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A075892 Average of squares of successive primes: a(n) = (prime(n+1)^2 + prime(n)^2)/2, with n >= 2.

%I A075892 #36 Jul 27 2022 11:44:27
%S A075892 17,37,85,145,229,325,445,685,901,1165,1525,1765,2029,2509,3145,3601,
%T A075892 4105,4765,5185,5785,6565,7405,8665,9805,10405,11029,11665,12325,
%U A075892 14449,16645,17965,19045,20761,22501,23725,25609,27229,28909,30985,32401
%N A075892 Average of squares of successive primes: a(n) = (prime(n+1)^2 + prime(n)^2)/2, with n >= 2.
%C A075892 a(n) is prime for n in A240749. - _Robert Israel_, Jul 06 2017
%C A075892 If p and q are primes such that p > q > 3, then ((p^2 - q^2)/2, p*q, (p^2 + q^2)/2) is a primitive Pythagorean triple. - _César Aguilera_, Jun 02 2022
%H A075892 Zak Seidov, <a href="/A075892/b075892.txt">Table of n, a(n) for n = 2..10001</a>
%F A075892 a(n)^2 = A124434(n)^2 + A006094(n)^2. - _César Aguilera_, Jun 02 2022
%e A075892 a(2)=17 because (prime(3)^2 + prime(2)^2)/2 = (5^2 + 3^2)/2 = 17.
%p A075892 seq((ithprime(i)^2 + ithprime(i+1)^2)/2, i=2..100); # _Robert Israel_, Jul 06 2017
%t A075892 Table[(Prime[n + 1]^2 + Prime[n]^2)/2, {n, 2, 50}] (* _Vincenzo Librandi_, Mar 07 2015 *)
%t A075892 p=2;q=3;Table[p=q;q=NextPrime[q];(q^2+p^2)/2,{100}] (* _Zak Seidov_, Jul 06 2017 *)
%o A075892 (PARI) a(n) = (prime(n+1)^2+prime(n)^2)/2; \\ _Michel Marcus_, Oct 03 2013
%o A075892 (Magma) [(NthPrime(n+1)^2+NthPrime(n)^2)/2: n in [2..50]]; // _Vincenzo Librandi_, Mar 07 2015
%Y A075892 Cf. A006094, A124434, A143850, A240749.
%K A075892 nonn,easy
%O A075892 2,1
%A A075892 _Zak Seidov_, Oct 17 2002