OFFSET
1,1
COMMENTS
It is conjectured that a(n) > 0 for all n, and for infinitely many terms, a(n) = prime(n+1).
a(n) = prime(n+1) for n = 9, 100, 508, 627, 752, 835, 889, ... (that is, for p = 23, 541, 3631, 4643, 5711, 6421, 6911, ...) - Derek Orr, Aug 25 2014
We have a(n) - prime(n) == 0 (mod 6) for all n > 2. Indeed, suppose p = 6k + 1, then q - p = 6n + 2 would imply that q is divisible by 3, and q - p = 6n + 4 would imply that p*(q-p)+q is divisible by 3. A similar reasoning applies for p = 6k - 1: here q - p = 6n + 4 entails 3|q, and q - p = 6n + 2 yields 3 | p*(q-p)-q.
LINKS
Chai Wah Wu, Table of n, a(n) for n = 1..1000
W. Sindelar, Certain Pairs of Consecutive Prime Numbers, in yahoo group "primenumbers", Jan 20 2011.
W. Sindelar, David Broadhurst, Certain Pairs of Consecutive Prime Numbers, digest of 2 messages in primenumbers Yahoo group, Jan 20 - Jan 21, 2011.
MATHEMATICA
sp[p_]:=Module[{q=NextPrime[p]}, While[AnyTrue[{p(q-p)+q, p(q-p)-q, q(q-p)+p, q(q-p)-p}, CompositeQ], q=NextPrime[q]]; q]; Join[{11, 23}, Table[sp[p], {p, Prime[Range[3, 60]]}]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Mar 23 2021 *)
PROG
(PARI) A180481(p)={ forprime( q=1+p=prime(p), default(primelimit), isprime(p*(q-p)+q)||next; isprime(p*(q-p)-q)||next; isprime(q*(q-p)+p)||next; isprime(q*(q-p)-p)||next; return(q)) }
(Python)
from sympy import prime, isprime
def A180481(n):
p = prime(n)
n += 1
q = prime(n)
while q < 10**14: # note: search limit
if isprime(p*(q-p)+q) and isprime(p*(q-p)-q) and isprime(q*(q-p)+p) and isprime(q*(q-p)-p):
return(q)
n += 1
q = prime(n)
return(0) # limit in search for q was reached. A180481(n) may be > 0
# Chai Wah Wu, Aug 24 2014
CROSSREFS
KEYWORD
nonn
AUTHOR
M. F. Hasler, Jan 20 2011
STATUS
approved