

A128832


Number of ntuples where each entry is chosen from the subsets of {1,2,3,4} such that the intersection of all n entries is empty.


2



1, 81, 2401, 50625, 923521, 15752961, 260144641, 4228250625, 68184176641, 1095222947841, 17557851463681, 281200199450625, 4501401006735361, 72040003462430721, 1152780773560811521, 18445618199572250625
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OFFSET

1,2


COMMENTS

The general formula where each entry is chosen from the subsets of {1,...,k} is (2^n1)^k. This may be shown by exhibiting a bijection to a set whose cardinality is obviously (2^n1)^k, namely the set of all ktuples with each entry chosen from the 2^n1 proper subsets of {1,...,n}, i.e., for of the k entries {1,...,n} is forbidden. The bijection is given by (X_1,...,X_n) > (Y_1,...,Y_k) where for each j in {1,...,k} and each i in {1,...,n}, i is in Y_j if and only if j is in X_i. Sequence A060867 is the case where the entries are chosen from subsets of {1,2}.


REFERENCES

Stanley, R. P.: Enumerative Combinatorics: Volume 1: Wadsworth & Brooks: 1986: p. 11


LINKS



FORMULA

a(n) = (2^n  1)^4.
G.f.: x*(4*x+1)*(16*x^2+46*x+1)/((x1)*(2*x1)*(4*x1)*(8*x1)*(16*x1)). [Colin Barker, Nov 17 2012]


EXAMPLE

a(1) = (2^1  1)^4 = 1 because only one tuple of length one, namely ({}), has an empty intersection of its sole entry.


MAPLE

for k from 1 to 20 do (2^k1)^4; od;
with (combinat):seq(mul(stirling2(n, 2), k=1..4), n=2..17); # Zerinvary Lajos, Dec 16 2007


MATHEMATICA

LinearRecurrence[{31, 310, 1240, 1984, 1024}, {1, 81, 2401, 50625, 923521}, 20] (* Harvey P. Dale, Mar 30 2019 *)


CROSSREFS



KEYWORD

easy,nonn


AUTHOR

Peter C. Heinig (algorithms(AT)gmx.de), Apr 13 2007


STATUS

approved



