A280941 - OEIS

%I #8 Jan 16 2017 14:07:34

%S 2,3,10,0,33,45,0,294,98,0,296,262,0,428,984,0,1456,3086,0,2343,1878,

%T 0,14938,8422,0,2809,4259,0,7809,13819,0,51036,45506,0,15782,30764,0,

%U 57764,24553,0,23282,51942,0,44902,34214,0,1242641,95929,0,66761

%N Least integer k such that prime(k+1) - prime(k) = 2 and prime(k+2) - prime(k+1) = 2n, or 0 if no such k exists.

%C Or least integer k such that prime(k+2) - prime(k+1) = 2n where prime(k) is in A001359 (lesser of twin primes).

%C The corresponding prime(k) are 3, 5, 29, 137, 197, 1931, 521, 1949, 1667, 2969, 7757, 12161, 28349, 20807, ...

%C a(n) is a subsequence of A029707(n) or subsequence of A107770(n) - 1.

%C a(n) = 0 for n == 1 mod 3 for n > 1.

%C Proof: prime(k+1) - prime(k) = 2 => prime(k+1) == 1 mod 6 and prime(k) == -1 mod 6. If prime(k+2) - prime(k+1) = 2n, then prime(k+2) = 2(n+1) + prime(k). Combining n == 1 mod 3 and prime(k) == -1 mod 6 we obtain prime(k+2) == 3 mod 6, a contradiction because prime(k+2) == +-1 mod 6. Hence, a(n) = 0.

%e a(3) = 10 because prime(11) - prime(10) = 31 - 29 = 2 and prime(12) - prime(11) = 37 - 31 = 6 = 2*3.

%e a(11) = 296 because prime(297) - prime(296) = 1951 - 1949 = 2 and prime(298) - prime(297) = 1973 - 1951 = 22 = 2*11.

%p nn:=50:m:=10^5:

%p for n from 1 to 50 do:

%p ii:=0:

%p   for k from 1 to m while(ii=0) do:

%p    p1:=ithprime(k):p2:=ithprime(k+1):p3:=ithprime(k+2):

%p     if p2-p1 = 2 and p3-p2 = 2*n

%p     then

%p     ii:=1:printf(`%d %d \n`,n,k):

%p     else

%p     fi:

%p    od:

%p     if ii=0 then printf(`%d %d \n`,n,0):

%p     else

%p     fi:

%p od:

%t Table[If[And[n > 1, Mod[n, 3] == 1], 0, k = 1; While[Nand[# - Prime@ k == 2, Prime[k + 2] - # == 2 n] &@ Prime[k + 1], k++]; k], {n, 40}] (* _Michael De Vlieger_, Jan 14 2017 *)

%Y Cf. A000040, A001359, A006512, A029707, A107770.

%K nonn

%O 1,1

%A _Michel Lagneau_, Jan 11 2017