OFFSET
1,2
COMMENTS
In other words, if a(n_1) = ... = a(n_k) with n_1 < ... < n_k then 1/n_1 + ... + 1/n_k <= 1.
Each positive integer appears in the sequence, a finite number of times. This is a consequence of the fact that the greedy algorithm for Egyptian fractions terminates in a finite number of steps for any rational starting value.
LINKS
EXAMPLE
The first terms, alongside the sums of the inverses of the indices so far where the sequence has the same value, are:
n a(n) Sums
-- ---- ---------
1 1 1
2 2 1/2
3 2 5/6
4 3 1/4
5 3 9/20
6 2 1
7 3 83/140
8 3 201/280
9 3 2089/2520
10 3 2341/2520
PROG
(PARI) { b = vector(6); for (n = 1, 87, for (v = 1, oo, if (b[v] + 1/n <= 1, b[v] += 1/n; print1 (v", "); break; ); ); ); }
(Python)
from fractions import Fraction
from itertools import count, islice
from collections import defaultdict
def agen(): # generator of terms
invsum, mink = defaultdict(int), 1
for n in count(1):
an = next(k for k in count(mink) if invsum[k] + Fraction(1, n) <= 1)
yield an
invsum[an] += Fraction(1, n)
while invsum[mink] == 1: mink += 1
print(list(islice(agen(), 87))) # Michael S. Branicky, Aug 31 2024
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Rémy Sigrist, Aug 29 2024
STATUS
approved