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A011847
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Triangle of numbers read by rows: T(n,k) = floor( C(n,k)/(k+1) ), where k=0..n-1 and n >= 1.
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15
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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
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OFFSET
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1,8
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COMMENTS
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When k+1 is a prime >= 2, then T(n,k) = floor(C(n,k)/(k+1)) is the number of aperiodic necklaces of n+1 beads of 2 colors such that k+1 of them are black and n-k of them are white. This is not true when k+1 is a composite >= 4. For more details, see the comments for sequences A032168 and A032169. - Petros Hadjicostas, Aug 27 2018
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LINKS
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FORMULA
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The rows are palindromic: T(n, k) = T(n, n-k-1) for n >= 1 and 0 <= k <= n-1.
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EXAMPLE
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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, ....
For k+1 = 2 and n >= k+1 = 2, the n-th element of column k=1 above, [0, 1, 1, 2, 2, 3, 3, 4, 4, ...] (i.e., the number A008619(n-2) = floor(n/2)), gives the number of aperiodic necklaces of n+1 beads of 2 colors such that 2 of them are black and n-1 of them are white. (The offset of sequence A008619 is 0.)
For k+1 = 3 and n >= k+1 = 3, the n-th element of column k=2 above, [0, 0, 1, 2, 3, 5, 7, 9, 12, ...] (i.e., the number A001840(n-2) = 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 n-2 of them are white. (The offset of sequence A001840 is 0.)
For k+1 = 5 and n >= k+1 = 5, the n-th 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 n-4 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 n-3 of them are white is (1/4)*Sum_{d|4} mu(d)*I(d|n+1)*C(floor((n+1)/d) - 1, 4/d - 1) = (1/4)*(C(n, 3) - I(2|n+1)*floor((n-1)/2)), where I(a|b) = 1 if integer a divides integer b, and 0 otherwise. For n odd >= 9, the number (1/4)*(C(n, 3) - I(2|n+1)*floor((n-1)/2)) = A006918(n-3) is not equal to floor(C(n,3)/4) = A011842(n).
(End)
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MATHEMATICA
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Table[Floor[Binomial[n, k]/(k+1)], {n, 20}, {k, 0, n-1}]//Flatten (* Harvey P. Dale, Jan 09 2019 *)
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PROG
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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