

A070925


Number of subsets of A = {1,2,...,n} that have the same center of gravity as A, i.e., (n+1)/2.


9



1, 1, 3, 3, 7, 7, 19, 17, 51, 47, 151, 137, 471, 427, 1519, 1391, 5043, 4651, 17111, 15883, 59007, 55123, 206259, 193723, 729095, 688007, 2601639, 2465133, 9358943, 8899699, 33904323, 32342235, 123580883, 118215779, 452902071, 434314137, 1667837679, 1602935103
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OFFSET

1,3


COMMENTS

Also the number of nonempty subsets of {0..n} with mean n/2. The a(0) = 1 through a(5) = 7 subsets are:
{0} {0,1} {1} {0,3} {2} {0,5}
{0,2} {1,2} {0,4} {1,4}
{0,1,2} {0,1,2,3} {1,3} {2,3}
{0,2,4} {0,1,4,5}
{1,2,3} {0,2,3,5}
{0,1,3,4} {1,2,3,4}
{0,1,2,3,4} {0,1,2,3,4,5}
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LINKS



FORMULA

(End)


EXAMPLE

Of the 32 (2^5) sets which can be constructed from the set A = (1,2,3,4,5} only the sets {3}, {2, 3, 4}, {2, 4}, {1, 2, 4, 5}, {1, 2, 3, 4, 5}, {1, 3, 5}, {1, 5} give an average of 3.


MATHEMATICA

Needs["DiscreteMath`Combinatorica`"]; f[n_] := Block[{s = Subsets[n], c = 0, k = 2}, While[k < 2^n + 1, If[ (Plus @@ s[[k]]) / Length[s[[k]]] == (n + 1)/2, c++ ]; k++ ]; c]; Table[ f[n], {n, 1, 20}]
(* second program *)
Table[Length[Select[Subsets[Range[0, n]], Mean[#]==n/2&]], {n, 0, 10}] (* Gus Wiseman, Apr 15 2023 *)


CROSSREFS

Including the empty set gives A222955.
A327481 counts subsets by integer mean.


KEYWORD

nonn


AUTHOR

Sharon Sela (sharonsela(AT)hotmail.com), May 20 2002


EXTENSIONS



STATUS

approved



