A219108 - OEIS

%I #45 Sep 18 2023 06:17:54

%S 0,1,3,5,17,59,101,563,2617,9299,22109,132989,364979,1970869,23515229,

%T 109258049,831731339

%N a(n) = k is the smallest number such that k^3 + 1 has exactly n distinct prime factors.

%C This is to exponent 3 as A180278 is to exponent 2.

%F a(n) = MIN{k>=0 such that A001221(A001093(k)) = omega(k^3 + 1) = n}.

%e a(0) = 0 because 0^3 + 1 = 1 has no prime factors (being a unit).

%e a(1) = 1 because 1^3 + 1 = 2 has one prime factor (being a prime).

%e a(2) = 3 because 3^3 + 1 = 28 has two distinct prime factors {2, 7}.

%e a(3) = 5 because 5^3 + 1 = 126 has three distinct prime factors {2, 3, 7}.

%e a(4) = 17 because 17^3 + 1 = 4914 has four distinct prime factors {2, 3, 7, 13}.

%e a(5) = 59 because 59^3 + 1 = 205380 has five distinct prime factors {2, 3, 5, 7, 163}.

%t k = 1; t = Table[0, {15}]; While[k < 30000001, a = PrimeNu[k^3 + 1]; If[ t[[a]] == 0, t[[a]] = k; Print[{a, k}]]; k++]; t (* _Robert G. Wilson v_, Dec 12 2012 *)

%o (PARI) a(n)=for(k=0, oo, if(omega(k^3+1) == n, return(k))) \\ _Andrew Howroyd_, Sep 12 2023

%Y Cf. A001093, A001221, A180278.

%K nonn,more

%O 0,3

%A _Jonathan Vos Post_, Nov 11 2012

%E a(13)-a(14) from _Robert G. Wilson v_, Dec 12 2012

%E a(15)-a(16) from _Giovanni Resta_, May 10 2017

%E Name clarified and incorrect program removed by _Pontus von Brömssen_, Sep 12 2023