A348257 Number of ways we can write [n] as the union of 2 sets of sizes i, j which intersect in exactly 2 elements (2 < i,j < n; i = j allowed).
6, 30, 105, 315, 868, 2268, 5715, 14025, 33726, 79794, 186277, 429975, 982920, 2228088, 5013351, 11206485, 24903490, 55050030, 121110297, 265289475, 578813676, 1258290900, 2726297275, 5888802465, 12683574918, 27246198378, 58384711245, 124822486575, 266287971856, 566935682544
Offset: 4
Examples
a(4) = 6 since we can write [4] as the following unions: {1,2,3} U {1,2,4}, {1,2,3} U {1,3,4}, {1,2,3} U {2,3,4}, {1,2,4} U {1,3,4}, {1,2,4} U {2,3,4}, {1,3,4} U {2,3,4}.
Links
- Index entries for linear recurrences with constant coefficients, signature (9,-33,63,-66,36,-8).
Programs
-
Mathematica
nterms=50;Table[Binomial[n,2]*StirlingS2[n-2,2],{n,4,nterms+3}] (* Paolo Xausa, Nov 20 2021 *)
Formula
a(n) = (Sum_{j=3..n/2} binomial(n,j)*binomial(j,2)) + (1/2)*binomial(n,n/2+1) * binomial(n/2+1,2), if n is even.
a(n) = Sum_{j=3..ceiling(n/2)} binomial(n,j)*binomial(j,2), if n is odd.
G.f.: x^4*(6 - 24*x + 33*x^2 - 18*x^3 + 4*x^4)/((1 - x)^3*(1 - 2*x)^3). - Stefano Spezia, Oct 09 2021
a(n) = A000554(n)/2. - Enrique Navarrete, Nov 16 2021
a(n) = binomial(n,2) * Stirling2(n-2,2). - Alois P. Heinz, Nov 16 2021
Comments