

A050186


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


15



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
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OFFSET

0,5


LINKS

Table of n, a(n) for n=0..76.
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


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


EXAMPLE

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;


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 (* JeanFrançois Alcover, Jul 16 2022 *)


PROG

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


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.
Sequence in context: A053202 A188122 A341841 * A334218 A342984 A342985
Adjacent sequences: A050183 A050184 A050185 * A050187 A050188 A050189


KEYWORD

nonn,tabl,nice


AUTHOR

Clark Kimberling


STATUS

approved



