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%I #19 Feb 18 2018 05:48:02
%S 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,
%T 448,273,105,35,0,1,0,0,1,0,155,1085,6293,27776,105183,327050,876525,
%U 2011776,4032015,7048811,10855425,14721280,17678835,18771864
%N 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).
%H G. C. Greubel, <a href="/A022619/b022619.txt">Table of n, a(n) for the first 10 rows, flattened</a>
%H <a href="/index/Bo#Boolean">Index entries for sequences related to Boolean functions</a>
%F 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.
%e Triangle begins:
%e [0,1,0],
%e [0,1,0,1,0],
%e [0,1,0,7,7,7,0,1,0],
%e ...;
%e 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.
%t 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];
%t Table[T[n, k], {n, 1, 5}, {k, 0, 2^n}] // Flatten (* _G. C. Greubel_, Feb 15 2018 *)
%Y Row sums give A051502.
%Y Cf. A054724.
%K nonn,tabf
%O 1,12
%A _Vladeta Jovovic_, Jul 13 2000
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