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A219108 a(n) = k is the smallest number such that k^3 + 1 has exactly n distinct prime factors.

%I A219108 #45 Sep 18 2023 06:17:54
%S A219108 0,1,3,5,17,59,101,563,2617,9299,22109,132989,364979,1970869,23515229,
%T A219108 109258049,831731339
%N A219108 a(n) = k is the smallest number such that k^3 + 1 has exactly n distinct prime factors.
%C A219108 This is to exponent 3 as A180278 is to exponent 2.
%F A219108 a(n) = MIN{k>=0 such that A001221(A001093(k)) = omega(k^3 + 1) = n}.
%e A219108 a(0) = 0 because 0^3 + 1 = 1 has no prime factors (being a unit).
%e A219108 a(1) = 1 because 1^3 + 1 = 2 has one prime factor (being a prime).
%e A219108 a(2) = 3 because 3^3 + 1 = 28 has two distinct prime factors {2, 7}.
%e A219108 a(3) = 5 because 5^3 + 1 = 126 has three distinct prime factors {2, 3, 7}.
%e A219108 a(4) = 17 because 17^3 + 1 = 4914 has four distinct prime factors {2, 3, 7, 13}.
%e A219108 a(5) = 59 because 59^3 + 1 = 205380 has five distinct prime factors {2, 3, 5, 7, 163}.
%t A219108 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 A219108 (PARI) a(n)=for(k=0, oo, if(omega(k^3+1) == n, return(k))) \\ _Andrew Howroyd_, Sep 12 2023
%Y A219108 Cf. A001093, A001221, A180278.
%K A219108 nonn,more
%O A219108 0,3
%A A219108 _Jonathan Vos Post_, Nov 11 2012
%E A219108 a(13)-a(14) from _Robert G. Wilson v_, Dec 12 2012
%E A219108 a(15)-a(16) from _Giovanni Resta_, May 10 2017
%E A219108 Name clarified and incorrect program removed by _Pontus von Brömssen_, Sep 12 2023