OFFSET
1,8
COMMENTS
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
Differs from A245558 from row n = 9, k = 4 on. - M. F. Hasler, Sep 29 2018
LINKS
G. C. Greubel, Rows n = 1..100 of the triangle, flattened
FORMULA
The rows are palindromic: T(n, k) = T(n, n-k-1) for n >= 1 and 0 <= k <= n-1.
Sum_{k=0..floor((n-1)/2)} T(n-k, k) = A095719(n). - G. C. Greubel, Oct 20 2024
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;
...
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 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)
MATHEMATICA
Table[Floor[Binomial[n, k]/(k+1)], {n, 20}, {k, 0, n-1}]//Flatten (* Harvey P. Dale, Jan 09 2019 *)
PROG
(PARI) A011847(n, k)=binomial(n, k)\(k+1) \\ M. F. Hasler, Sep 30 2018
(Magma)
A011847:= func< n, k | Floor(Binomial(n+1, k+1)/(n+1)) >;
[A011847(n, k): k in [0..n-1], n in [1..20]]; // G. C. Greubel, Oct 20 2024
(SageMath)
def A011847(n, k): return binomial(n+1, k+1)//(n+1)
flatten([[A011847(n, k) for k in range(n)] for n in range(1, 21)]) # G. C. Greubel, Oct 20 2024
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
STATUS
approved