A022619 Triangle T(n,k)of numbers of asymmetric Boolean functions of n variables with exactly k = 0..2^n nonzero values (atoms) under action of complementing group C(n,2).

0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 7, 7, 7, 0, 1, 0, 0, 1, 0, 35, 105, 273, 448, 715, 750, 715, 448, 273, 105, 35, 0, 1, 0, 0, 1, 0, 155, 1085, 6293, 27776, 105183, 327050, 876525, 2011776, 4032015, 7048811, 10855425, 14721280, 17678835, 18771864

Offset

1,12

Links

G. C. Greubel, Table of n, a(n) for the first 10 rows, flattened

Index entries for sequences related to Boolean functions

Formula

T(n, k) = coefficient of x^k in (1/2^n)*Sum_{j = 0..n} (-1)^j*2^C(j, 2)*[n, j]*(1+x^(2^j))^(2^(n-j)), where [n, j] is Gaussian 2-binomial coefficient; k = 0..2^n.

Example

Triangle begins:
  [0,1,0],
  [0,1,0,1,0],
  [0,1,0,7,7,7,0,1,0],
  ...;
T(5,k) = coefficient of x^k in (1/32)*((1+x)^32-31*(1+x^2)^16+310*(1+x^4)^8-1240*(1+x^8)^4+1984*(1+x^16)^2-1024*(1+x^32)),k = 0..32.

Mathematica

T[n_, 0]:=0; T[n_, k_] := (1/2^n)*Coefficient[Sum[(-1)^j*2^(Binomial[j, 2])* QBinomial[n, j, 2]*(1 + x^(2^j))^(2^(n - j)), {j, 0, n}], x^k];
Table[T[n, k], {n, 1, 5}, {k, 0, 2^n}] // Flatten (* G. C. Greubel, Feb 15 2018 *)

Crossrefs

Row sums give A051502.

Cf. A054724.

Sequence in context: A124930 A195202 A252799 * A131685 A019860 A011422

Adjacent sequences: A022616 A022617 A022618 * A022620 A022621 A022622

Keyword

nonn,tabf

Author

Vladeta Jovovic, Jul 13 2000

Status

approved