So, described algorithm really provides different answers to different quizzes, but I don’t get the way we can find real answers to quiz from these numbers.

From the first 2 questions we know that we have only one T and that both answers to A group are F.

Now we got the answers to our ‘sum’ attempt which are 2 for both quizzes, and to our ‘diff’ attempt, which are 3 and 1.

You say: The parity of the sum of “false” answers in A − Ai + Bi and Ai + Bi + Ci is the same as in A plus Ci. But we know A’s score from the second test. Hence we can derive Ci.

And here is where I have a difficulty to follow. For FFFFT quiz where C=T we get an odd sum (2+3) and for FFFTF case where C=F we get an odd sum (2+1). So, how can you derive from the parity of the sum (3 and 5 in our cases) the value of Ci bit?

Than you. ]]>

For two questions the two tests are TT (base) and TF. A is the first two, B is the second two, C is the last question.

Hence, for five questions.

The first test is base: TTTTT.

The second test establishing the base for A: FFTTT.

The Sum test is: TFTFF.

The Diff test is: FTTFT

As you can see this is almost the same solution as I gave: you just need to reverse test 3.

]]>I tried to do this myself and failed. Here is how I was thinking:

For 2 questions A1=TF,B1=TF,AA1=FT,C1=T. So after TT,TT,T and TT,FF,F we will have TF,TF,T and FT,TF,? What should be in the last position for the last question? If we have T in the last position, then for questions 3 and 4, parity of the sum would be equal to parity of A, and if we have F in the last position, then parity of the sum will be equal to parity of A+1. None of this approaches gives us the value of C bit. Can you help me with this problem?

Thank you. ]]>

https://oeis.org/A000788

a(n) = A000788(n-1)+1

]]>The sequence is interesting in itself. The fact that it appeared in your paper and in my blog is more than enough motivation for submitting it.

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