Sorry, Fibonacci.

Notice that the consecutive integers $1, 2, 3$ appear in the Fibonacci sequence. Therefore, $d$ cannot be $1, 2$ or $3$ ; otherwise, the sequence would contain one of the integers.

Now note that all terms in the arithmetic progression are congruent modulo $d.$ So, if the Fibonacci sequence contains every possible residue modulo $d,$ then it is impossible to find any such progression. We check the first few values of $d$ by taking the Fibonacci sequence mod $d$ :

• In mod $4,$ the Fibonacci sequence is $1,1, 2, 3, 1, 0,\ldots$ which contains all residues.

• In mod $5,$ the Fibonacci sequence is $1, 1, 2, 3, 0, 3, 3, 1, 4, \ldots$ which contains all residues.

• In mod $6,$ the Fibonacci sequence is $1, 1, 2, 3, 5, 2, 1, 3, 4, 1, 5, 0, \ldots$ which contains all residues.

• In mod $7,$ the Fibonacci sequence is $1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, \ldots$ which contains all residues.

• In mod $8,$ the Fibonacci sequence is $1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, 0, 1, 1, \ldots$ Notice that this sequence has now repeated, without containing the residues $4$ or $6.$ Therefore, $\boxed{d=8}$ is the minimum possible value of $d.$ $\blacksquare$

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Note: A possible progression would be $4, 12, 20, 28, \ldots$ since all terms are $4 \pmod8.$

Let $b_{n}$ be a sequence with minimal $d$ subject to the constraints given. For each $N = 1, 2, 3, ...$ consider the sequence $(a_{N,n})$ defined $a_{N,n} := F_{n} Mod N$ .

Claim: If there exists integers $i$ and $j$ , with $i < j$ such that $a_i = a_j$ and $a_{i+1} = a_{j+1}$ , then sequence (\ (a_N,n ) repeats the sequence $a_i, a_{i+1}, ... , a_{j-1}$ repeats indefinitely.

Proof: This follows from how $(a_{N,n})$ is defined as well as the properties of Mod and the fact that the Fibonacci sequence is a recursion relation with two seed values.

Claim: There is a repeating sequence for each N.

Proof: Since each sequence $(a_{N,n})$ can take on at most $N$ values, there are only $N^2$ distinct pairs of successive terms $a_i$ and $a_{i+1}$ , therefore there must eventually be an occurrence of the situation described in the previous claim.

Consider $(b_1)$ of the desired sequence. If d is the common difference then, $a_{d,n}$ must contain only finitely many of some $m$ in ${0, 1, 2, ... d-1}$ . Otherwise $b_n$ will hit every Fibonacci number greater than $b_1$ such that $b_1 Mod d = 0$ .

Thus the smallest $N$ such that the repeating part of $(a_{N,n})$ is missing one of the values $0, 1, 2, ..., N-1$ is the $d$ we are looking for. We begin listing the sequence $a_{N,n}$ until repetition is discovered.

We start with N = 2: 1, 1, 0 , 1, 1,...

N = 3: 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, ...

N = 4: 1, 1, 2, 3, 1, 0, 1, 1, ....

N = 5: 1, 1, 2, 3, 0, 3, 3, 1, 4, 0, 4, 4, 3, 2, 0, 2, 2, 4, 1, 0, 1, 1, ...

N = 6: 1, 1, 2, 3, 5, 1, 0, 1, 1, ...

N = 7: 1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, 2, 6, 1, 0, 1, 1, ...

N = 8: 1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, 0, 1, 1,...

We see that $N = 8$ has $(a_{N,n}$ with no instances of 4 or 6. So $d = 8$ .

Let $(a_n)$ be an arithmetic progression that satisfies the given conditions. Let $d$ be its common difference. Assume that the set of congruence classes of $(F_n)$ modulo $d$ consists of all integers $0,1,\dots, d-1.$ Since each residue is obtained from the residues of the previous terms (by using the recurrence relation of the Fibonacci sequence), each residue occurs infinitely many times with a fixed pattern. For example, the Fibonacci numbers mod $3$ are $1, 1, 2, 0, 2, 2, 1, 0, 1, 1, 2, 0, 2, 2, 1, 0, \overline{1,1,2,0,2,2,1,0,} \dots.$ This allows us to choose a positive integer $m$ such that $F_m \ge a_1 \quad \text{and} \quad F_m \equiv a_1 \pmod{d}.$ This implies $F_m-a_1 = kd$ or $F_m = a_1 + kd$ for some nonnegative integer $k$ , which is the $(k+1)$ st term of the arithmetic progression $(a_n)$ . This is a contradiction to condition 3.

Hence we must look for the smallest integer $d$ where the set of congruence classes of $(F_n)$ modulo $d$ is not complete. After some numerical experiments, we find that such smallest number is $\boxed{d=8},$ where the residues of $(F_n)$ mod $d$ are $0,1,2,3,5,$ and $7.$ Therefore, if we, for example, let $(a_n)$ be defined by $a_n = 4+(n-1)8 = 8n-4,$ then $(a_n)$ is an arithmetic progression with no term of the progression appearing in the Fibonacci sequence, since $a_n \equiv 4 \pmod{8}$ for all $n$ but $F_n \not\equiv 4 \pmod{8}$ for all $n$ .

I tried to write down the remainder when Fibonacci numbers is divided by d (d>2) with d=8 the sequence of remainder is periodic and in one period the sequence lacks of " 4" and "6" It is 1 1 2 3 5 0 5 5 2 7 1 0 1 1 2 3..... From d=2 to 7 the period contains all the number less than d So if we choose the starting point as 4 or 6 the sequece of remainder contains no zero. That is no fibonacci number -6 (or 4) is divisible by 8 Or every Fibonacci numbers \neq 6(or 4) + 8\times n So the answer is 8

So I am assuming that the progression is infinite, otherwise there would be no definite answer.

Note that if we have an arithmetic progression, there exists $a$ and $b$ such that all the terms in that progression is congruent to $a mod b$ . Note that if we consider the terms of the fibonacci sequence $mod b$ , a each term that appears at least once would appear infinitely many times. This is because there is a finite number of possibilities for $2$ consecutive terms and each term depends solely on the $2$ previous terms. So we just have to look for the lowest $d$ such that there are no terms in the fibonacci sequences that are congruent to $a mod d$ for some integer $a$ .

We need not to consider $mod 1, 2, 3$ because the first $3$ terms are $1, 2, 3$ .

mod 4: 1, 1, 2, 3, 1, 0. all exists. mod 5: 1, 1, 2, 3, 0, 3, 3, 1, 4, all exists. mod 6: 1, 1, 2, 3, 5, 2, 7, 1, all exists. mod 7: 1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, 2, all exists.

But for mod 8, 1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, 0, 1, 1. Note that the 1, 1 already repeated but there still has no 6. Thus the 6 will never come in the sequence because the 1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, 0, pattern will keep on repeating as each term is dependent solely on the previous 2 terms.

Our sequence can be written as $a+dn$ $(n \in \mathbb{N}$ ).

We are looking for the smallest $d$ so that there is an $a$ that does not appear in the Fibonacci sequence modulo $d$ .

This can be done with trial and error. The Fibonacci sequence modulo $8$ is

$1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, 0, 1, 1, ...$

Both $4$ and $6$ don't appear in this sequence, so the sequences defined by $4+8n$ $(n \in \mathbb{N})$ and $6+8n$ $(n \in \mathbb{N})$ both don't contain a number from the Fibonacci sequence.

All of the terms of an arithmetic progression with common difference $d$ leave the same remainder when divided by $d$ . Using that fact, if we find a $d$ such that the Fibonacci sequence modulo $d$ enters a cycle that skips at least one of the possible remainders, there is an arithmetic sequence with common difference $d$ that satisfies the problem's constraints.

By trial and error, we can see that the first $d$ with these properties is 8.