The language is $L = \{a^{i} b^{j} c^{k} \;|\; k \neq 2j\}$. I'm trying to write a grammar for this language, what I have so far is:$S \rightarrow AT_{1} \;|\; AT_{2} \;|\; AT_{3} \;|\; AB \;|\; AC$$A \rightarrow aA \;|\; \varepsilon$ $B \rightarrow bB \;|\; \varepsilon$$C \rightarrow cC \;|\; \varepsilon$$T_{1} \rightarrow bbB'T_{1}c \;|\; \varepsilon $   (for $2j > k$)(1)$B' \rightarrow bB' \;|\; b$$T_{2} \rightarrow bT_{2}ccC'\;|\; \varepsilon$ (for $2j < k$)$C' \rightarrow cC' \;|\; c$$T_{3} \rightarrow bT_{3}c \;|\; \varepsilon$ (for $j = k$)the problem that I am having is, the string $bbccc$  can't be generated although valid, in that case $j = 2$ and $k = 3$  so $2\times 2 > 3$  corresponding to production rule (1), how can I fix this?
Context Free Grammar for language
formal languages;formal grammars;context free
Production for $\{b^jc^k\; |\; j\neq 2k \}$ can be written as $B\rightarrow bBcc\;|\;bB_1\;|\;cB_2$$B_1\rightarrow bB_1\;|\;b\;|\;c\;|\;\epsilon$$B_2\rightarrow cB_2\;|\;c\;|\;\epsilon$You can see that it can't accept $bbcccc$. We can use $B\rightarrow bBcc$ twice but the final $B$ would have to be substituted with either b or c.It accepts $bbccc$ as $B\rightarrow (bBcc) \rightarrow b(bB_1)cc \rightarrow bb(c)cc$For even number of cs $B\rightarrow bBcc$ can be used. For odd number of cs, an extra $B_1\rightarrow c$ can be used.So $\{a^ib^jc^k\; |\; j\neq 2k \}$ has the following grammar$S\rightarrow AB$$A\rightarrow aA \;|\; \epsilon$$B\rightarrow bBcc\;|\;bB_1\;|\;cB_2$$B_1\rightarrow bB_1\;|\;b\;|\;c\;|\;\epsilon$$B_2\rightarrow cB_2\;|\;c\;|\;\epsilon$Note: i can be 0 but j and k can not be 0 simultaneously.