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Let $k$ be a positive integer.

Let $n$ be an integer such that $n=6k-1$

Let $r$ be the remainder of the division of $(n-1)!-n$ by $(n+2)$

Conjecture: if $6k+1$ is prime $r=3k+2$

For example the first 25 values of $r$ are:

${5,8,11,2,17,20,23,2,2,32,35,38,41,2,2,50,53,56,2,2,65,2,71,2,77}$

And we have:

$8+5=13=6(2)+1$
$11+8=19=6(3)+1$
$20+17=37=6(6)+1$
$23+20=43=6(7)+1$
$35+32=67=6(11)+1$
$38+35=73=6(12)+1$
$41+38=79=6(13)+1$
$53+50=103=6(17)+1$
$56+53=109=6(18)+1$

I have already proved that if $6k+1$ is a composite number $r=2$ but I failed to prove this conjecture.

Is it hard to prove it?

Thanks.

Craw Craw
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    It's not true that $r$ alternates in parity. The first $25$ values are ${5, 8, 11, 2, 17, 20, 23, 2, 2, 32, 35, 38, 41, 2, 2, 50, 53, 56, 2, 2, 65, 2, 71, 2, 77}$. – lulu Jun 30 '22 at 00:44
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    eyeballintg the first $500$ values, it does appear to be true that no consecutive values are both odd, but consecutive even values are common. And not just two consecutive terms, at one point there is a string of $11$ consecutive evens. – lulu Jun 30 '22 at 00:46
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    Typo? Clement's theorem should be $4[(n-1)!+1]+\color{#c00}n \equiv 0 \pmod {n^2+2n}\ \ $ – Bill Dubuque Jun 30 '22 at 02:17
  • @lulu : I state that for a given natural number $n$ such that $2n+2 \equiv 0 \pmod {12}$, $k$ and $r$ are both odd or both even. Bill Dubuque: thanks, I edit. – Craw Craw Jun 30 '22 at 04:16
  • I edited because the conjecture seems to be true when $r \ne 2$. – Craw Craw Jun 30 '22 at 05:18
  • Based on the comment of lulu I found an interesting thing about the first 25 values of $r$... If we sum two consecutive values of $r$ and when $r \ne 2$, we have a prime number. Is always true? – Craw Craw Jun 30 '22 at 06:46
  • No...again, there are long blocks of consecutive even $r$. And even if you exclude consecutive evens, $25$ occurs as the sum $23+2$. – lulu Jun 30 '22 at 09:47
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    Important point: if you are making a numerical conjecture, test it. Always. And not for a couple of values. Check a lot of values. For instance, there are lots of small primes so it's very misleading to test conjectures about primes for very small numbers. Do a proper search. – lulu Jun 30 '22 at 09:51
  • I think we do not calculate 23+2 because I supposed $r \ne 2$ for my conjecture. Thanks for your advice I will check my conjecture for larger numbers. – Craw Craw Jun 30 '22 at 10:04
  • I suspect that the Wilson theorem would be useful but I’m not sure. – Craw Craw Jun 30 '22 at 15:24

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$$n+2=6k+1$$ by Wilson's theorem $$(n+1)!\equiv -1\pmod {n+2}$$ it follows that $$n!\equiv 1\pmod{n+2}$$ and that $$(n-1)!\equiv -(2^{-1})\equiv 3k\pmod {n+2}$$ and $$-n\equiv 2\pmod {n+2}$$ so together we have $$3k+2$$ as the remainder $\blacksquare$

As to the addition chains $$(3k+2)+(3(k+1)+2)=6(k+1)+1$$ so what.

Roddy MacPhee
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