login
Doudna-tree reduced modulo 4: a(n) = A005940(1+n) mod 4.
3

%I #17 Dec 01 2017 18:51:16

%S 1,2,3,0,1,2,1,0,3,2,3,0,1,2,3,0,3,2,1,0,3,2,1,0,1,2,3,0,1,2,1,0,1,2,

%T 1,0,3,2,3,0,1,2,1,0,3,2,3,0,1,2,3,0,1,2,1,0,3,2,3,0,1,2,3,0,1,2,3,0,

%U 1,2,3,0,3,2,1,0,3,2,1,0,3,2,3,0,1,2,3,0,3,2,1,0,3,2,1,0,1,2,3,0,1,2,1,0,3,2,3,0,1,2,3,0,3,2,1,0,3,2,1,0,1

%N Doudna-tree reduced modulo 4: a(n) = A005940(1+n) mod 4.

%H Antti Karttunen, <a href="/A292603/b292603.txt">Table of n, a(n) for n = 0..16383</a>

%H <a href="/index/Pri#prime_indices">Index entries for sequences computed from indices in prime factorization</a>

%F a(n) = A010873(A005940(1+n)).

%F a(n) + 4*A292602(n) = A005940(1+n).

%F a(2n+1) = 2*a(n) mod 4.

%F a(A004767(n)) = 0.

%F a(A016813(n)) = 2.

%F a(2*A156552(A246261(n))) = 1.

%F a(2*A156552(A246263(n))) = 3.

%F a(n * 2^(1+A246271(A005940(1+n)))) = 1.

%e The first six levels of the binary tree (compare also to the illustrations given at A005940 and A292602):

%e 1

%e |

%e 2

%e ............../ \..............

%e 3 0

%e ....../ \...... ....../ \......

%e 1 2 1 0

%e / \ / \ / \ / \

%e / \ / \ / \ / \

%e 3 2 3 0 1 2 3 0

%e / \ / \ / \ / \ / \ / \ / \ / \

%e 3 2 1 0 3 2 1 0 1 2 3 0 1 2 1 0

%o (Scheme) (define (A292603 n) (modulo (A005940 (+ 1 n)) 4))

%Y Cf. A003961, A005940, A292602.

%Y Cf. A004767 (gives the positions of 0's), A016813 (of 2's).

%Y Cf. also A246261, A246263, A246271, A292271, A292274, A292375, A292377, A292381, A292383, A292384, A292583.

%K nonn

%O 0,2

%A _Antti Karttunen_, Dec 01 2017