A332877 Arrange the first n primes in a circle in any order. a(n) is the minimum value of the largest product of two consecutive primes out of all possible orders.

6, 15, 21, 35, 55, 77, 91, 143, 187, 221, 253, 323, 391, 493, 551, 667, 713, 899, 1073, 1189, 1271, 1517, 1591, 1763, 1961, 2183, 2419, 2537, 2773, 3127, 3233, 3599, 3953, 4189, 4331, 4757, 4897, 5293, 5723, 5963, 6499, 6887, 7171, 7663, 8051, 8633, 8989, 9797, 9991, 10403, 10807, 11303

Offset: 2

Author

Bobby Jacobs, Apr 11 2020

Keywords

• nonn

Comments

It might appear that all terms are either the product of two consecutive primes or two primes with a prime in between (A333747). However, 253=11*23 is the first term that is not in that sequence.

The easiest optimal permutation of n primes is probably {p_1, p_n, p_2, p_n-1, …, p_ceiling(n/2)}. - Ivan N. Ianakiev, Apr 20 2020

Examples

			Here are the different ways to arrange the first 4 primes in a circle.
  2-3
  | |  Products: 6, 21, 35, 10. Largest product: 35.
  5-7
.
  2-3
  | |  Products: 6, 15, 35, 14. Largest product: 35.
  7-5
.
  2-5
  | |  Products: 10, 15, 21, 14. Largest product: 21.
  7-3
The minimum largest product is 21, so a(4)=21.
		

Links

• Giovanni Resta, Examples for a(2)-a(22)

Crossrefs

Cf. A064796, A332765, A333747.

Programs

Mathematica

primes[n_]:=Prime/@Range[n];
partition[n_]:=Partition[primes[n],UpTo[Ceiling[n/2]]];
riffle[n_]:=Riffle[partition[n][[1]],Reverse[partition[n][[2]]]];
a[n_]:=Max[Table[riffle[n][[i]]*riffle[n][[i+1]],{i,1,n-1}]];
a/@Range[2,60] (* Ivan N. Ianakiev, Apr 20 2020 *)

PARI

a(n) = {my(x = oo); for (k=1, (n-1)!, my(vp = Vec(numtoperm(n, k-1))); vp = apply(x->prime(x), vp); x = min(x, max(vp[1]*vp[n-1], vecmax(vector(n-1, j, vp[j]*vp[j+1]))));); x;} \\ Michel Marcus, Apr 14 2020

Formula

Probably a(n) = A332765(n+1) for n > 4.

Extensions

a(12)-a(13) from Michel Marcus, Apr 14 2020

a(14) from Alois P. Heinz, Apr 15 2020

a(15)-a(22) from Giovanni Resta, Apr 19 2020

More terms from Ivan N. Ianakiev, Apr 20 2020