A145293 a(n) is the smallest nonnegative x such that the Euler polynomial x^2 + x + 41 has exactly n distinct prime proper divisors.
0, 41, 420, 2911, 38913, 707864, 6618260, 78776990, 725005500
Offset: 1
Examples
a(1)=0 because when x=0 then x^2+x+41=41 (1 distinct prime divisor); a(2)=41 because when x=41 then x^2+x+41=1763=41*43 (2 distinct prime divisors); a(3)=420 because when x=420 then x^2+x+41=176861=47*53*71 (3 distinct prime divisors); a(4)=2911 because when x=2911 then x^2+x+41=8476873=41*47*53*83 (4 distinct prime divisors); a(5)=38913 because when x=38913 then x^2+x+41=1514260523=43*47*61*71*173 (5 distinct prime divisors); a(6)=707864 because when x=707864 then x^2+x+41=501072150401=41*43*47*53*71*1607 (6 distinct prime divisors); a(7)=6618260 because when x=6618260 then x^2+x+41=43801372045901=41*43*47*61*83*131*797 (7 distinct prime divisors); a(8)=78776990 because when x=78776990 then x^2+x+41=6205814232237131=41*43*61*71*97*131*167*383 (8 distinct prime divisors). a(9)=725005500: a(9)^2 + a(9) + 41 = 525632975755255541 = 41*43*47*53*61*71*151*397*461. - _Hugo Pfoertner_, Mar 05 2018
Crossrefs
Programs
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Mathematica
a = {}; Do[x = 1; While[Length[FactorInteger[x^2 + x + 41]] < k - 1, x++ ]; AppendTo[a, x]; Print[x], {k, 2, 10}]; a
Extensions
Corrected and edited, a(8) added by Zak Seidov, Jan 31 2016
Example for a(8) corrected by Hugo Pfoertner, Mar 02 2018
a(9) from Hugo Pfoertner, Mar 05 2018
Comments