Qusetion: Write transition functions for accepting the language 
a^1 b^j c^k, with i=2j or j=2k

I presume i,j,k not= 0.

Part a
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i=2j only.

Start state q1

1. d(q1, a, Z) = (q2, AZ),	// change state
2. d(q2, a, A) = (q1, A),  // do not push anything on stack but toggle state
3. d(q1, a, A) = (q2, AA), //toogle + push on stack
4. d(q1, b, A) = (q1, epsilon), // b clears stack
5. d(q1, c, Z) = (q1, Z), // first c must see bottom of stack
6. d(q1, epsilon, Z) = (q1, epsilon) // emty stack on end of string

Part b
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j=2k added

Create new start state q0, to have two paths in the automaton
7. d(q0, epsilon, Z) = {(q1, Z), (q3, Z)} // q3 for the new language part b

8. d(q3, a, Z) = (q3, Z)  // ignore all a's do not touch stack
9. d(q3, b, Z) = (q4, BZ) //  first b encountered
10. d(q4, b, Z) = (q3, B) //ignore every second b
11. d(q3, b, B) = (q4, BB) // toggle and push stack symbol
12. d(q3, c, B) = (q3, epsilon) // c's take off stack symbol
13. d(q3, epsilon, Z) = (q3, epsilon)

Example: aabc
(q0, aabc, Z) [rule 7]: (q1, aabc, Z) [1]: (q2, abc, AZ) [2]:
(q1, bc, AZ) [4]: (q1, c, Z) [5]: (q1, epsilon, Z) [6]: (q1, epsilon, epsilon)
Stack empty at the end of the string, the string is accepted