A377351 Lexicographically earliest sequence of positive integers such that the means of consecutive terms are all distinct.

1, 2, 4, 7, 5, 10, 12, 18, 13, 16, 28, 23, 14, 25, 48, 32, 17, 46, 30, 45, 67, 60, 27, 71, 39, 85, 68, 99, 78, 44, 102, 87, 126, 57, 118, 69, 121, 74, 125, 119, 112, 136, 107, 110, 170, 120, 175, 142, 194, 75, 222, 152, 164, 180, 177, 184, 188, 135, 255, 210

Offset

1,2

Comments

In other words, for any distinct nonempty intervals t..u and v..w, Sum_{i = t..u} a(i)/(u-t+1) <> Sum_{j = v..w} a(j)/(w-v+1).

This sequence corresponds essentially to the first differences of A033808.

By necessity, all terms are distinct.

Links

Rémy Sigrist, Table of n, a(n) for n = 1..10000

Rémy Sigrist, C++ program

Formula

a(n) = A033808(n) - A033808(n-1).

Example

The first terms, alongside the means of consecutive terms ending with a(n), are:
  n  a(n)  Corresponding means
  -  ----  ------------------------------------------
  1     1  1
  2     2  3/2, 2
  3     4  7/3, 3, 4
  4     7  7/2, 13/3, 11/2, 7
  5     5  19/5, 9/2, 16/3, 6, 5
  6    10  29/6, 28/5, 13/2, 22/3, 15/2, 10
  7    12  41/7, 20/3, 38/5, 17/2, 9, 11, 12
  8    18  59/8, 58/7, 28/3, 52/5, 45/4, 40/3, 15, 18

Prog

(C++) // See Links section.
(Python)
from fractions import Fraction
from itertools import count, islice
def agen(): # generator of terms
    alst, means_seen = [1], {1}
    while True:
        yield alst[-1]
        for k in count(1):
            if k in means_seen: continue
            mk, failed, sk = {k}, False, k
            for j in range(1, len(alst)+1):
                sk += alst[-j]
                m = Fraction(sk, j+1)
                if m in means_seen or m in mk: failed = True; break
                mk.add(m)
            if not failed: break
        means_seen |= mk
        alst.append(k)
print(list(islice(agen(), 60))) # Michael S. Branicky, Oct 26 2024, Oct 28 2024

Crossrefs

Cf. A033808, A377388.

Sequence in context: A035311 A182310 A244591 * A299324 A261076 A302991

Adjacent sequences: A377348 A377349 A377350 * A377352 A377353 A377354

Keyword

nonn

Author

Rémy Sigrist, Oct 26 2024

Status

approved