$\binom{16}{8}$O (n,$\alpha$)$\binom{13}{6}$C and $\binom{9}{4}$Be ($\alpha$,n)$\binom{12}{6}$C

I had initially thought the products & reactants may be O-16 $\to$ C-13 + He-2 + p$^+$ from alpha decay, but I'm leaning towards this form O-16 + C-13 $\to$ ... now. Any help clarifying & determining products would be much appreciated.

The given expression

$\ce{^{16}_{8}O}\,(\mathrm{n},\alpha)\,\ce{^{13}_{6}C}$

translates as

$\ce{^{16}_{8}O ->[+\ \mathrm{n}][-\ \alpha] ^{13}_{6}C}$

or

$\ce{^{16}O + n -> ^{13}C + ^4He}$

and the expression

$\ce{^{9}_{4}Be}\,(\alpha,\mathrm{n})\,\ce{^{12}_{6}C}$

translates as

$\ce{^{9}_{4}Be ->[+\ \alpha][-\ \mathrm{n}] ^{12}_{6}C}$

or

$\ce{^9Be + ^4He -> ^{12}C + n}$