A382699 First member of the least set of 4 consecutive primes such that the sum of each pair of consecutive primes in this set is a multiple of n.

2, 3, 5, 23, 157, 5, 977, 53, 5171, 157, 33871, 137, 159293, 977, 2969, 541, 406873, 5171, 471313, 6047, 166739, 33871, 2112193, 5309, 520763, 159293, 207869, 5443, 2404471, 2969, 1531487, 88919, 2673791, 406873, 6056569, 95737, 8480357, 471313, 561829, 73477

Offset: 1

Author

Paolo P. Lava, Apr 04 2025

Keywords

• nonn

Examples

			a(4) = 23. The least 4 consecutive primes are 23, 29, 31, 37:
  23 + 29 = 52 and 52/4 = 13;
  29 + 31 = 60 and 60/4 = 15;
  31 + 37 = 68 and 68/4 = 17.
a(37) = 8480357. The least 4 consecutive primes are 8480357, 8480369, 8480431, 8480443:
  8480357 + 8480369 = 16960726 and 16960726/37 = 458398;
  8480369 + 8480431 = 16960800 and 16960800/37 = 458400;
  8480431 + 8480443 = 16960874 and 16960874/37 = 458402.
		

Links

• Paolo P. Lava, Table of n, a(n) for n = 1..100
• Carlos Rivera, Conjecture 92. For any integer m there is at least one set of consecutive primes..., The Prime Puzzles and Problems Connection.

Crossrefs

Cf. A254862 (2 consecutive), A382698 (3 consecutive), A382700 (5 consecutive).

Programs

Maple

P:=proc(q) local a,b,c,d,n,v; v:=[];for n from 1 to 30 do a:=2; b:=3; c:=5; d:=7;
while true do if frac((a+b)/n)=0 and frac((b+c)/n)=0 and frac((c+d)/n)=0 then v:=[op(v),a]; break;
else a:=b; b:=c; c:=d; d:=nextprime(d); fi; od; od; op(v); end: P(2*10^6);

Mathematica

Do[p=0;Until[Mod[Prime[p]+Prime[p+1],n]==0&&Mod[Prime[p+1]+Prime[p+2],n]==0&&Mod[Prime[p+2]+Prime[p+3],n]==0,p++];a[n]=Prime[p],{n,45}];Array[a,40] (* James C. McMahon, Apr 09 2025 *)