A075894 Average of four successive primes squared, (prime(n)^2 + prime(n+1)^2 + prime(n+2)^2 + prime(n+3)^2)/4, n>=2.

51, 91, 157, 235, 337, 505, 673, 925, 1213, 1465, 1777, 2137, 2587, 3055, 3625, 4183, 4645, 5275, 5875, 6595, 7615, 8605, 9535, 10417, 11035, 11677, 13057, 14485, 16207, 17845, 19363, 20773, 22243, 24055, 25477, 27259, 29107, 30655, 32803

Offset

2,1

Comments

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]

Links

Harvey P. Dale, Table of n, a(n) for n = 2..1000

Formula

a(n) = (prime(n)^2 + prime(n+1)^2 + prime(n+2)^2 + prime(n+3)^2)/4, n>=2.

Example

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.

Mathematica

Mean[#]&/@(Partition[Prime[Range[2, 50]], 4, 1]^2) (* Harvey P. Dale, Dec 14 2011 *)
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 *)

Prog

(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
(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
(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

Crossrefs

Sequence in context: A351026 A020180 A049328 * A229274 A044140 A044521

Adjacent sequences: A075891 A075892 A075893 * A075895 A075896 A075897

Keyword

easy,nonn

Author

Zak Seidov, Oct 17 2002

Status

approved