OFFSET
1,1
COMMENTS
Let a, b and 10 be pairwise coprime, with A001222(a) = A001222(b) = 4. There exists c such that c == 5 (mod a) and c == -5 (mod b). Dickson's conjecture implies that (c+k*a*b-5)/a and (c+k*a*b+5)/b are prime for infinitely many k; for such k, c+k*a*b is in the sequence. - Robert Israel, Mar 22 2020
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
MAPLE
N:= 10^4: # for terms <= N
T5:= select(t -> numtheory:-bigomega(t)=5, {$1..N+5}):
S:= T5 intersect map(`+`, T5, 10):
sort(convert(map(`-`, S, 5), list)); # Robert Israel, Mar 22 2020
MATHEMATICA
q=5; lst={}; Do[If[Plus@@Last/@FactorInteger[n-q]==q&&Plus@@Last/@FactorInteger[n+q]==q, AppendTo[lst, n]], {n, 8!}]; lst
SequencePosition[PrimeOmega[Range[5100]], {5, _, _, _, _, _, _, _, _, _, 5}][[All, 1]]+5 (* Harvey P. Dale, Sep 23 2021 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Joseph Stephan Orlovsky, Mar 01 2009
STATUS
approved