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a(n) is the least k such that there are exactly n pairs of consecutive primes (p,q) such that (2*k-q,2*k-p) are consecutive primes.
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%I #8 Feb 08 2021 02:57:17

%S 1,4,5,9,33,12,24,39,51,30,63,72,84,45,57,102,123,60,297,351,162,120,

%T 252,231,519,150,285,195,345,105,360,165,255,483,510,462,1182,552,210,

%U 1158,330,315,1023,1197,525,570,690,675,1353,858,750,645,585,987,1533,420,900,825,1110,660,630,960

%N a(n) is the least k such that there are exactly n pairs of consecutive primes (p,q) such that (2*k-q,2*k-p) are consecutive primes.

%C a(n) is the least k such that A341237(k) = n.

%H Robert Israel, <a href="/A341252/b341252.txt">Table of n, a(n) for n = 0..300</a>

%e a(3) = 9 as there are 3 pairs (p,q) = (5,7), (7,11), (11,13) of consecutive primes such that (18-q,18-p) = (11,13), (7,11), (5,7) are pairs of consecutive primes, and 9 is the least number for which there are 3 such pairs.

%p f:= proc(n) local p, q, count;

%p q:= 2: count:= 0:

%p while q < 2*n -2 do

%p p:= q; q:= nextprime(q);

%p if isprime(2*n-p) and prevprime(2*n-p)=2*n-q then count:= count+1 fi;

%p od;

%p count

%p end proc:

%p V:= Array(0..100): count:= 0:

%p for n from 1 while count < 101 do

%p v:= f(n);

%p if v <= 100 and V[v] = 0 then count:= count+1; V[v]:= n fi

%p od:

%p convert(V,list);

%Y Cf. A341237, A341238.

%K nonn

%O 0,2

%A _J. M. Bergot_ and _Robert Israel_, Feb 07 2021