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A259342
Irregular triangle read by rows: T(n,k) = number of equivalence classes of binary sequences of length n containing exactly 2k ones.
1
1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 3, 3, 1, 1, 3, 4, 1, 1, 4, 8, 4, 1, 1, 4, 10, 7, 1, 1, 5, 16, 16, 5, 1, 1, 5, 20, 26, 10, 1, 1, 6, 29, 50, 29, 6, 1, 1, 6, 35, 76, 57, 14, 1, 1, 7, 47, 126, 126, 47, 7, 1, 1, 7, 56, 185, 232, 111, 19, 1, 1, 8, 72, 280, 440, 280, 72, 8, 1
OFFSET
1,7
LINKS
W. D. Hoskins and Anne Penfold Street, Twills on a given number of harnesses, J. Austral. Math. Soc. Ser. A 33 (1982), no. 1, 1-15.
W. D. Hoskins and A. P. Street, Twills on a given number of harnesses, J. Austral. Math. Soc. (Series A), 33 (1982), 1-15. (Annotated scanned copy)
FORMULA
Theorem 1 of Hoskins-Street gives an explicit formula.
EXAMPLE
Triangle begins:
1;
1, 1;
1, 1;
1, 2, 1;
1, 2, 1;
1, 3, 3, 1;
1, 3, 4, 1;
1, 4, 8, 4, 1;
1, 4, 10, 7, 1;
1, 5, 16, 16, 5, 1;
1, 5, 20, 26, 10, 1;
1, 6, 29, 50, 29, 6, 1;
1, 6, 35, 76, 57, 14, 1;
1, 7, 47, 126, 126, 47, 7, 1;
...
MAPLE
with(numtheory):
T:= (n, k)-> (add(`if`(irem(2*k*d, n)=0, phi(n/d)
*binomial(d, 2*k*d/n), 0), d=divisors(n))
+n*binomial(iquo(n, 2), k))/(2*n):
seq(seq(T(n, k), k=0..n/2), n=1..20); # Alois P. Heinz, Jun 28 2015
MATHEMATICA
T[n_, k_] := (DivisorSum[n, If[Mod[2k*#, n]==0, EulerPhi[n/#]*Binomial[#, 2k*#/n], 0]&] + n*Binomial[Quotient[n, 2], k])/(2n); Table[T[n, k], {n, 1, 20}, { k, 0, n/2}] // Flatten (* Jean-François Alcover, Feb 28 2017, after Alois P. Heinz *)
CROSSREFS
Row sums are (essentially) A000011.
Sequence in context: A029292 A343662 A152198 * A258280 A159255 A376647
KEYWORD
nonn,tabf
AUTHOR
N. J. A. Sloane, Jun 27 2015
STATUS
approved