A000236

[Max]

David Wilson wrote:
> So, how does one effectively compute these values?

I'm not sure yet.
There is something wrong with understanding what A000236 is.

>>> The author defines another function $\Lambda^*(k,m)$ obtained by 
>>> deleting the condition of divisibility in (1)
>>> so that one asks only for the first appearance of $m$ consecutive 
>>> integers whose
>>> $k$th power characters are identical.
>>
>> As I understood, these m consecutive integers are all either k-powers 
>> or k-nonpowers modulo p.
>>
>>> $\Lambda^*(k,2)$ turns out to be very much smaller. The author has 
>>> established that $\Lambda^*(2,2)=3$, $\Lambda^*(3,2)=8$, 
>>> $\Lambda^*(4,2)=20$,
>>> $\Lambda^*(5,2)=44$, $\Lambda^*(6,2)=80$, $\Lambda^*(7,2)=343$. The 
>>> proofs are short enough to be done "by hand", although the last two 
>>> proofs are omitted to save space.
>>
>> This is A000236 and its definition again is very cryptic.
>> I suggest to replace it with something like ``A000236(n)=minimum t 
>> such that for every prime p there exists a pair (r,r+1) of consecutive 
>> n-powers or n-nonpowers modulo p with r<=t''.

If prime p deliver the longest sequence of consecutive residues not containing two adjacent n-powers or n-nonpowers (i.e., 1,2,...,A000236(n)) then this sequence must alternate n-powers and n-nonpowers.
Since 1 is a n-power modulo p (for any n,p), all odd numbers in the sequence 1,2,...,A000236(n) must be n-powers and all even numbers must be n-nonpowers modulo p.
But the number 2^n  is obviously n-power implying that A000236(n) < 2^n which is not the case.

What's wrong?

Max