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

0, 1, 3, 5, 17, 59, 101, 563, 2617, 9299, 22109, 132989, 364979, 1970869, 23515229, 109258049, 831731339

Offset

0,3

Comments

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

Links

Table of n, a(n) for n=0..16.

Formula

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

Example

a(0) = 0 because 0^3 + 1 = 1 has no prime factors (being a unit).
a(1) = 1 because 1^3 + 1 = 2 has one prime factor (being a prime).
a(2) = 3 because 3^3 + 1 = 28 has two distinct prime factors {2, 7}.
a(3) = 5 because 5^3 + 1 = 126 has three distinct prime factors {2, 3, 7}.
a(4) = 17 because 17^3 + 1 = 4914 has four distinct prime factors {2, 3, 7, 13}.
a(5) = 59 because 59^3 + 1 = 205380 has five distinct prime factors {2, 3, 5, 7, 163}.

Mathematica

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 *)

Prog

(PARI) a(n)=for(k=0, oo, if(omega(k^3+1) == n, return(k))) \\ Andrew Howroyd, Sep 12 2023

Crossrefs

Cf. A001093, A001221, A180278.

Sequence in context: A357442 A049540 A097144 * A176964 A085749 A281623

Adjacent sequences: A219105 A219106 A219107 * A219109 A219110 A219111

Keyword

nonn,more

Author

Jonathan Vos Post, Nov 11 2012

Extensions

a(13)-a(14) from Robert G. Wilson v, Dec 12 2012

a(15)-a(16) from Giovanni Resta, May 10 2017

Name clarified and incorrect program removed by Pontus von Brömssen, Sep 12 2023

Status

approved