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A260873 Lexicographically first sequence of positive integers, every nonempty subset of which has a distinct mean.

%I A260873 #36 Aug 22 2023 17:38:49
%S A260873 1,2,4,8,16,32,104,321,1010,3056,9477,29437,91060,286574,919633,
%T A260873 2967499,10043936,40000426
%N A260873 Lexicographically first sequence of positive integers, every nonempty subset of which has a distinct mean.
%C A260873 The seeming pattern a(n) = 2^(n-1) is broken at a(7)=104. Can the value of lim_{n->inf.} a(n)/a(n-1) be determined?
%H A260873 Manfred Scheucher, <a href="/A260873/a260873.sage.txt">Sage Script</a>
%H A260873 Manfred Scheucher, <a href="/A260873/a260873_1.sage.txt">Sage Script#2</a>
%e A260873 {1} has only 1 nonempty subset, {1}; its mean is 1.
%e A260873 {1,2} has 3 nonempty subsets, {1}, {2}, and {1,2}; their means are 1, 2, and 3/2, respectively.
%e A260873 {1,2,3} has 7 nonempty subsets, not all of which have distinct means: {2}, {1,3}, and {1,2,3} all have a mean of 2. Therefore, a(3) > 3.
%e A260873 {1,2,4} has 7 nonempty subsets, {1}, {2}, {4}, {1,2}, {1,4}, {2,4} and {1,2,4}, all of which have distinct means, so a(3)=4.
%e A260873 For the set {1,2,4,5}, the subsets {1,5} and {2,4} have the same mean; for {1,2,4,6}, {4} and {2,6} have the same mean; and for {1,2,4,7}, {4} and {1,7} have the same mean; but all nonempty subsets of {1,2,4,8} are distinct, so a(4)=8.
%e A260873 For each k in 9 <= k <= 15, there are at least two subsets of {1,2,4,8,k} having the same mean, but all nonempty subsets of {1,2,4,8,16} have distinct means, so a(5)=16.
%o A260873 (Python)
%o A260873 from copy import copy
%o A260873 from fractions import Fraction
%o A260873 from itertools import chain, combinations
%o A260873 def powerset(s):
%o A260873   return chain.from_iterable(combinations(s, r) for r in range(len(s)+1))
%o A260873 def distinct_means(means, lst, t):
%o A260873   newmeans = copy(means)
%o A260873   for subset in powerset(lst):
%o A260873     sm = Fraction(t+sum(subset), len(subset)+1)
%o A260873     if sm in newmeans: return False, means
%o A260873     else: newmeans.add(sm)
%o A260873   return True, newmeans
%o A260873 def aupto(n):
%o A260873   largest = 0
%o A260873   alst = []
%o A260873   prevmeans = set()
%o A260873   for k in range(n):
%o A260873     t = largest + 1
%o A260873     passes, means = distinct_means(prevmeans, alst, t)
%o A260873     while not passes:
%o A260873       t += 1
%o A260873       passes, means = distinct_means(prevmeans, alst, t)
%o A260873     alst.append(t)
%o A260873     largest = t
%o A260873     prevmeans = means
%o A260873   return alst
%o A260873 print(aupto(10)) # _Michael S. Branicky_, Jan 02 2021
%Y A260873 Cf. A259544.
%K A260873 nonn,more,hard
%O A260873 1,2
%A A260873 _Jon E. Schoenfield_, Aug 01 2015
%E A260873 a(11)-a(13) from _Manfred Scheucher_, Aug 04 2015
%E A260873 a(14)-a(15) from _Manfred Scheucher_, Aug 09 2015
%E A260873 a(16)-a(17) from _Michael S. Branicky_, Aug 05 2023
%E A260873 a(18) from _Michael S. Branicky_, Aug 22 2023