login
The OEIS is supported by the many generous donors to the OEIS Foundation.

 

Logo
Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A341590 a(n) = (Sum_{j=1..3} StirlingS1(3,j)*(2^j-1)^n)/3!. 2
0, 0, 4, 44, 360, 2680, 19244, 136164, 957520, 6715760, 47049684, 329465884, 2306615480, 16147371240, 113034787324, 791253077204, 5538800238240, 38771687761120, 271402072608164, 1899815283098124, 13298709306209800 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,3
COMMENTS
Number of 3-element subsets of the powerset P([n]) such that their union is equal to the universe [n] = {1,2, ... ,n}.
StirlingS1(k,j) is a signed Stirling number of the first kind (cf. A048994).
In general, the number of k-element subsets of P([n]) such that their union is equal to [n] is (Sum_{j=0..k} StirlingS1(k,j)*(2^j-1)^n)/k!. That can be expressed also as (-1)^n*(Sum_{j=0..n} binomial(n,j)*(-1)^j*binomial(2^j,k)). See the below link to Mathematics Stack Exchange for proofs. The case k = 2 is A003462.
LINKS
FORMULA
a(n) = (Sum_{j=1..3} A048994(3,j)*(2^j-1)^n)/3!.
a(n) = (2 - 3^(1+n) + 7^n)/6.
a(n) = (-1)^n*(Sum_{j=0..n} binomial(n,j)*(-1)^j*binomial(2^j,3)).
G.f.: 4*x^2/(1 - 11*x + 31*x^2 - 21*x^3). - Stefano Spezia, Feb 15 2021
a(n) = 4 * A016212(n-2) for n >= 2. - Alois P. Heinz, Feb 15 2021
EXAMPLE
For n = 2 and [n] = [2] = {1,2} the a(2) = 4 solutions are {{},{1},{2}}, {{},{1},{1,2}}, {{},{2},{1,2}}, {{1},{2},{1,2}}.
CROSSREFS
Sequence in context: A035014 A259989 A030987 * A297384 A178294 A043039
KEYWORD
nonn,easy
AUTHOR
Fabio Visonà, Feb 15 2021
STATUS
approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified March 28 16:58 EDT 2024. Contains 371254 sequences. (Running on oeis4.)