Regarding the inradius sequence A120062

[Jonathan Post]

Regarding the inradius sequence A120062, which makes me wonder what else we
know about A057721(n) =
n^4+3*n^2+1, I just added the sequence:

%I A000001
%S A000001 0, 1, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 3, 3, 3, 1, 2, 3,
1, 2, 2, 1, 2, 2, 2, 2, 3, 2, 1, 3, 1
%N A000001 The number of prime factors (with multiplicity) of n^4 +
3*n^2 + 1 (A057721).
%C A000001 A120062 It is conjectured that the longest possible side c
of a triangle with integer sides and inradius n is given by A057721(n) =
n^4+3*n^2+1. Primes in A057721(n) (i.e. this sequence has a(n) = 1)
include 5, 29, 109, 701, 2549, 4289, 10301, 21169, 84389, 161201, 281429,
812701, 1051649. Semiprimes include 305 = 5 * 61, 1405 = 5 * 281, 6805
= 5 * 1361, 15005 = 5 * 3001, 29069 = 41 * 709, 105949 = 101 * 1049,
195805 = 5 * 39161, 235709 = 41 * 5749, 333505 = 5 * 66701, 392501 = 389
* 1009, 459005 = 5 * 91801, 533629 = 29 * 18401, 709805 = 5 * 141961.
%F A000001 a(n) = Bigomega(n^4 + 3*n^2 + 1) = A001222(A057721(n)).
%Y A000001 Cf. A001222, A057721, A120062.
%O A000001 0,5
%K A000001 ,easy,nonn,
%A A000001 Jonathan Vos Post
(jvospost2 at yahoo.com<http://us.f365.mail.yahoo.com/ym/Compose?To=jvospost2@yahoo.com&YY=88111&order=down&sort=date&pos=0&view=a&head=b>),
Jun 11 2006
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