
EXAMPLE

Triangle begins:
[1]
[1, 1]
[1, 1, 1]
[1, 2, 2, 1]
[1, 2, 3, 2, 1]
[1, 3, 5, 5, 3, 1]
[1, 3, 7, 8, 7, 3, 1]
[1, 4, 9, 14, 14, 9, 4, 1]
[1, 4, 12, 21, 25, 21, 12, 4, 1]
[1, 5, 15, 30, 42, 42, 30, 15, 5, 1]
[1, 5, 18, 41, 66, 77, 66, 41, 18, 5, 1]
[1, 6, 22, 55, 99, 132, 132, 99, 55, 22, 6, 1]
[1, 6, 26, 71, 143, 214, 245, 214, 143, 71, 26, 6, 1]
[1, 7, 30, 91, 200, 333, 429, 429, 333, 200, 91, 30, 7, 1]
[1, 7, 35, 113, 273, 500, 715, 804, 715, 500, 273, 113, 35, 7, 1]
[1, 8, 40, 140, 364, 728, 1144, 1430, 1430, 1144, 728, 364, 140, 40, 8, 1]
[1, 8, 45, 170, 476, 1031, 1768, 2431, 2701, 2431, 1768, 1031, 476, 170, 45, 8, 1]
...
More than the usual number of rows are shown in order to distinguish this triangle from A245558, from which it differs in rows 9, 11, 13, ....
From Petros Hadjicostas, Aug 27 2018: (Start)
For k+1 = 2 and n >= k+1 = 2, the nth element of column k=1 above, [0, 1, 1, 2, 2, 3, 3, 4, 4, ...] (i.e., the number A008619(n2) = floor(n/2)), gives the number of aperiodic necklaces of n+1 beads of 2 colors such that 2 of them are black and n1 of them are white. (The offset of sequence A008619 is 0.)
For k+1 = 3 and n >= k+1 = 3, the nth element of column k=2 above, [0, 0, 1, 2, 3, 5, 7, 9, 12, ...] (i.e., the number A001840(n2) = floor(C(n,2)/3)), gives the number of aperiodic necklaces of n+1 beads of 2 colors such that 3 of them are black and n2 of them are white. (The offset of sequence A001840 is 0.)
For k+1 = 5 and n >= k+1 = 5, the nth element of column k=4 above, [0, 0, 0, 0, 1, 3, 7, 14, 25, 42, ... ] (i.e., the number A011795(n) = floor(C(n,4)/5)), gives the number of aperiodic necklaces of n+1 beads of 2 colors such that 5 of them are black and n4 of them are white. (The offset of sequence A011795 is 0.)
Counterexample for k+1 = 4: It can be proved that, for n >= k+1 = 4, the number of aperiodic necklaces of n+1 beads of 2 colors such that 4 of them are black and n3 of them are white is (1/4)*Sum_{d4} mu(d)*I(dn+1)*C(floor((n+1)/d)  1, 4/d  1) = (1/4)*(C(n, 3)  I(2n+1)*floor((n1)/2)), where I(ab) = 1 if integer a divides integer b, and 0 otherwise. For n odd >= 9, the number (1/4)*(C(n, 3)  I(2n+1)*floor((n1)/2)) = A006918(n3) is not equal to floor(C(n,3)/4) = A011842(n).
(End)
