A138769 - OEIS

%I #22 Oct 12 2024 01:58:07

%S 1,3,9,33,201,1125,5259,98481,1176579,4970985,83471355,607500315,

%T 20298622815,302065005093,2979977447571,46728566085441,

%U 541457057096937,13094093041014057,231069516389617197,5992213273680818217

%N a(n) = least positive integer k such that k^2+3 is divisible by at least n distinct primes.

%C For n<=20, a(n)^2+3 happens to be divisible by exactly n distinct primes. - _Max Alekseyev_, Oct 10 2024

%H Elgin H. Johnston, <a href="http://www.matematikolimpiyatokulu.com/images/Mag_Abr_2009.pdf">Problem 1792</a>, Matematikolimpiyatokulu, Solution, 2009, p. 149. [broken link]

%H H. A. ShahAli, <a href="https://www.jstor.org/stable/27643098">Problem 1792</a>, Math. Magazine, vol. 81, No. 2, 2008, p. 155.

%e a(3)=9 because 1^2+3=2*2, 2^2+3=7, 3^2+3=2*2*3, 4^2+3=19, 5^2+3=2*2*7, 6^2+3=3*13, 7^2+3=2*2*13, 8^2+3=67 have at most 2 distinct prime divisors, while 9^2+3=2*2*3*7 has 3 distinct prime divisors.

%p n:=7: with(numtheory): for k while nops(factorset(k^2+3)) < n do end do: a[n]:=k; A[n]:=factorset(k^2+3); # yields a(7) as well as its 7 prime divisors; change the value of n to obtain other terms.

%t a[n_] := Block[{k=1}, While[PrimeNu[k^2 + 3] != n, k++]; k]; Array[a, 8] (* _Giovanni Resta_, Nov 29 2019 *)

%Y Cf. A001221, A180278, A257366.

%K nonn,more

%O 1,2

%A _Emeric Deutsch_, Apr 03 2008

%E a(11)-a(12) from _Donovan Johnson_, Aug 31 2008

%E a(13)-a(14) from _Giovanni Resta_, Nov 29 2019

%E a(15)-a(20) from _Max Alekseyev_, Oct 10 2024