

A363764


a(1)=1, and thereafter, a(n) is the number of terms in the sequence thus far that appear with a frequency not equal to that of a(n1).


1



1, 0, 0, 1, 0, 2, 5, 5, 4, 7, 7, 5, 6, 10, 10, 9, 12, 12, 10, 10, 16, 16, 14, 18, 18, 15, 20, 20, 16, 20, 18, 16, 24, 26, 26, 27, 28, 28, 28, 24, 30, 33, 33, 31, 35, 35, 32, 37, 37, 33, 32, 35, 31, 37, 30, 39, 48, 48, 40, 50, 50, 41, 52, 52, 42, 54, 54, 43, 56, 56, 44, 58, 58
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OFFSET

1,6


COMMENTS

The first time a number appears for the 8th time is a(1491228545) = 953950324.  Pontus von Brömssen, Jul 06 2023


LINKS



EXAMPLE

a(8)=5 occurs two times, so a(9) is the number of terms which do not occur two times, which is 4 (there are three 0s and one 2).


MATHEMATICA

a[1] = 1; a[n_] := a[n] = Total[Select[Tally[v = Array[a, n  1]][[;; , 2]], # != Count[v, a[n  1]] &]]; Array[a, 100] (* Amiram Eldar, Jun 30 2023 *)


PROG

(Python)
from itertools import count, islice
from collections import defaultdict
x = 1
freq = defaultdict(int)
freq[x] = f0 = 1
freqfreq = defaultdict(int)
freqfreq[1] = 1
for n in count(1):
yield x
x = nf0*freqfreq[f0]
freq[x] = f0 = freq[x]+1
if f0 != 1: freqfreq[f01] = 1
freqfreq[f0] += 1
(MATLAB)
s = zeros(1, max_n); a = 1; s(2) = 1;
for n = 2:max_n
a(n) = length(find(s(a+1)~=s(a(n1)+1)));
s(a(n)+1) = s(a(n)+1)+1;
end


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



