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(In collaboration with Z. Chase.) A Fibonacci number $F_{n}$ is never sandwiched between two twin primes $(p,p+2)$.

This is because this would require $F_{n}+1$ to be a prime, but that can only happen iff $n=1,2,3$, and one can check that $F_{n}-1$ is not a prime in these cases.

The fact that $F_{n}+1$ is a prime iff $n=1,2,3$ is probably quite old. One reference is this OEIS page, where (the) Richard Guy shows

1. $F_{4n}+1 = F_{2n-1} L_{2n-1}$,
2. $F_{4n+1}+1 = F_{2n+1} L_{2n}$,
3. $F_{4n+2}+1 = F_{2n+2} L_{2n}$,
4. $F_{4n+3}+1 = F_{2n+1} L_{2n+2}$

where $L_n$ is the $n$th Lucas number.

In fact, we only need the first of these identities for your question, because if $F_{n}$ is sandwiched between twin primes $(p,p+2)$ then $F_n \equiv 0 \bmod 6$ implying $n \equiv 0 \bmod 12$.

(In collaboration with Z. Chase.) A Fibonacci number $F_{n}$ is never sandwiched between two twin primes $(p,p+2)$.

This is because this would require $F_{n}+1$ to be a prime, but that can only happen iff $n=1,2,3$, and one can check that $F_{n}-1$ is not a prime in these cases.

The fact that $F_{n}+1$ is a prime iff $n=1,2,3$ is probably quite old. One reference is this OEIS page, where (the) Richard Guy shows

1. $F_{4n}+1 = F_{2n-1} L_{2n+1}$,
2. $F_{4n+1}+1 = F_{2n+1} L_{2n}$,
3. $F_{4n+2}+1 = F_{2n+2} L_{2n}$,
4. $F_{4n+3}+1 = F_{2n+1} L_{2n+2}$

where $L_n$ is the $n$th Lucas number.

In fact, we only need the first of these identities for your question, because if $F_{n}$ is sandwiched between twin primes $(p,p+2)$ then $F_n \equiv 0 \bmod 6$ implying $n \equiv 0 \bmod 12$.

(In collaboration with Z. Chase.) A Fibonacci number $F_{n}$ is never sandwiched between two twin primes $(p,p+2)$.

This is because this would require $F_{n}+1$ to be a prime, but that can only happen iff $n=1,2,3$, and one can check that $F_{n}-1$ is not a prime in these cases.

The fact that $F_{n}+1$ is a prime iff $n=1,2,3$ is probably quite old. One reference is this OEIS page, where (the) Richard Guy shows

1. $F_{4n}+1 = F_{2n-1} L_{2n-1}$,
2. $F_{4n+1}+1 = F_{2n+1} L_{2n}$,
3. $F_{4n+2}+1 = F_{2n+2} L_{2n}$,
4. $F_{4n+3}+1 = F_{2n+1} L_{2n+2}$

where $L_n$ is the $n$th Lucas number.

In fact, we only need the first of these identities, because if $F_{n}$ is sandwiched between twin primes $(p,p+2)$ then $F_n \equiv 0 \bmod 6$ implying $n \equiv 0 \bmod 12$.

(In collaboration with Z. Chase.) A Fibonacci number $F_{n}$ is never sandwiched between two twin primes $(p,p+2)$.

This is because this would require $F_{n}+1$ to be a prime, but that can only happen iff $n=1,2,3$, and one can check that $F_{n}-1$ is not a prime in these cases.

The fact that $F_{n}+1$ is a prime iff $n=1,2,3$ is probably quite old. One reference is this OEIS page, where (the) Richard Guy shows

1. $F_{4n}+1 = F_{2n-1} L_{2n-1}$,
2. $F_{4n+1}+1 = F_{2n+1} L_{2n}$,
3. $F_{4n+2}+1 = F_{2n+2} L_{2n}$,
4. $F_{4n+3}+1 = F_{2n+1} L_{2n+2}$

where $L_n$ is the $n$th Lucas number.

In fact, we only need the first of these identities for your question, because if $F_{n}$ is sandwiched between twin primes $(p,p+2)$ then $F_n \equiv 0 \bmod 6$ implying $n \equiv 0 \bmod 12$.

(In collaboration with Z. Chase.) A Fibonacci number $F_{n}$ is never sandwiched between two twin primes $(p,p+2)$.

This is because this would require $F_{n}+1$ to be a prime, but that can only happen iff $n=1,2,3$, and one can check that $F_{n}-1$ is not a prime in these cases.

The fact that $F_{n}+1$ is a prime iff $n=1,2,3$ is probably quite old. One reference is this OEIS page, where (the) Richard Guy shows

1. $F_{4n}+1 = F_{2n-1} L_{2n-1}$,
2. $F_{4n+1}+1 = F_{2n+1} L_{2n}$,
3. $F_{4n+2}+1 = F_{2n+2} L_{2n}$,
4. $F_{4n+3}+1 = F_{2n+1} L_{2n+2}$

where $L_n$ is the $n$th Lucas number.

(In collaboration with Z. Chase.) A Fibonacci number $F_{n}$ is never sandwiched between two twin primes $(p,p+2)$.

This is because this would require $F_{n}+1$ to be a prime, but that can only happen iff $n=1,2,3$, and one can check that $F_{n}-1$ is not a prime in these cases.

The fact that $F_{n}+1$ is a prime iff $n=1,2,3$ is probably quite old. One reference is this OEIS page, where (the) Richard Guy shows

1. $F_{4n}+1 = F_{2n-1} L_{2n-1}$,
2. $F_{4n+1}+1 = F_{2n+1} L_{2n}$,
3. $F_{4n+2}+1 = F_{2n+2} L_{2n}$,
4. $F_{4n+3}+1 = F_{2n+1} L_{2n+2}$

where $L_n$ is the $n$th Lucas number.

In fact, we only need the first of these identities, because if $F_{n}$ is sandwiched between twin primes $(p,p+2)$ then $F_n \equiv 0 \bmod 6$ implying $n \equiv 0 \bmod 12$.