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Let $L_n$ be the $n$th Lucas number. I tested whether $L_n-1$ is prime for all $n<100000$ and found that it is prime only for $n=2,3,6,24,48,96$. Are there any other prime numbers? Also, is there a reason why most of these $n$s are in the form of $2^m\cdot3$?

We can show that $n$ must be even by these formulas for $n>3$, but they may not be useful to show that it is of the form $2^m\cdot3$: $$L_{4k}-1=L_{6k}/L_{2k}$$ $$L_{4k+1}-1=5F_{2k}\cdot F_{2k+1}$$ $$L_{4k+2}-1=F_{6k+3}/F_{2k+1}$$ $$L_{4k+3}-1=L_{2k+1}\cdot L_{2k+2}$$

dodicta
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    If you don't get an answer soon it might be because nobody knows. – DanielWainfleet Sep 11 '21 at 10:10
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    It's easy to show that we need $3\mid n$ (otherwise, $L_n-1$ is even) – Mastrem Sep 11 '21 at 11:30
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    I extended the search limit to $190\ 000$. No further primes. – Peter Sep 12 '21 at 08:14
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    Still no further prime upto $n=243\ 000$ – Peter Sep 13 '21 at 17:23
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    '2, 1, 3, 4, 7, 11, 18, 29, 17, 16, 3, 19, 22, 11, 3, 14, 17, 1, 18, 19, 7, 26, 3, 29,' is modulo 30. Might have messed up my earlier comment ... – Roddy MacPhee Sep 14 '21 at 23:14
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    By the way, the analogue number $L_n+1$ has very often "small" factors. I found no prime for $n>18$ in this case. In both cases , I passed $n=300\ 000$ and aborted shortly after this limit. – Peter Sep 16 '21 at 13:35
  • @Peter Thank you for expanding the range of search. For $L_n+1$, $L_{4k}+1=F_{6k}/F_{2k}=(L_{2k}-1)(L_{2k}+1),\ L_{4k+1}+1=L_{2k}L_{2k+1},\ L_{4k+2}+1=L_{6k+3}/L_{2k+1},\ L_{4k+3}+1=5F_{2k+1}F_{2k+2}$ are hold. Because of the first formula, I think they have many small factors. – dodicta Sep 17 '21 at 01:48
  • @dodicta It would be nice if you would prove your idendities since I could find them nowhere in the internet. – Peter Sep 17 '21 at 12:13
  • @Peter Some of them are from: https://www.ams.org/journals/mcom/1988-50-181/S0025-5718-1988-0917832-6/home.html They can be shown by substituting $L_n=\phi^n+(1-\phi)^n, F_n=(\phi^n-(1-\phi)^n)/\sqrt{5}$. – dodicta Sep 18 '21 at 01:04
  • @Peter I tried to do some numerical checks, but it wound up being very slow. What software were you using, and did you do something other than list the Lucas numbers and check for primality one by one? – Julian Rosen Sep 18 '21 at 18:36
  • I used PFGW and brute force. – Peter Sep 19 '21 at 14:07
  • $L_{2k-1}-1=L_kL_{k-1}$ for $k$ even and $L_{2k-1}-1=5F_kF_{k-1}$ for $k$ odd (see OEIS A000032 John Blythe Dobson's 2007 note). So $n$ must be even as you already know. Also, as already commented, $n$ must be divisible by $3$ otherwise $L_{2k}-1=5F_{k}^2+2(-1)^k-1$ would be even. Therefore we need to check only $n=6k$. – Fabius Wiesner Oct 13 '21 at 20:31

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