

A174734


Prime numbers n such that 2n1 and 3n2 are prime.


5



3, 7, 37, 211, 271, 307, 331, 337, 601, 727, 1171, 1237, 1297, 1531, 1657, 2221, 2281, 2557, 3037, 3061, 3067, 4261, 4447, 4801, 4951, 5227, 5581, 5851, 6151, 6361, 6691, 6841, 6967, 7621, 7681, 7687, 7867, 8017, 8167, 8191, 8287, 8521, 8527, 8647, 8941
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OFFSET

1,1


COMMENTS

If n, 2n1 and 3n2 are prime numbers, and if n >= 5, then n*(2*n1)*(3*n2) is a Carmichael number (A033502).
Proof: there exist numbers m such that n=6m+1 is prime (if n=6m+5, then 2n1 = 12m+9 is composite). Let p=(6m+1)(12m+1)(18m+1) = a*b*c. Then p1 = 6*12*18*m^3 + (6*12 + 6*18 + 12*18)*m^2 + (6 + 12 + 19)*m, so p1 is divisible by a1=6m, by b1=12m, and by c1=18m; thus p is a Carmichael number.


REFERENCES

R. K. Guy, Unsolved Problems in Number Theory, A13.


LINKS



EXAMPLE

For n=3, 2n1 = 5, 3n2 = 7.
For n=7, 2n1 = 13, 3n2 = 19 and 7*13*19 = 1729 (a Carmichael number).
For n=37, 2n1 = 73, 3n2 = 109 and 37*73*109 = 294409 (a Carmichael number).


MAPLE

with(numtheory): for n from 2 to 15000 do: if type(n, prime)=true and type(2*n1, prime)=true and type(3*n2, prime)=true then print (n):else fi:od:


MATHEMATICA



PROG

(Magma) [ n: n in PrimesUpTo(10000)  IsPrime(2*n1) and IsPrime(3*n2) ];
(PARI) forprime(p=3, 10^3, isprime(2*p1) && isprime(3*p2) && print1(p, ", ")); \\ Joerg Arndt, Nov 29 2014


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



