Question Number 298 by 123456 last updated on 25/Jan/15
$${a}\left(\mathrm{n},{m}\right)=\begin{cases}{{n}+{m}}&{{n}\leqslant\mathrm{0}}\\{{b}\left({n}\right)+{m}}&{{n}>\mathrm{0}\wedge{m}\leqslant\mathrm{0}}\\{{b}\left({n}\right)+{b}\left({m}\right)}&{{n}>\mathrm{0}\wedge{m}>\mathrm{0}}\end{cases} \\ $$$${b}\left({n}\right)=\begin{cases}{\mathrm{0}}&{{n}\leqslant\mathrm{0}}\\{{b}\left({n}−\mathrm{1}\right)}&{\mathrm{0}<{n}<\mathrm{5}}\\{{b}\left({n}−\mathrm{1}\right)+{b}\left({n}+\mathrm{1}\right)}&{{n}=\mathrm{5}}\\{{b}\left({n}+\mathrm{1}\right)}&{\mathrm{5}<{n}\leqslant\mathrm{10}}\\{\mathrm{1}}&{{n}>\mathrm{10}}\end{cases} \\ $$$$\mathrm{find} \\ $$$${a}\left(\mathrm{5},\mathrm{5}\right)+{a}\left(\mathrm{4},\mathrm{6}\right) \\ $$
Answered by prakash jain last updated on 19/Dec/14
$${a}\left(\mathrm{5},\mathrm{5}\right)={b}\left(\mathrm{5}\right)+{b}\left(\mathrm{5}\right) \\ $$$${a}\left(\mathrm{4},\mathrm{6}\right)={b}\left(\mathrm{4}\right)+{b}\left(\mathrm{6}\right) \\ $$$${b}\left(\mathrm{4}\right)={b}\left(\mathrm{3}\right)={b}\left(\mathrm{2}\right)={b}\left(\mathrm{1}\right)={b}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${b}\left(\mathrm{6}\right)={b}\left(\mathrm{7}\right)={b}\left(\mathrm{8}\right)={b}\left(\mathrm{9}\right)={b}\left(\mathrm{10}\right)={b}\left(\mathrm{11}\right)=\mathrm{1} \\ $$$${b}\left(\mathrm{5}\right)={b}\left(\mathrm{4}\right)+{b}\left(\mathrm{6}\right)=\mathrm{1} \\ $$$${a}\left(\mathrm{5},\mathrm{5}\right)=\mathrm{2} \\ $$$${a}\left(\mathrm{4},\mathrm{6}\right)=\mathrm{1} \\ $$$${a}\left(\mathrm{5},\mathrm{5}\right)+{a}\left(\mathrm{4},\mathrm{6}\right)=\mathrm{3} \\ $$