

A136191


Primes p such that 2p3 and 2p+3 are both prime (A092110), with last decimal being 3.


4



13, 43, 53, 113, 193, 223, 283, 563, 613, 643, 743, 773, 1033, 1193, 1453, 1483, 1543, 1583, 1663, 1733, 2143, 2393, 2503, 2843, 3163, 3413, 3433, 3793, 3823, 4133, 4463, 4483, 4523, 4603, 4673, 4813, 5443, 5743, 5953, 6073, 6133, 6163, 6553, 6733, 6863
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OFFSET

1,1


COMMENTS

Except for p=5, the decimals in A092110 end in 3 or 7.
Theorem: If in the triple (2n3,n,2n+3) all numbers are primes then n=5 or the decimal representation of n ends in 3 or 7. Proof: Consider Q=(2n3)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 2n3=5 or n=5 or 2n+1=5 : this means n=4 (not prime); or n=5 (included); or n=2 (impossible, because 2n3=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



PROG

(PARI) isok(n) = (n % 10 == 3) && isprime(n) && isprime(2*n3) && isprime(2*n+3); \\ Michel Marcus, Sep 02 2013


CROSSREFS



KEYWORD

base,nonn


AUTHOR



STATUS

approved



