OFFSET
1,1
COMMENTS
Except for p=5, the decimals in A092110 end in 3 or 7.
Theorem: If in the triple (2n-3,n,2n+3) all numbers are primes then n=5 or the decimal representation of n ends in 3 or 7. Proof: Consider Q=(2n-3)n(2n+3), by hypothesis factorized into primes. If n is prime, n=10k+r with r=1,3,7 or 9. We want to exclude r=1 and r=9. Case n=10k+1. Then Q=5(-1+6k+240k^2+800k^3) and 5 is a factor; thus 2n-3=5 or n=5 or 2n+1=5 : this means n=4 (not prime); or n=5 (included); or n=2 (impossible, because 2n-3=1). Case n=10k+9. Then Q=5(567+1926k+2160k^2+800k^3) and 5 is a factor; the arguments, for the previous case, also hold.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
MATHEMATICA
bpQ[n_]:=Last[IntegerDigits[n]]==7&&And@@PrimeQ[2n+{3, -3}]; Select[Prime[ Range[1200]], bpQ] (* Harvey P. Dale, Sep 25 2013 *)
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Carlos Alves, Dec 20 2007
EXTENSIONS
Definition clarified by Harvey P. Dale, Sep 25 2013
STATUS
approved