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A138769
a(n) = least positive integer k such that k^2+3 is divisible by at least n distinct primes.
0
1, 3, 9, 33, 201, 1125, 5259, 98481, 1176579, 4970985, 83471355, 607500315, 20298622815, 302065005093, 2979977447571, 46728566085441, 541457057096937, 13094093041014057, 231069516389617197, 5992213273680818217
OFFSET
1,2
COMMENTS
For n<=20, a(n)^2+3 happens to be divisible by exactly n distinct primes. - Max Alekseyev, Oct 10 2024
LINKS
Elgin H. Johnston, Problem 1792, Matematikolimpiyatokulu, Solution, 2009, p. 149. [broken link]
H. A. ShahAli, Problem 1792, Math. Magazine, vol. 81, No. 2, 2008, p. 155.
EXAMPLE
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.
MAPLE
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.
MATHEMATICA
a[n_] := Block[{k=1}, While[PrimeNu[k^2 + 3] != n, k++]; k]; Array[a, 8] (* Giovanni Resta, Nov 29 2019 *)
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Emeric Deutsch, Apr 03 2008
EXTENSIONS
a(11)-a(12) from Donovan Johnson, Aug 31 2008
a(13)-a(14) from Giovanni Resta, Nov 29 2019
a(15)-a(20) from Max Alekseyev, Oct 10 2024
STATUS
approved