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

1, 2, 4, 8, 16, 32, 104, 321, 1010, 3056, 9477, 29437, 91060, 286574, 919633, 2967499, 10043936, 40000426

Offset: 1

Author

Jon E. Schoenfield, Aug 01 2015

Keywords

• nonn
• more
• hard

Comments

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?

Examples

			{1} has only 1 nonempty subset, {1}; its mean is 1.
{1,2} has 3 nonempty subsets, {1}, {2}, and {1,2}; their means are 1, 2, and 3/2, respectively.
{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.
{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.
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.
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.
		

Links

• Manfred Scheucher, Sage Script
• Manfred Scheucher, Sage Script#2

Crossrefs

Cf. A259544.

Programs

Python

from copy import copy
from fractions import Fraction
from itertools import chain, combinations
def powerset(s):
  return chain.from_iterable(combinations(s, r) for r in range(len(s)+1))
def distinct_means(means, lst, t):
  newmeans = copy(means)
  for subset in powerset(lst):
    sm = Fraction(t+sum(subset), len(subset)+1)
    if sm in newmeans: return False, means
    else: newmeans.add(sm)
  return True, newmeans
def aupto(n):
  largest = 0
  alst = []
  prevmeans = set()
  for k in range(n):
    t = largest + 1
    passes, means = distinct_means(prevmeans, alst, t)
    while not passes:
      t += 1
      passes, means = distinct_means(prevmeans, alst, t)
    alst.append(t)
    largest = t
    prevmeans = means
  return alst
print(aupto(10)) # Michael S. Branicky, Jan 02 2021

Extensions

a(11)-a(13) from Manfred Scheucher, Aug 04 2015

a(14)-a(15) from Manfred Scheucher, Aug 09 2015

a(16)-a(17) from Michael S. Branicky, Aug 05 2023

a(18) from Michael S. Branicky, Aug 22 2023