login
A105551
Number of distinct prime factors of n^3 + n^2 + 71.
2
1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 2, 1, 1, 1, 1, 3, 2, 2, 2, 2, 1, 3, 3, 2, 1, 2, 1, 1, 3, 2, 3, 1, 2, 1, 1, 3, 2, 1, 2, 3, 2, 2, 2, 1, 3, 1, 1, 3, 2, 1, 2
OFFSET
0,6
COMMENTS
This cubic equation with small positive coefficients is strangely rich in primes and semiprimes. The first 44 consecutive values, for n = 0, 1, 2, ..., 43, are all either prime (23 of them) or semiprime (21 of them), before the first 3-almost prime value is encountered.
LINKS
Ulrich Abel and Hartmut Siebert, Sequences with Large Numbers of Prime Values, Am. Math. Monthly 100, 167-169, 1993.
Robin Forman, Sequences with Many Primes, Amer. Math. Monthly 99, 548-557, 1992.
Betty Garrison, Polynomials with Large Numbers of Prime Values, Amer. Math. Monthly 97, 316-317, 1990.
Eric Weisstein's World of Mathematics, Prime-Generating Polynomial.
FORMULA
a(n) = A001221(n^3 + n^2 + 71).
EXAMPLE
a(0) = 1 because 0^3 + 0^2 + 71 = 71 is prime.
a(1) = 1 because 1^3 + 1^2 + 71 = 73 is prime.
a(2) = 1 because 2^3 + 2^2 + 71 = 83 is prime.
a(3) = 1 because 3^3 + 3^2 + 71 = 107 is prime.
a(4) = 1 because 3^3 + 3^2 + 71 = 151 is prime.
a(5) = 2 because 3^3 + 3^2 + 71 = 221 = 13 * 17 is the first semiprime.
a(44) = 3 because 44^3 + 44^2 + 71 = 87191 = 13 * 19 * 353 is the first 3-almost prime for nonnegative integers n.
MATHEMATICA
Table[PrimeNu[n^3+n^2+71], {n, 0, 90}] (* Harvey P. Dale, Oct 09 2012 *)
PROG
(PARI) a(n)=omega(n^3 + n^2 + 71) \\ Charles R Greathouse IV, Jan 31 2017
KEYWORD
easy,nonn
AUTHOR
Jonathan Vos Post, May 03 2005
EXTENSIONS
More terms from Robert G. Wilson v, May 21 2005
STATUS
approved