A050186 Triangular array T read by rows: T(h,k) = number of binary words of k 1's and h-k 0's which are not a juxtaposition of 2 or more identical subwords.

1, 1, 1, 0, 2, 0, 0, 3, 3, 0, 0, 4, 4, 4, 0, 0, 5, 10, 10, 5, 0, 0, 6, 12, 18, 12, 6, 0, 0, 7, 21, 35, 35, 21, 7, 0, 0, 8, 24, 56, 64, 56, 24, 8, 0, 0, 9, 36, 81, 126, 126, 81, 36, 9, 0, 0, 10, 40, 120, 200, 250, 200, 120, 40, 10, 0, 0, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11

Offset: 0

Author

Clark Kimberling

Keywords

• nonn
• tabl
• nice

Examples

			For example, T(4,2) counts 1100,1001,0011,0110; T(2,1) counts 10, 01 (hence also counts 1010, 0101).
Rows:
  1;
  1,  1;
  0,  2,  0;
  0,  3,  3,  0;
  0,  4,  4,  4,  0;
  0,  5, 10, 10,  5,  0;
		

Links

• M. F. Hasler, A050186, rows 0..50, Sep 27 2018
• M. F. Hasler, A050186, rows 0..100, Sep 27 2018
• M. F. Hasler, A050186, rows 0..200, Sep 27 2018
• N. J. A. Sloane, Transforms
• Index entries for triangles and arrays related to Pascal's triangle

Crossrefs

Same triangle as A053727 except this one includes column 0.

T(2n, n), T(2n+1, n) match A007727, A001700, respectively. Row sums match A027375.

Programs

Mathematica

T[n_, k_] := If[n == 0, 1, DivisorSum[GCD[k, n], MoebiusMu[#] Binomial[n/#, k/#]&]];
Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 16 2022 *)

PARI

A050186(n,k)=sumdiv(gcd(n+!n,k),d,moebius(d)*binomial(n/d,k/d)) \\ M. F. Hasler, Sep 27 2018

Formula

MOEBIUS transform of A007318 Pascal's Triangle.

If rows n > 1 are divided by n, this yields the triangle A051168, which equals A245558 surrounded by 0's (except for initial terms). This differs from A011847 from row n = 9 on. - M. F. Hasler, Sep 29 2018