A011847 Triangle of numbers read by rows: T(n,k) = floor( C(n,k)/(k+1) ), where k=0..n-1 and n >= 1.

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

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

Table of n, a(n) for n=1..82.

Formula

The rows are palindromic: T(n, k) = T(n, n-k-1) for n >= 1 and 0 <= k <= n-1.

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 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

Crossrefs

Columns include A008619, A001840, A011842, A011795, A011843, A011797, A011844, A011845, A011846, A032169. Row sums are in A095718.

Cf. A245558.

Sequence in context: A370062 A169623 A245558 * A091325 A193596 A275420

Adjacent sequences: A011844 A011845 A011846 * A011848 A011849 A011850

Keyword

nonn,tabl

Author

N. J. A. Sloane

Status

approved