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%I #19 Sep 08 2022 08:45:07
%S 51,91,157,235,337,505,673,925,1213,1465,1777,2137,2587,3055,3625,
%T 4183,4645,5275,5875,6595,7615,8605,9535,10417,11035,11677,13057,
%U 14485,16207,17845,19363,20773,22243,24055,25477,27259,29107,30655,32803
%N Average of four successive primes squared, (prime(n)^2 + prime(n+1)^2 + prime(n+2)^2 + prime(n+3)^2)/4, n>=2.
%C Unlike the average of four successive primes, the average of four successive primes squared is integer for all n>=2. [corrected by _Zak Seidov_, Jul 07 2017]
%H Harvey P. Dale, <a href="/A075894/b075894.txt">Table of n, a(n) for n = 2..1000</a>
%F a(n) = (prime(n)^2 + prime(n+1)^2 + prime(n+2)^2 + prime(n+3)^2)/4, n>=2.
%e a(2)=51 because (prime(2)^2 + prime(3)^2 + prime(4)^2 + prime(5)^2)/4 = (3^2 + 5^2 + 7^2 + 11^2)/4 = 51.
%t Mean[#]&/@(Partition[Prime[Range[2,50]],4,1]^2) (* _Harvey P. Dale_, Dec 14 2011 *)
%t Table[(Prime[n]^2 + Prime[n+1]^2 + Prime[n+2]^2 + Prime[n+3]^2)/4, {n, 2, 50}] (* _Vincenzo Librandi_, Jul 07 2017 *)
%o (Magma) [(NthPrime(n)^2+NthPrime(n+1)^2+NthPrime(n+2)^2+ NthPrime(n+3)^2)/4: n in [2..50]]; // _Vincenzo Librandi_, Jul 07 2017
%o (PARI) a(n,p=prime(n))=my(q=nextprime(p+1),r=nextprime(q+1),s=nextprime(r+1)); norml2([p,q,r,s])/4 \\ _Charles R Greathouse IV_, Jul 07 2017
%o (PARI) first(n)=my(p=primes(n+3)); vector(n-1,i, (p[i+1]^2 + p[i+2]^2 + p[i+3]^2 + p[i+4]^2)/4) \\ _Charles R Greathouse IV_, Jul 07 2017
%K easy,nonn
%O 2,1
%A _Zak Seidov_, Oct 17 2002
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