A382700 -id:A382700 - cp's OEIS frontend

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

Paolo P. Lava, Apr 04 2025

			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.

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.

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

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);

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 *)

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.

Original entry on oeis.org

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

Paolo P. Lava, Apr 04 2025

nonn

			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.

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.

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

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);

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 *)

cp's OEIS Frontend

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.

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