A365706 - OEIS

%I #39 Oct 22 2023 17:03:07

%S 3,2393,5,827,53,271,1063,23993,197,29,193,2143,359,6829,397,17,433,

%T 661,2837,25171,13597,563,10301,1814233,51427,6781,316817,7477,71,

%U 238919,11491,3109,42293,38653,6263,13043,474497,21433,13,21419,16963,5119,705209,183761

%N For n >= 1, a(n) is the least prime p such that the arithmetic mean of (n + 1) consecutive primes starting with p is a perfect square, or a(n) = -1 if no such p exists.

%C Does a(n) exists for all n >= 1 ?

%C From _David A. Corneth_, Oct 18 2023: (start)

%C Let s(n) be the sum of n + 1 consecutive primes starting with a(n). Then s(n)/(n+1) = m^2 for some positive integer m.

%C This means s(n) = (n + 1) * m^2. If n is even then m is odd if a(n) > 2.

%C As s(n) >= A007504(n) we have m^2 >= s(n)/(n+1) >= A007504(n)/(n+1) i.e. m >= sqrt(A007504(n)/(n+1)). So for some m we can see if m^2 * (n+1) is the sum of n+1 consecutive primes and if so a(n) is the smallest prime of these n+1 primes after testing all candidates up to m. (End)

%C From _Ctibor O. Zizka_, Oct 18 2023: (start)

%C s(n) = (n + 1)* a(n) + Sum_{i=0..(n-1)} (n-i)*g(i+1), thus we have Sum_{i=0..(n-1)} (n-i)*g(i+1) = (m^2 - a(n)) * (n + 1),  g(j) are the n gaps between n + 1 consecutive primes. (End)

%e n = 2: we search for the least prime(i) such that (prime(i) + prime(i + 1) + prime(i + 2))/3 = m^2, m an integer. This is valid for (2393 + 2399 + 2411)/3 = 49^2 thus a(2) = 2393.

%o (PARI) isok(x) = (denominator(x)==1) && issquare(x);

%o a(n) = my(k=1); while (!isok((vecsum(primes(k+n))-vecsum(primes(k-1)))/(n+1)), k++); prime(k); \\ _Michel Marcus_, Oct 16 2023

%o (PARI) a(n) = {my(m = n + 1, ps = vector(m, i, prime(i)), s); forprime(p = ps[m] + 1, , s = vecsum(ps); if(!(s % m) && issquare(s/m), return(ps[1])); ps = concat(vecextract(ps, "^1"), p));} \\ _Amiram Eldar_, Oct 18 2023

%Y Cf. A000040, A007504, A053001, A225195.

%K nonn

%O 1,1

%A _Ctibor O. Zizka_, Oct 15 2023

%E More terms from _Amiram Eldar_, Oct 18 2023