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Irregular triangle read by rows of the entries of the Collatz tree A088975 modulo 6, starting with entry 8 == 2 (mod 6).
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%I #10 Feb 16 2019 19:29:59

%S 2,4,5,2,4,4,3,2,3,2,0,4,0,4,0,1,2,0,1,2,0,2,4,0,2,4,0,4,5,2,0,4,5,2,

%T 0,5,2,4,4,0,5,2,4,4,0,4,4,5,2,3,2,0,4,4,5,2,3,2,0,5,2,3,2,4,4,0,4,0,

%U 3,2,3,2,4,4,0,4

%N Irregular triangle read by rows of the entries of the Collatz tree A088975 modulo 6, starting with entry 8 == 2 (mod 6).

%C The length of row l of this irregular triangle is A005186(n+3), n >= 0.

%C The entries of the Collatz tree A088975 modulo 6 are interesting because each 4 (mod 6) entry belongs to a vertex with outdegree 2 and all other vertices have outdegree 1. See a comment in A088975. The root 8 is chosen because the vertex 4 of the preceding level does not obey this rule (otherwise a tree repetiton would occur).

%C The number of entries of level n congruent to 4 modulo 6 are given by A176866(n+4), for n >= 0.

%F T(n, k) = A088975(n+3, k) (mod 6), k = 1..A005186(n+3), n >= 0.

%e The irregular triangle T begins:

%e n\k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 ... A005186(n+3)

%e 0: 2 1

%e 1: 4 1

%e 2: 5 2 2

%e 3: 4 4 2

%e 4: 3 2 3 2 4

%e 5: 0 4 0 4 4

%e 6: 0 1 2 0 1 2 6

%e 7: 0 2 4 0 2 4 6

%e 8: 0 4 5 2 0 4 5 2 8

%e 9: 0 5 2 4 4 0 5 2 4 4 10

%e 10: 0 4 4 5 2 3 2 0 4 4 5 2 3 2 14

%e 11: 0 5 2 3 2 4 4 0 4 0 3 2 3 2 4 4 0 4 18

%e ...

%Y Cf. A005186, A088975, A176866.

%K nonn,tabf,easy

%O 0,1

%A _Wolfdieter Lang_, Feb 14 2019