0,1

Inverse binomial transform of A001146.

Number of nondegenerate Boolean functions of n variables.

Twice the number of covers of an n-set S (A003465). That is, the number of subsets of the power set of S whose union is S. [corrected by Manfred Boergens, May 02 2024]

From David P. Moulton, Nov 11 2010: (Start)

To see why the formula in the definition gives the number of covers of an n-set we use inclusion-exclusion.

The set S has n elements and T, the power set of S, has 2^n elements.

Let U be the power set of T; we want to know how many elements of U have union S.

For any element i of S, let U_i be the subset of U whose unions do not contain i, so we want to compute the size of the complement of the union of the U_i s.

Write U_I for the union of U_i for i in I. Then U_I consists of all subsets of T whose union is disjoint from I, so it consists of all subsets of the power set of S - I. The power set of S - I has 2^(n - #I) elements, so U_I has size 2^2^(n - #I).

Then the basic inclusion-exclusion formula says that our answer is

#(U - union_{i in S} U_i) = Sum_{I subseteq S} (-1)^#I #U_I = Sum_{j=0..n} (-1)^j Sum_{#I = j} #U_I = Sum_{j=0..n} (-1)^j binomial(n,j)*2^2^(n-j), as required.

(End)

Here is Comtet's proof: Let P'(S) be the power set of nonempty subsets of S. Then |P'(P'(S))| = 2^(2^n-1)-1 = Sum_k binomial(n,k)*a(k). Apply the inverse binomial transform to get a(n) = Sum_k (-1)^k*binomial(n,k)*2^(2^(n-k)-1). - N. J. A. Sloane, May 19 2011

For disjoint subsets of the power set see A186021. For disjoint nonempty subsets of the power set see A000110. - Manfred Boergens, May 02 2024 and Apr 09 2025

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 165.

M. A. Harrison, Introduction to Switching and Automata Theory. McGraw Hill, NY, 1965, p. 170.

S. Muroga, Threshold Logic and Its Applications. Wiley, NY, 1971, p. 38 and 214.

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

C. G. Wagner, Covers of finite sets, Proc. 4th S-E Conf. Combin., Graph Theory, Computing, Congress. Numer. 8 (1973), 515-520.

R. Baumann and H. Strass, On the number of bipolar Boolean functions, 2014, preprint.

R. Baumann and H. Strass, On the number of bipolar Boolean functions, Journal of Logic and Computation, 27(8) (2017), 2431-2449.

Thomas M. A. Fink, Regulatory motifs: structural and functional building blocks of genetic computation, arXiv:2208.14996 [q-bio.MN], 2022.

Goto, Eiichi, and Hidetosi Takahasi, Some Theorems Useful in Threshold Logic for Enumerating Boolean Functions, in Proceedings International Federation for Information Processing (IFIP) Congress, 1962, pp. 747-752. [Annotated scans of certain pages]

R. K. Guy, Letter to N. J. A. Sloane, Mar 1974

T. Hearne and C. G. Wagner, Minimal covers of finite sets, Discr. Math. 5 (1973), 247-251.

M. Klazar, Extremal problems for ordered hypergraphs, arXiv:math/0305048 [math.CO], 2003.

A. J. Macula, Covers of a finite set, Math. Mag., 67 (1994), 141-144.

S. Muroga, Threshold Logic and Its Applications, Wiley, NY, 1971 [Annotated scans of a few pages]

N. J. A. Sloane, Transforms

Eric Weisstein's World of Mathematics, Cover

The coefficient of x^k in the polynomial p_n(x) = Sum_{j=0..n} (-1)^j binomial(n,j) * (x+1)^2^(n-j) gives the number of covers of a set of size n where the covers have k elements. Also, there is a recurrence: f_n(k) = k, if n = 0, and f_n(k) = f_{n-1}(k^2) - f_{n-1}(k), if n > 0, that gives a(n) = f_n(2) and p_n(x) = f_n(x+1). - David W. Wilson, Nov 11 2010

E.g.f.: Sum(exp((2^n-1)*x)*log(2)^n/n!, n=0..infinity). - Vladeta Jovovic, May 30 2004

For n > 0, a(n) = A076078(A002110(n)). - Matthew Vandermast, Nov 14 2010

a(n) ~ 2^(2^n). - Charles R Greathouse IV, Jan 02 2012

a(n) = 2*A003465(n). - Maurizio De Leo, Feb 27 2015

Let n = 2, S = {a,b}, and P = {0,a,b,ab}. There are ten subsets of P whose union is S: {ab}, {a,b}, {a,ab}, {b,ab}, {a,b,ab}, and the empty set together with the same five. - Marc LeBrun, Nov 10 2010

f:=n->add((-1)^(n-k)*binomial(n, k)*2^(2^k), k=0..n);

Table[Sum[(-1)^(n-k) Binomial[n, k]2^(2^k), {k, 0, n}], {n, 0, 10}] (* Harvey P. Dale, Oct 17 2011 *)

(PARI) a(n)=sum(k=0, n, (-1)^(n-k)*binomial(n, k)<<(2^k)) \\ Charles R Greathouse IV, Jan 02 2012

(PARI) a(n) = sum(k=0, n, (-1)^k*n!/k!/(n-k)!*2^(2^(n-k))); \\ Altug Alkan, Dec 29 2015

(Magma) [&+[(-1)^(n-k)*Binomial(n, k)*2^(2^k): k in [0..n]]: n in [0..10]]; // Vincenzo Librandi, Dec 28 2015

Equals twice A003465.

Row sums of A163353.

Diagonal of A381683.

Cf. A001146, A002110, A076078, A055154.

Cf. A000110, A186021.

Sequence in context: A005617 A011248 A069240 * A081088 A236369 A001038

Adjacent sequences: A000368 A000369 A000370 * A000372 A000373 A000374

nonn,easy,nice

Since this sequence arises in several different contexts, I replaced the old definition with an explicit formula. - N. J. A. Sloane, Nov 23 2010

approved