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A000371
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a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n,k)*2^(2^k).
(Formerly M0385 N0145)
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7
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2, 2, 10, 218, 64594, 4294642034, 18446744047940725978, 340282366920938463334247399005993378250, 115792089237316195423570985008687907850547725730273056332267095982282337798562
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,1
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COMMENTS
| Inverse binomial transform of A001146.
Number of nondegenerate Boolean functions of n variables.
Number of covers of an n-set. That is, the number of subsets of an n-element set S whose union is S.
Comments 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 (n choose j) 2^2^(n-j), as required. (End)
Comments from N. J. A. Sloane, May 19 2011: 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).
For n>0, a(n)=A076078(A002110(n)). - Matthew Vandermast (ghodges14(AT)comcast.net), Nov 14 2010
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REFERENCES
| L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 165.
M. A. Harrison, Introduction to Switching and Automata Theory. McGraw Hill, NY, 1965, p. 170.
T. Hearne and C. G. Wagner, Minimal covers of finite sets, Discr. Math. 5 (1973), 247-251.
A. J. Macula, Covers of a finite set, Math. Mag., 67 (1994), 141-144.
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.
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LINKS
| M. Klazar, Extremal problems for ordered hypergraphs
N. J. A. Sloane, Transforms
Eric Weisstein's World of Mathematics, Cover
Index entries for sequences related to Boolean functions
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FORMULA
| The coefficient of x^k in the polynomial p_n(x) = sum_{j=0}^n (-1)^j (n choose 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, = f_{n-1}(k^2) - f_{n-1}(k) if k > 0 that gives a(n) = f_n(2) and p_n(x) = f_n(x+1). - David Wilson, Nov 11 2010.
E.g.f.: Sum(exp((2^n-1)*x)*ln(2)^n/n!, n=0..infinity). - Vladeta Jovovic (vladeta(AT)eunet.rs), May 30 2004
a(n) ~ 2^2^n. [Charles R Greathouse IV, Jan 02 2012]
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EXAMPLE
| Let n=2, S={a,b}, 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.
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MAPLE
| f:=n->add((-1)^(n-k)*binomial(n, k)*2^(2^k), k=0..n);
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MATHEMATICA
| Table[Sum[(-1)^(n-k) Binomial[n, k]2^(2^k), {k, 0, n}], {n, 0, 10}] (* From Harvey P. Dale, Oct 17 2011 *)
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PROG
| (PARI) a(n)=sum(k=0, n, (-1)^(n-k)*binomial(n, k)<<(2^k)) \\ Charles R Greathouse IV, Jan 02 2012
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CROSSREFS
| Equals twice A003465.
Cf. A001146, A002110, A076078, A055154.
Sequence in context: A002250 A011248 A069240 * A081088 A001038 A027623
Adjacent sequences: A000368 A000369 A000370 * A000372 A000373 A000374
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KEYWORD
| nonn,easy,nice
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com).
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EXTENSIONS
| Since this sequences arises in several different contexts, I replaced the old definition by an explicit formula. - N. J. A. Sloane, Nov 23 2010
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