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A134045 Let P(A) be the power set of an n-element set A. Then a(n) = the number of pairs of elements {x,y} of P(A) for which either 0) x and y are disjoint and for which either x is a subset of y or y is a subset of x, or 1) x and y are intersecting but for which x is not a subset of y and y is not a subset of x, or 2) x = y. 0

%I #18 Nov 30 2015 08:36:20

%S 1,3,7,18,61,258,1177,5358,23821,103338,439297,1838598,7605781,

%T 31191618,127100617,515462238,2083142941,8396683098,33779525137,

%U 135697396278,544529307301,2183340065778,8749036112857,35043186680718,140313902770861,561679137947658,2247987249823777,8995761328275558

%N Let P(A) be the power set of an n-element set A. Then a(n) = the number of pairs of elements {x,y} of P(A) for which either 0) x and y are disjoint and for which either x is a subset of y or y is a subset of x, or 1) x and y are intersecting but for which x is not a subset of y and y is not a subset of x, or 2) x = y.

%H Ross La Haye, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL12/LaHaye/lahaye5.html">Binary Relations on the Power Set of an n-Element Set</a>, Journal of Integer Sequences, Vol. 12 (2009), Article 09.2.6. [_Ross La Haye_, Feb 22 2009]

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (10,-35,50,-24).

%F a(n) = (1/2)(4^n - 3^(n+1) + 7*2^n - 3) = 3*StirlingS2(n+1,4) + 2*StirlingS2(n+1,2) + 1.

%F G.f.: (1-7*x+12*x^2+3*x^3)/((1-x)*(1-2*x)*(1-3*x)*(1-4*x)). [_Colin Barker_, Jul 29 2012]

%e a(2) = 7 because for P(A) = {{},{1},{2},{1,2}} we have for case 0 {{},{1}}, {{},{2}}, {{},{1,2}} and we have for case 2 {{},{}}, {{1},{1}}, {{2},{2}}, {{1,2},{1,2}}. There are 0 {x,y} of P(A) in this example that fall under case 1.

%t Table[3 StirlingS2[n + 1, 4] + 2 StirlingS2[n + 1, 2] + 1, {n, 0, 27}] (* _Michael De Vlieger_, Nov 30 2015 *)

%o (PARI) a(n) = (4^n - 3^(n+1) + 7*2^n - 3)/2; \\ _Michel Marcus_, Nov 30 2015

%Y Cf. A000225, A032263, A000079.

%K nonn,easy

%O 0,2

%A _Ross La Haye_, Jan 11 2008

%E More terms from _Michael De Vlieger_, Nov 30 2015

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