A simple combinatorics Le Monde mathematical puzzle:

N is a golden number if the sequence {1,2,…,N} can be reordered so that the sum of any consecutive pair is a perfect square. What are the golden numbers between 1 and 25?

Indeed, from an R programming point of view, all I have to do is to go over all possible permutations of {1,2,..,N} until one works or until I have exhausted all possible permutations for a given N. However, 25!=10²⁵ is a wee bit too large… Instead, I resorted once again to brute force simulation, by first introducing possible neighbours of the integers

  perfect=(1:trunc(sqrt(2*N)))^2
  friends=NULL
  le=1:N
  for (perm in 1:N){
    amis=perfect[perfect>perm]-perm
    amis=amis[amis<N]
    le[perm]=length(amis)
    friends=c(friends,list(amis)) #,recursive=TRUE)
    }

and then proceeding to construct the permutation one integer at time by picking from its remaining potential neighbours until there is none left or the sequence is complete

orderin=0*(1:N)
t=1
orderin[1]=sample((1:N),1)
for (perm in 1:N){
  friends[[perm]]=friends[[perm]]
              [friends[[perm]]!=orderin[1]]
  le[perm]=length(friends[[perm]])
  }
while (t<N){
  if (length(friends[[orderin[t]]])==0)
        break()
  if (length(friends[[orderin[t]]])>1){
    orderin[t+1]=sample(friends[[orderin[t]]],1)}else{
    orderin[t+1]=friends[[orderin[t]]]
    }
  for (perm in 1:N){
    friends[[perm]]=friends[[perm]]
      [friends[[perm]]!=orderin[t+1]]
    le[perm]=length(friends[[perm]])
    }
  t=t+1}

and then repeating this attempt until a full sequence is produced or a certain number of failed attempts has been reached. I gained in efficiency by proposing a second completion on the left of the first integer once a break occurs:

while (t<N){
  if (length(friends[[orderin[1]]])==0)
        break()
  orderin[2:(t+1)]=orderin[1:t]
  if (length(friends[[orderin[2]]])>1){
    orderin[1]=sample(friends[[orderin[2]]],1)}else{
    orderin[1]=friends[[orderin[2]]]
    }
  for (perm in 1:N){
    friends[[perm]]=friends[[perm]]
       [friends[[perm]]!=orderin[1]]
    le[perm]=length(friends[[perm]])
    }
  t=t+1}

(An alternative would have been to complete left and right by squared taken at random…) The result of running this program showed there exist permutations with the above property for N=15,16,17,23,25,26,…,49.  Here is the solution for N=49:

25 39 10 26 38 43 21 4 32 49 15 34 30 6 3 22 42 7 9 27 37 12 13 23 41 40 24 1 8 28 36 45 19 17 47 2 14 11 5 44 20 29 35 46 18 31 33 16 48

As an aside, the authors of Le Monde puzzle pretended (in Tuesday, Nov. 12, edition) that there was no solution for N=23, while this sequence

22 3 1 8 17 19 6 10 15 21 4 12 13 23 2 14 11 5 20 16 9 7 18/p>

sounds fine enough to me… I more generally wonder at the general principle behind the existence of such sequences. It sounds quite probable that they exist for N>24. (The solution does not bring any light on this issue.)