigma _{7}(n)-{\frac {1}{480}}\sigma (n).}$

${\displaystyle \sum _{i=1}^{n-1}\sigma _{3}(i)\sigma _{5}(n-i)={\frac {11}{5040}}\sigma _{9}(n)-{\frac {1}{240}}\sigma _{5}(n)+{\frac {1}{504}}\sigma _{3}(n).}$

${\displaystyle \sum _{i=1}^{n-1}\sigma (i)\sigma _{11}(n-i)={\frac {691}{65520}}\sigma _{13}(n)+\left({\frac {1}{24}}-{\frac {1}{24}}n\right)\sigma _{11}(n)-{\frac {691}{65520}}\sigma (n).}$

${\displaystyle \sum _{i=1}^{n-1}\sigma _{3}(i)\sigma _{9}(n-i)={\frac {1}{2640}}\sigma _{13}(n)-{\frac {1}{240}}\sigma _{9}(n)+{\frac {1}{264}}\sigma _{3}(n).}$

${\displaystyle \sum _{i=1}^{n-1}\sigma _{5}(i)\sigma _{7}(n-i)={\frac {1}{10080}}\sigma _{13}(n)+{\frac {1}{504}}\sigma _{7}(n)-{\frac {1}{480}}\sigma _{5}(n).}$

These are the only ${\displaystyle S_{e,f}(n)}$ that can be evaluated in terms of divisor sums and polynomials in ${\displaystyle n}$. For odd integers ${\displaystyle e<=f,\;\;e+f=10,14,16,18,20,24}$, evaluating the sum requires the Ramanujam function ${\displaystyle \tau (n)}$. For example:

${\displaystyle \sum _{i=1}^{n-1}\sigma _{5}(i)\sigma _{5}(n-i)={\frac {65}{174132}}\sigma _{11}(n)+{\frac {1}{252}}\sigma _{5}(n)-{\frac {3}{691}}\tau (n).}$

There are many other similar formulas. For example:

${\displaystyle \sum _{\begin{aligned}r+s+t=n\\r,\;s,\;t>0\end{aligned}}\sigma (r)\sigma (s)\sigma (t)={\frac {7}{192}}\sigma _{5}(n)+\left({\frac {5}{96}}-{\frac {5}{12}}n\right)\sigma _{3}(n)-\left({\frac {1}{192}}-{\frac {1}{16}}n+{\frac {1}{8}}n^{2}\right)\sigma (n).}$

See Eisenstein series for a discussion of the series and functional identities involved in these formulas. ^[1]

${\displaystyle \sigma _{3}(n)={\frac {1}{5}}\left\{6n\sigma _{1}(n)-\sigma _{1}(n)+12\sum _{0<k<n}\sigma _{1}(k)\sigma _{1}(n-k)\right\}.}$     ^[2]

${\displaystyle \sigma _{5}(n)={\frac {1}{21}}\left\{10(3n-1)\sigma _{3}(n)+\sigma _{1}(n)+240\sum _{0<k<n}\sigma _{1}(k)\sigma _{3}(n-k)\right\}.}$     ^[3]

${\displaystyle {\begin{aligned}\sigma _{7}(n)&={\frac {1}{20}}\left\{21(2n-1)\sigma _{5}(n)-\sigma _{1}(n)+504\sum _{0<k<n}\sigma _{1}(k)\sigma _{5}(n-k)\right\}\\&=\sigma _{3}(n)+120\sum _{0<k<n}\sigma _{3}(k)\sigma _{3}(n-k).\end{aligned}}}$     ^{[3] [4]}

${\displaystyle {\begin{aligned}\sigma _{9}(n)&={\frac {1}{11}}\left\{10(3n-2)\sigma _{7}(n)+\sigma _{1}(n)+480\sum _{0<k<n}\sigma _{1}(k)\sigma _{7}(n-k)\right\}\\&={\frac {1}{11}}\left\{21\sigma _{5}(n)-10\sigma _{3}(n)+5040\sum _{0<k<n}\sigma _{3}(k)\sigma _{5}(n-k)\right\}.\end{aligned}}}$     ^{[2] [5]}

${\displaystyle \tau (n)={\frac {65}{756}}\sigma _{11}(n)+{\frac {691}{756}}\sigma _{5}(n)-{\frac {691}{3}}\sum _{0<k<n}\sigma _{5}(k)\sigma _{5}(n-k),}$     where τ(n) is Ramanujan's function.     ^{[6] [7]}

Since σ_k(n) (for natural number k) and τ(n) are integers, the above formulas can be used to prove congruences ^[8] for the functions. See Ramanujan tau function for some examples.

Extend the domain of the partition function by setting p(0) = 1.

${\displaystyle p(n)={\frac {1}{n}}\sum _{1\leq k\leq n}\sigma (k)p(n-k).}$     ^[9]   This recurrence can be used to compute p(n).