OFFSET
1,2
COMMENTS
The general formula where each entry is chosen from the subsets of {1,..,k} is (2^n-1)^k. This may be shown by exhibiting a bijection to a set whose cardinality is obviously (2^n-1)^k, namely the set of all k-tuples with each entry chosen from the 2^n-1 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
R. P. Stanley, Enumerative Combinatorics, Volume 1, Wadsworth & Brooks 1986 p. 11.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (15,-70,120,-64).
FORMULA
a(n) = (2^n-1)^3.
G.f.: x*(8*x^2+12*x+1)/((x-1)*(2*x-1)*(4*x-1)*(8*x-1)). [Colin Barker, Nov 17 2012]
EXAMPLE
a(1)=(2^1-1)^3=1 because only one tuple of length one, namely ({}) has an empty intersection of its sole entry.
a(2)=27 because the valid 2-tuples are: ({},{}), ({},{1}), ({},{2}), ({},{3}), ({},{1,2}), ({},{1,3}), ({},{2,3}), ({},{1,2,3}), ({1},{}), ({2},{}), ({3},{}), ({1,2},{}), ({1,3},{}), ({2,3},{}), ({1,2,3},{}), ({1},{2}), ({1},{3}), ({1},{2,3}), ({2},{1}), ({2},{3}), ({2},{1,3}), ({3},{1}), ({3},{2}), ({3},{1,2}), ({1,2},{3}), ({1,3},{2}), ({2,3},{1})
MAPLE
for k from 1 to 20 do (2^k-1)^3; od;
MATHEMATICA
Table[(2^n - 1)^3, {n, 30}] (* Vincenzo Librandi, Mar 04 2018 *)
PROG
(Magma) [(2^n-1)^3: n in [1..20]]; // Vincenzo Librandi, Mar 04 2018
(PARI) a(n) = (2^n-1)^3; \\ Altug Alkan, Mar 04 2018
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Peter C. Heinig (algorithms(AT)gmx.de), Apr 13 2007
STATUS
approved