cp's OEIS Frontend

A382700 First member of the least set of 5 consecutive primes such that the sum of each pair of consecutive primes in this set is a multiple of n.

%I A382700 #22 Aug 09 2025 15:27:02
%S A382700 2,3,5,47,3593,5,10487,523,38377,3593,1796671,409,947423,10487,60383,
%T A382700 62501,18164651,38377,15095579,32633,3272567,1796671,116863451,67819,
%U A382700 65835479,947423,7005239,1165217,1154953243,60383,800037461,7442557,15442121,18164651,771405431
%N A382700 First member of the least set of 5 consecutive primes such that the sum of each pair of consecutive primes in this set is a multiple of n.
%H A382700 Paolo P. Lava, <a href="/A382700/b382700.txt">Table of n, a(n) for n = 1..50</a>
%H A382700 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 A382700 a(4) = 47. The least 5 consecutive primes are 47, 53, 59, 61, 67:
%e A382700   47 + 53 = 100 and 100/4 = 25;
%e A382700   53 + 59 = 112 and 112/4 = 28;
%e A382700   59 + 61 = 120 and 120/4 = 30;
%e A382700   61 + 67 = 128 and 128/4 = 32.
%e A382700 a(27) = 7005239. The least 5 consecutive primes are 7005239, 7005277, 7005293, 7005331, 7005347:
%e A382700   7005239 + 7005277 = 14010516 and 14010516/27 = 518908;
%e A382700   7005277 + 7005293 = 14010570 and 14010570/27 = 518910;
%e A382700   7005293 + 7005331 = 14010624 and 14010624/27 = 518912;
%e A382700   7005331 + 7005347 = 14010678 and 14010678/27 = 518914.
%Y A382700 Cf. A254862 (2 consecutive), A382698 (3 consecutive), A382699 (4 consecutive).
%K A382700 nonn
%O A382700 1,1
%A A382700 _Paolo P. Lava_, Apr 04 2025