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Arithmetic mean of next a(n) successive squares of positive integers is prime.
2

%I #8 May 02 2018 22:35:43

%S 5,11,29,43,17,131,13,7,17,7,53,19,25,35,65,59,17,35,113,43,25,35,5,7,

%T 5,11,89,23,17,35,29,43,5,31,109,71,65,7,41,31,61,35,25,107,25,11,41,

%U 47,25,175,41,35,29,23,17,43,197,91,13,95,17,47,5,7,25,11

%N Arithmetic mean of next a(n) successive squares of positive integers is prime.

%C Corresponding primes (arithmetic means) :

%C 11, 131, 1031, 4643, 9433, 30671, 59063, 64013, 70249, 76733, 94483, 117679, 133277, 156127, 198377, 257339, 297049, 326143, 417089, 522883, 573101, 619471, 651251, 660973, 670763

%C All terms are in A007310. Do all members of A007310 except 1 occur? - _Robert Israel_, May 02 2018

%H Robert Israel, <a href="/A214451/b214451.txt">Table of n, a(n) for n = 1..10000</a>

%e (1+4+9+16+25)/5 = 11, so a(1)=5. The next set of successive squares with prime arithmetic mean: (6^2 + 7^2 + ... + 16^2)/11=131, so a(2)=11.

%p r:= 0:

%p for n from 1 to 100 do

%p t:= 0:

%p for j from r+1 do

%p t:= t + j^2;

%p s:= t/(j-r);

%p if s::integer and isprime(s) then

%p A[n]:= j-r;

%p r:= j;

%p break

%p fi

%p od;

%p od:

%p seq(A[i],i=1..100); # _Robert Israel_, May 02 2018

%Y Cf. A000040, A000290, A007310, A214442, A214444.

%K nonn

%O 1,1

%A _Alex Ratushnyak_, Jul 18 2012