For a variation on the bilinear recursion relation in the OP that involves only sums up to $n/2$, see [1,2].

Expanding the question to interesting and little known recursion relations for the zeta function I could mention this linear relation [3,4]:

$$(-1)^n\frac{\pi^{2n}n}{(2n+1)!}+\sum_{k=0}^{n-1}(-1)^k\frac{\pi^{2k}}{(2k+1)!}\zeta(2n-2k)=0.$$

[1] H.-T. Kuo, A recurrence formula for $\zeta(2n)$ (1949).
[2] L. Carlitz, A recurrence formula for $\zeta(2n)$ (1961).
[3] I. Song, A recursive formula for even order harmonic series (1987).
[4] M. Merca, On the Song recurrence relation for the Riemann zeta function (2016).

For a variation on the bilinear recursion relation in the OP that involves only sums up to $\lfloor n/2\rfloor$, rather than up to $n-1$, see [1,2].

Expanding the question to interesting and little known recursion relations for the zeta function I could mention this linear relation [3,4]:

$$(-1)^n\frac{\pi^{2n}n}{(2n+1)!}+\sum_{k=0}^{n-1}(-1)^k\frac{\pi^{2k}}{(2k+1)!}\zeta(2n-2k)=0.$$

[1] H.-T. Kuo, A recurrence formula for $\zeta(2n)$ (1949).
[2] L. Carlitz, A recurrence formula for $\zeta(2n)$ (1961).
[3] I. Song, A recursive formula for even order harmonic series (1987).
[4] M. Merca, On the Song recurrence relation for the Riemann zeta function (2016).

This is not an answer to the specific question in the OP, but expanding the question to interesting and little known recursion relations for the zeta function I could mention this linear relation [1,2]:

$$(-1)^n\frac{\pi^{2n}n}{(2n+1)!}+\sum_{k=0}^{n-1}(-1)^k\frac{\pi^{2k}}{(2k+1)!}\zeta(2n-2k)=0.$$

[1] I. Song, A recursive formula for even order harmonic series (1987).
[2] M. Merca, On the Song recurrence relation for the Riemann zeta function (2016).

For a variation on the bilinear recursion relation in the OP that involves only sums up to $n/2$, see [1,2].

Expanding the question to interesting and little known recursion relations for the zeta function I could mention this linear relation [3,4]:

$$(-1)^n\frac{\pi^{2n}n}{(2n+1)!}+\sum_{k=0}^{n-1}(-1)^k\frac{\pi^{2k}}{(2k+1)!}\zeta(2n-2k)=0.$$

[1] H.-T. Kuo, A recurrence formula for $\zeta(2n)$ (1949).
[2] L. Carlitz, A recurrence formula for $\zeta(2n)$ (1961).
[3] I. Song, A recursive formula for even order harmonic series (1987).
[4] M. Merca, On the Song recurrence relation for the Riemann zeta function (2016).