cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A218384 Number of nonempty subsets S of the powerset of a set of size n that have the even intersection property.

Original entry on oeis.org

1, 7, 71, 3071, 1966207, 270499994623, 2342736474457787596799, 86772003564839307784895323681111305093119, 59169757600268575861444773339439520883460632949720404019392912099891777942585343
Offset: 1

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Author

Michel Marcus, Oct 27 2012

Keywords

Comments

If A is a set, an element S of P(P(A)) \ {{}} has the even intersection property (eip) if there exists a set B (necessarily nonempty) included in A with |B∩S| even for each s in S.
For instance for S={{},{1}} of A={1,2}, we can take B={2}, and then |{}∩{2}|=0 (even) and |{1}∩{2}|=0 (even), so S has eip.

Examples

			For |A|=2, A = {1,2} and P(A) = {{}, {1}, {2}, {1,2}}
S can be
  {{}, {1}, {2}, {1,2}}
  {{}, {1}, {2}}
  {{}, {1}, {1,2}}
  {{}, {2}, {1,2}}
  {{1}, {2}, {1,2}}
  {{}, {1}} has eip, with B={2}
  {{}, {2}} has eip, with B={1}
  {{}, {1,2}} has eip, with B={1,2}
  {{1}, {1,2}}
  {{2}, {1,2}}
  {{1}, {2}}
  {{}} has eip, with B={1,2}
  {{1}} has eip, with B={2}
  {{2}} has eip, with B={1}
  {{1,2}} has eip, with B={1,2}
So we have 7 S with eip.
		

Crossrefs

Programs

  • Maple
    A218384:=n->1+2*add((-1)^(n-i-1)*(2^(2^i-1)-1)* product((2^(n-j+1)-1)/(2^j-1), j=1..i)*2^binomial(n-i, 2), i=0..n-1): seq(A218384(n), n=1..10); # Wesley Ivan Hurt, Dec 11 2015
  • Mathematica
    Table[1 + 2 Sum[((-1)^(n - i - 1)) (2^(2^i - 1) - 1) Product[(2^(n - j + 1) - 1)/(2^j - 1), {j, 1, i}] 2^Binomial[n - i, 2], {i, 0, n - 1}], {n, 9}] (* Michael De Vlieger, Dec 11 2015 *)
  • PARI
    e(m) = {for (n=1, m, v = 1+2*sum(i=0, n-1, ((-1)^(n-i-1))*(2^(2^i-1)-1)* prod(j=1,i,(2^(n-j+1)-1)/(2^j-1))*2^binomial(n-i,2));print1(v, ", "););}

Formula

a(n) = 1 + 2*Sum_{i=0..n-1} (-1)^(n-i-1)*(2^(2^i-1)-1)*(Product_{j=1..i} (2^(n-j+1)-1)/(2^j-1)) * 2^binomial(n-i,2).
a(n) ~ 2^(n + 2^(n-1)). - Vaclav Kotesovec, Apr 03 2021
Apparently, a(n) = A051179(n) - A305737(n). - Tilman Piesk, Jan 26 2025
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