[seqfan] Re: Q about A152926

[Benoît Jubin]

Just compute 19n+{2,4,8,10} mod 15. You don't want it to equal
0,3,5,6,9,10,12 (which are not prime with 15). The only solution is
n=6 mod15.

Benoit


On Mon, Dec 15, 2008 at 8:30 AM, zak seidov <zakseidov at yahoo.com> wrote:
> %C A152926 All terms == 6 (mod 15)
> - but why?
> thx, zak
>
> %I A152926
> %S A152926 171,3801,5781,8721,8781,17601,18231,19011,24741,28251,40431,48951,
> %T A152926 49371,58821,70521,79401,79701,83391,87321,95781,96501,99501,102861,
> %U A152926 109431,123171,125061,137091,177201,220311,224511,225561,229551,242451
> %N A152926 Numbers n with property that 19n+{2,4, 8,10} are two subsequent twin primes.
> %C A152926 All terms == 6 (mod 15).
> %e A152926 19*171+{2,4}={3251,3253} and 19*171+{8,10}={3257,3259} are 85th and 86th twin primes.
> %e A152926 19*3801+{2,4}={72221,72223} and 19*3801+{8,10}={72227,72229} are 935-th and 936-th twin primes.
> %Y A152926 A001359 Lesser of twin primes.
> %K A152926 nonn
> %O A152926 1,1
> %A A152926 Zak Seidov (zakseidov(AT)yahoo.com), Dec 15 2008
>
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