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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.

%I A382698 #23 Apr 15 2025 08:26:56
%S A382698 2,3,5,3,43,5,977,53,313,43,787,137,9587,977,2473,541,3967,313,28979,
%T A382698 947,3121,787,72823,283,47441,9587,81463,4363,61153,2473,478001,21617,
%U A382698 160243,3967,132763,8017,227873,28979,218279,12163,1772119,3121,3070187,57413,841459
%N 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.
%H A382698 Paolo P. Lava, <a href="/A382698/b382698.txt">Table of n, a(n) for n = 1..100</a>
%H A382698 Carlos Rivera, <a href="https://www.primepuzzles.net/conjectures/conj_092.htm">Conjecture 92. For any integer m there is at least one set of consecutive primes...</a>, The Prime Puzzles and Problems Connection.
%e A382698 a(5) = 43. The least 3 consecutive primes are 43, 47, 53:
%e A382698   43 + 47 = 90 and 90/5 = 18;
%e A382698   47 + 53 = 100 and 100/5 = 20.
%e A382698 a(41) = 1772119. The least 3 consecutive primes are 1772119, 1772167, 1772201:
%e A382698   1772119 + 1772167 = 3544286 and 3544286/41 = 86446;
%e A382698   1772167 + 1772201 = 3544368 and 3544368/41 = 86448.
%p A382698 P:=proc(q) local a,b,c,n,v; v:=[]; for n from 1 to 45 do a:=2; b:=3; c:=5;
%p A382698 while true do if frac((a+b)/n)=0 and frac((b+c)/n)=0 then v:=[op(v),a]; break;
%p A382698 else a:=b; b:=c; c:=nextprime(c); fi; od; od; op(v); end: P(2*10^6);
%t A382698 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 *)
%Y A382698 Cf. A254862 (2 consecutive), A382699 (4 consecutive), A382700 (5 consecutive).
%K A382698 nonn,easy
%O A382698 1,1
%A A382698 _Paolo P. Lava_, Apr 04 2025