[seqfan] Re: Q about A152926
f.firoozbakht at sci.ui.ac.ir
f.firoozbakht at sci.ui.ac.ir
Wed Dec 17 09:39:43 CET 2008
Dear Zak,
We can show the stronger assertion: all terms == 21 (mod 30).
Proof:
19n+2 is prime so n is odd, namely
n = 1 (mod 2) (I).
19n+2, 19n+4 are two primes greater than 3 and 19n+2 = n-1 <>0(mod 3) ,
19n+4 = n-2<>0 (mod 3) so n<>1 (mod 3) and n<>2 (mod 3) hence,
n=0 (mod 3) (II).
Also since 19n+2, 19n+4, 19n+8 & 19n+10 are four primes greater than 5
and 19n+2 = 2-n (mod 5) , 19n+4 = 4-n (mod 5)
19n+8 = 3-n (mod 5) & 19n+10 = -n (mod 5) hence n = 1 (mod 5) (III).
From (I) & (III) we conclude that n = 1 (mod 10) (IV).
And from (IV) and (II) we conclude that n is of the form 30k+21 namely
n = 21 (mod 30).
Farideh
TheQuoting zak seidov <zakseidov at yahoo.com>:
> %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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