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

1,1

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.

Example

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.

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 *)

Crossrefs

Cf. A254862 (2 consecutive), A382699 (4 consecutive), A382700 (5 consecutive).

Sequence in context: A366671 A060444 A002587 * A152814 A280319 A021981

Adjacent sequences: A382695 A382696 A382697 * A382699 A382700 A382701

Keyword

nonn,easy

Author

Paolo P. Lava, Apr 04 2025

Status

approved