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Numbers n such that (prime(n)^2 + prime(n+1)^2 + prime(n+2)^2)/3 is prime (A084951).
3

%I #18 Dec 02 2018 02:23:50

%S 4,5,8,13,15,26,46,47,50,55,57,59,61,65,66,69,77,82,89,91,94,101,105,

%T 116,134,136,137,138,144,157,194,216,219,221,224,225,229,230,234,249,

%U 257,261,263,271,272,275,306,316,319,323

%N Numbers n such that (prime(n)^2 + prime(n+1)^2 + prime(n+2)^2)/3 is prime (A084951).

%C With the exception of the first two terms, all numbers in A133529 are divisible by 3.

%H Zak Seidov, <a href="/A133940/b133940.txt">Table of n, a(n) for n = 1..1000</a>

%e a(1)=4 because (prime(4)^2 + prime(5)^2 + prime(6)^2)/3 = 113 is prime.

%p select(n -> isprime((ithprime(n)^2 + ithprime(n+1)^2 + ithprime(n+2)^2)/3), [$3 .. 1000]); # _Robert Israel_, Apr 21 2015

%t b = {}; a = 2; Do[k = (Prime[n]^a + Prime[n + 1]^a + Prime[n + 2]^a)/3; If[PrimeQ[k], AppendTo[b, n]], {n, 1, 200}]; b

%o (PARI) is(n)=my(p=prime(n),q=nextprime(p+1),r=nextprime(q+1)); n>3 && isprime((p^2+q^2+r^2)/3) \\ _Charles R Greathouse IV_, Apr 21 2015

%Y Cf. A133529, A084951.

%K nonn

%O 1,1

%A _Artur Jasinski_, Sep 30 2007

%E Corrected and edited by _Zak Seidov_, Apr 21 2015