|
%I #24 Nov 15 2012 15:09:11
%S 1,2,2,3,2,4,2,3,3,3,2,4,2,3,3,3,2,5,2,4,3,3,2,4,2,3,4,4,2,4,2,4,3,3,
%T 2,4,2,3,3,3,2,5,2,3,3,3,2,6,2,4,3,3,2,4,2,4,3,3,2,6,2,4,3,3,3,4,2,3,
%U 4,5,2,4,2,3,3,5,2,4,2,3,3,3,2,6,2,3,3
%N Number of divisors of n^3 + 1 of the form a^3 + 1.
%H Alois P. Heinz, <a href="/A217984/b217984.txt">Table of n, a(n) for n = 0..10000</a>
%e a(0) = 1: 1.
%e a(1) = 2: 1, 2.
%e a(2) = 2: 1, 9.
%e a(3) = 3: 1, 2, 28.
%e a(4) = 2: 1, 65.
%e a(5) = 4: 1, 2, 9, 126.
%e a(719) = 10: 1, 2, 9, 28, 65, 126, 217, 2745, 4914, 371694960.
%p with (numtheory):
%p q:= proc(n) local t; t:=n-1;
%p is(iroot(t, 3)^3=t)
%p end:
%p a:= n-> nops(select(q, divisors(n^3+1))):
%p seq (a(n), n=0..120); # _Alois P. Heinz_, Nov 08 2012
%t Table[Count[Divisors[n^3+1],_?(IntegerQ[(#-1)^(1/3)]&)],{n,0,90}] (* _Harvey P. Dale_, Nov 15 2012 *)
%Y Cf. A001093.
%K nonn,easy
%O 0,2
%A _Jonathan Vos Post_, Oct 16 2012
%E More terms from _Alois P. Heinz_, Nov 08 2012
|
