A382698 First member of the least set of 3 consecutive primes such that the sum of each pair of consecutive primes in this set is a multiple of n.

2, 3, 5, 3, 43, 5, 977, 53, 313, 43, 787, 137, 9587, 977, 2473, 541, 3967, 313, 28979, 947, 3121, 787, 72823, 283, 47441, 9587, 81463, 4363, 61153, 2473, 478001, 21617, 160243, 3967, 132763, 8017, 227873, 28979, 218279, 12163, 1772119, 3121, 3070187, 57413, 841459

Offset: 1

Author

Paolo P. Lava, Apr 04 2025

Keywords

• nonn
• easy

Examples

			a(5) = 43. The least 3 consecutive primes are 43, 47, 53:
  43 + 47 = 90 and 90/5 = 18;
  47 + 53 = 100 and 100/5 = 20.
a(41) = 1772119. The least 3 consecutive primes are 1772119, 1772167, 1772201:
  1772119 + 1772167 = 3544286 and 3544286/41 = 86446;
  1772167 + 1772201 = 3544368 and 3544368/41 = 86448.
		

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), A382699 (4 consecutive), A382700 (5 consecutive).

Programs

Maple

P:=proc(q) local a,b,c,n,v; v:=[]; for n from 1 to 45 do a:=2; b:=3; c:=5;
while true do if frac((a+b)/n)=0 and frac((b+c)/n)=0 then v:=[op(v),a]; break;
else a:=b; b:=c; c:=nextprime(c); 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,p++];a[n]=Prime[p],{n,45}];Array[a,45] (* James C. McMahon, Apr 09 2025 *)