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A324035
Irregular triangle read by rows of the entries of the Collatz tree A088975 modulo 6, starting with entry 8 == 2 (mod 6).
0
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, 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, 3, 2, 3, 2, 4, 4, 0, 4
OFFSET
0,1
COMMENTS
The length of row l of this irregular triangle is A005186(n+3), n >= 0.
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).
The number of entries of level n congruent to 4 modulo 6 are given by A176866(n+4), for n >= 0.
FORMULA
T(n, k) = A088975(n+3, k) (mod 6), k = 1..A005186(n+3), n >= 0.
EXAMPLE
The irregular triangle T begins:
n\k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 ... A005186(n+3)
0: 2 1
1: 4 1
2: 5 2 2
3: 4 4 2
4: 3 2 3 2 4
5: 0 4 0 4 4
6: 0 1 2 0 1 2 6
7: 0 2 4 0 2 4 6
8: 0 4 5 2 0 4 5 2 8
9: 0 5 2 4 4 0 5 2 4 4 10
10: 0 4 4 5 2 3 2 0 4 4 5 2 3 2 14
11: 0 5 2 3 2 4 4 0 4 0 3 2 3 2 4 4 0 4 18
...
CROSSREFS
KEYWORD
nonn,tabf,easy
AUTHOR
Wolfdieter Lang, Feb 14 2019
STATUS
approved