Longest coinciding pair of integer sequences known 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.)
 A: For what it's worth, the OEIS has 99 sequences containing the string 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35, which is all I had the patience to type in. A153671, Minimal exponents m such that the fractional part of $(101/100)^m$ obtains a maximum (when starting with $m=1$), continues the pattern up to 69, then goes 110, 180, ....
A: A possibly unbeatable example: the sequence $a(n)$, where $a(n)$ is the periodicity of the first row in the Laver table based on the set $\left\lbrace 1,\ldots, 2^n\right\rbrace$:
The first values are $$1, 1, 2, 4, 4, 8, 8, 8, 8, 16, 16,\ldots$$ 
and then it stays at $16$ for ages:
Randall Dougherty showed that the first $n$ for which $a(n)$ can possibly differ from $16$ is A(9,A(8,A(8,255))), where A denotes the Ackermann function, a recursive function whose first values are already huge.
However, it can be shown that the sequence actually tends to infinity, under some additional axiom that is independent to the usual ones from ZFC.
A: Sorry not enough points to post a comment so had to make this an answer.
Not really the most natural sequences, but the sequences $a_n^{(k)}$ of positive integers which have at most $k$ distinct prime factors coincide a lot (among themselves and with the integers). 
The first term not in $a_n^{(k)}$ is $p_1\cdot p_2 .... \cdot p_{k+1}$. 
A: If you define the function $f$ of $h$ and $x$ by
 $f(1,x) = 1+ x $ and
 $ f(h+1,x) = (1+x) ^ {f(h,x)} $      
then the leading coefficents at the powers of x in the formal powerseries up to an index  k=1 ... h  are equal for f(h+j,x) and j>0
f(0,x) = 1 + x
f(1,x) = 1 + x + x^2 + 1/2*x^3 + 1/3*x^4 + 1/12*x^5 + 3/40*x^6 -  ...
f(2,x) = 1 + x + x^2 + 3/2*x^3 + 4/3*x^4 + 3/2*x^5 + 53/40*x^6 + ...
f(3,x) = 1 + x + x^2 + 3/2*x^3 + 7/3*x^4 + 3*x^5 + 163/40*x^6 + ....
f(4,x) = 1 + x + x^2 + 3/2*x^3 + 7/3*x^4 + 4*x^5 + 243/40*x^6 + ...    
So  if we use the coefficients of the powerseries of f(N,x) and f(N+1,x) the sought N can be arbitrarily high .
A: 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 $105, 106, 107, 108, 109, 115, 116, 117, 118, 119$.)
A: or another cheating example: positive integers and remainders of positive integers modulo 100000000.
A: The positive odd integers $n$ which pass the Euler-Jacobi primality test to base $2,$ $2^{(n-1)/2} \equiv \big(\frac2n\big) \mod n$ where the RHS is the Jacobi symbol, agree with the odd primes up to the inclusion of $561$. So, these sequences $3, 5, 7, 11, 13, ..., 557, 561, 563, ...$ and $3, 5, 7, 11, 13, ..., 557, 563, ...$ agree for $101$ terms.
A: It is a known example: sequence 1,2,3,5,7,11,13,$\dots$ of non-composite numbers coincides with the sequence of orders of finite simple groups until 60 appears (in this last sequence).
A: There are many natural examples of a sequence $a_{n,k}$ of two parameters such that $a_{n,k}$ approximates a sequence $a_n$ as $k \to \infty$ in the sense that the first $k$ terms of $a_n$ and $a_{n,k}$ agree.  Aaron Meyerowitz gives a good one; another example is the "partial Catalan" sequence $C_{n,k}$ of all parenthesizations using $n$ pairs of parentheses with parenthetical depth at most $k$.  So I don't think this is quite the questions you meant to ask.
(A nice commonality between Aaron Meyerowitz's example and this one is that for fixed $k$ the approximating sequences $a_{n,k}$ are regular, so their generating functions are rational.  So one can think of these generating functions as "rational approximations" to the generating function of $a_n$, which can in some cases be obtained by truncating a continued fraction expansion.  This is the case with my example; see this blog post.)
A: I know I am cheating :-)
A) $a_n = n + C \lfloor \frac{n}{N}\rfloor$
B) Integers of form $x+\prod_{1 \leq k \leq N}{(x-k)}$
EDIT: up to $N$ A and B coincide with $\mathbb{N}$ so it is a triple in a sense.
A: There is also, of course, what comes out of the answer which was given to this question:
Computer Algebra Errors
