## Le Monde puzzle [#840] Another number theory Le Monde mathematical puzzles:

Find 2≤n≤50 such that the sequence {1,…,n} can be permuted into a sequence such that the sum of two consecutive terms is a prime number.

Now this is a problem with an R code solution:

```library(pracma)
foundsol=TRUE
N=2
while (foundsol){

N=N+1
noseq=TRUE
uplim=10^6
t=0
while ((t<uplim)&&(noseq)){

randseq=sample(1:N)
sumseq=randseq[-1]+randseq[-N]
noseq=min(isprime(sumseq))==0
t=t+1
}

foundsol=!noseq
if (!noseq){
lastsol=randseq}else{ N=N-1}
}
```

which returns the solution as

```> N
 12
> lastsol
  6  7 12 11  8  5  2  1  4  3 10  9
```

and so it seems there is no solution beyond N=12…

However, reading the solution in the next edition of Le Monde, the authors claim there are solutions up to 50. I wonder why the crude search above fails so suddenly, between 12 and 13! So instead I tried a recursive program that exploits the fact that subchains are also verifying  the same property:

```findord=function(ens){

if (length(ens)==2){
sol=ens
foundsol=isprime(sum(ens))}
else{
but=sample(ens,1)
nut=findord(ens[ens!=but])
foundsol=FALSE
sol=ens
if (nut\$find){
tut=nut\$ord
foundsol=max(isprime(but+tut),
isprime(but+tut[length(tut)]))
sol=c(tut,but)
if (isprime(but+tut))
sol=c(but,tut)
}
}
list(find=foundsol,ord=sol)
}
```

And I ran the R code for N=13,14,…

```> stop=TRUE
> while (stop){
+   a=findord(1:N)
+   stop=!(a\$find)}
```

until I reached N=20 for which the R code would not return a solution. Maybe the next step would be to store solutions in N before moving to N+1. This is just getting  me too far from a mere Saturday afternoon break.

### 3 Responses to “Le Monde puzzle [#840]”

1. Cole Says:

A quick, iterative solution for N up to 750, https://gist.github.com/couthcommander/7761885

• xi'an Says:

Thank you, Cole. I checked and it works indeed. Quickly. Do you know of any theoretical work that would prove it holds for any n?

• Cole Says:

I wouldn’t be surprised if that held true, but I’m not familiar with a proof.

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