[Math] Longest coinciding pair of integer sequences known

integer-sequencessequences-and-series

There are arbitrarily many pairs of integer sequences (of arbitrary origins) that coincide upto an $N$ but differ for an $n > N$. I assume, the coincidence will be considered accidentally then by default, but I may be mistaken about that.

One is disadviced to draw any conclusions from coincidences of integer sequences unless its proven, that they coincide for all $n$. (Even then there may be no sensible conclusions, as I have learned here: Equivalence of families of objects with the same counting function.)

In any case, it is hard not to be entrapped to draw a conclusion when $N$ is very large. But what is "very large"? Thus my question:

What is the largest $N$ with two known
integer sequences coinciding upto $N$
but differing for an $n > N$?

(Can this information be captured from OEIS by an intelligent query?)

(I am aware of the fact that one can trivially define pairs of integer sequences which conincide for all $n$ but a single and arbitrarily large one. It should be clear that I am not interested in those but in pairs that are not adjusted to each other this way.)

Best Answer

The number of divisions of $\mathbb{R}^3$ by $k \ge 0$ planes in general position starts 1,2,4,8, then 15, etc. For $\mathbb{R}^6$ it is 1,2,4,8,16,32,64 then 127. In general for $\mathbb{R}^N$ it is the sum of the binomial coefficients from $\binom{k}{0}$ up to $\binom{k}{N}$ and hence it agrees with $2^k$ for terms 0,1,2, up to N before starting to fall off.

other answers Of course for prime p, $2^{p-1}=1 \mod{p}$ but there are only 2 known cases $p=1093$ and $3511$ where $2^{p-1}=1 \mod{p^2}$. SO primes and primes with $2^{p-1} \ne 1 \mod{ p^2}$ agree for the first 182 primes.

For "listed in the OEIS" there are a couple which go from 1 to 99 then skip 100: undulating numbers in base 10 and cents you can have in US coins without having change for a dollar (the latter being 1-99 along with $105, 106, 107, 108, 109, 115, 116, 117, 118, 119$.)

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