OFFSET
1,4
COMMENTS
Jud McCranie reports that he was able to find a solution for each n <= 225 (2n <= 450) in just a few seconds. - Jul 05 2002
Is there a proof that this can always be done?
The Mathematica program for this sequence uses backtracking to find all solutions for a given n. To verify that at least one solution exists for a given n, the backtracking function be made to stop when the first solution is found. Solutions have been found for n <= 48. - T. D. Noe, Jun 19 2002
This sequence is from the prime circle problem. There is no known proof that a(n) > 0 for all n. However, for many n (see A072618 and A072676), we can prove that a(n) > 0. Also, the sequence A072616 seems to imply that there are always solutions in which the odd (or even) numbers are in order around the circle. - T. D. Noe, Jul 01 2002
Prime circles can apparently be generated for any n using the Mathematica programs given in A072676 and A072184. - T. D. Noe, Jul 08 2002
The following seems to always produce a solution: Work around the circle starting with 1 but after that always choosing the largest remaining number that fits. For example, if n = 4 this gives 1, 6, 7, 4, 3, 8, 5, 2. See A088643 for a sequence on a related idea. - Paul Boddington, Oct 30 2007
See A228917 for a similar conjecture on twin primes. - Zhi-Wei Sun, Sep 08 2013
See A242527 for a similar problem on the set of numbers {0 through (n-1)}. - Stanislav Sykora, May 30 2014
James Tilley and Stan Wagon report that all terms up to n = 10^6 are nonzero. Charles R Greathouse IV, Feb 05 2016
REFERENCES
R. K. Guy, Unsolved Problems in Number Theory, second edition, Springer, 1994. See section C1.
LINKS
S. Sykora, On Neighbor-Property Cycles, Stan's Library, Volume V, 2014.
Eric Weisstein's World of Mathematics, Prime Circle.
EXAMPLE
One arrangement for 2n=6 is 1,4,3,2,5,6 and this is essentially unique, so a(3)=1.
MATHEMATICA
$RecursionLimit=500; try[lev_] := Module[{t, j}, If[lev>2n, (*then make sure the sum of the first and last is prime*) If[PrimeQ[soln[[1]]+soln[[2n]]]&&soln[[2]]<=soln[[2n]], (*Print[soln]; *) cnt++ ], (*else append another number to the soln list*) t=soln[[lev-1]]; For[j=1, j<=Length[s[[t]]], j++, If[ !MemberQ[soln, s[[t]][[j]]], soln[[lev]]=s[[t]][[j]]; try[lev+1]; soln[[lev]]=0]]]]; For[lst={}; n=1, n<=7, n++, s=Table[{}, {2n}]; For[i=1, i<=2n, i++, For[j=1, j<=2n, j++, If[i!=j&&PrimeQ[i+j], AppendTo[s[[i]], j]]]]; soln=Table[0, {2n}]; soln[[1]]=1; cnt=0; try[2]; AppendTo[lst, cnt]]; lst (* T. D. Noe *)
PROG
(C++) listed in the link (S. Sykora)
CROSSREFS
KEYWORD
nonn,nice,changed
AUTHOR
EXTENSIONS
a(14)-a(15) from Max Alekseyev, Sep 19 2013
STATUS
approved