## Le Monde puzzle [#840]

**A**nother 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.

**N**ow 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 [1] 12 > lastsol [1] 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[1]), isprime(but+tut[length(tut)])) sol=c(tut,but) if (isprime(but+tut[1])) 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.

December 3, 2013 at 1:46 am

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

December 3, 2013 at 12:02 pm

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?

December 3, 2013 at 5:01 pm

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