%I #21 Nov 02 2022 07:43:07
%S 1,4,2,6,4,2,6,2,2,2,2,2,2,6,10,4,4,16,2,12,4,18,2,4,6,2,2,12,2,4,2,2,
%T 2,24,10,2,8,2,2,8,12,2,16,2,6,2,2,30,4,2,4,8,2,2,4,2,6,2,6,2,24,20,2,
%U 4,6,36,2,6,4,6
%N Square root of largest square dividing A069482(n).
%C Analogous to A001223 with 2-norm.
%C a(n) is the square root of the square part of A069482(n).
%F Conjectures: (Start)
%F a(A068361(n)) = A001223(A068361(n)).
%F a(A068361(n)) = 2 for n>1.
%F These are the only a(n)=A001223(n).
%F (End)
%F a(n) = A000188(A069482(n)). - _Michel Marcus_, Apr 27 2016
%e sqrt(5)=1*sqrt(5), a(n)=1.
%e sqrt(16)=4*sqrt(1), a(n)=4.
%e sqrt(24)=2*sqrt(6), a(n)=2.
%t Table[Sqrt[Prime[n+1]^2-Prime[n]^2],{n,1,100}]/.Sqrt[_]->1
%o (PARI) a(n) = my(d=prime(n+1)^2 - prime(n)^2); sqrtint(d/core(d)); \\ _Michel Marcus_, Apr 27 2016
%Y Cf. A000188, A069482
%Y Cf. A001223, A068361.
%K nonn
%O 1,2
%A _Benedict W. J. Irwin_, Apr 20 2016