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A133224
Let P(A) be the power set of an n-element set A and let B be the Cartesian product of P(A) with itself. Remove (y,x) from B when (x,y) is in B and x <> y and let R35 denote the reduced set B. Then a(n) = the sum of the sizes of the union of x and y for every (x,y) in R35.
4
0, 2, 14, 78, 400, 1960, 9312, 43232, 197120, 885888, 3934720, 17307136, 75509760, 327182336, 1409343488, 6039920640, 25770065920, 109522223104, 463857647616, 1958507577344, 8246342451200
OFFSET
0,2
COMMENTS
A082134 is the analogous sequence if "union" is replaced by "intersection" and A002697 is the analogous sequence if "union" is replaced by "symmetric difference". Here, X union Y = Y union X are considered as the same Cartesian product [Relation (37): U_Q(n) in document of Ross La Haye in reference], if we want to consider that X Union Y and Y Union X are two distinct Cartesian products, see A212698. [Bernard Schott, Jan 11 2013]
LINKS
Ross La Haye, Binary Relations on the Power Set of an n-Element Set, Journal of Integer Sequences, Vol. 12 (2009), Article 09.2.6.
FORMULA
a(n) = n*(2^(n-2) + 3*2^(2*n-3)).
G.f.: 2*x*(7*x^2-5*x+1) / ((2*x-1)^2*(4*x-1)^2). [Colin Barker, Dec 10 2012]
E.g.f.: exp(2*x)*(1 + 3*exp(2*x))*x. - Stefano Spezia, Aug 04 2022
EXAMPLE
a(2) = 14 because for P(A) = {{},{1},{2},{1,2}} |{} union {1}| = 1, |{} union {2}| = 1, |{} union {1,2}| = 2, |{1} union {2}| = 2, |{1} union {1,2}| = 2 and |{2} union {1,2}| = 2, |{} union {}| = 0, |{1} union {1}| = 1, |{2} union {2}| = 1, |{1,2} union {1,2}| = 2, which sums to 14.
MATHEMATICA
LinearRecurrence[{12, -52, 96, -64}, {0, 2, 14, 78}, 30] (* Harvey P. Dale, Jan 24 2019 *)
PROG
(Magma) [n*(2^(n-2) + 3*2^(2*n-3)): n in [0..30]]; // Vincenzo Librandi, Jun 10 2011
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Ross La Haye, Dec 30 2007, Jan 03 2008
STATUS
approved