A350768 -id:A350768 - cp's OEIS frontend

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(n-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

Offset: 1

Neal Gersh Tolunsky, Jun 28 2023

nonn

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

			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).

Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10000
Thomas Scheuerle, Scatterplot of a(n)/n for the first 100000 values. It is remarkable that we see extreme values a(n)/n > 0.99 for n = {1, 20957, 22061, 24915, ...}.
Neal Gersh Tolunsky, Graph of first 500000 terms

Cf. A350768, A363301.

function a = A363764( max_n )
    s = zeros(1,max_n); a = 1; s(2) = 1;
    for n = 2:max_n
        a(n) = length(find(s(a+1)~=s(a(n-1)+1)));
        s(a(n)+1) = s(a(n)+1)+1;
    end
end % Thomas Scheuerle, Jun 30 2023

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 *)

from itertools import count,islice
from collections import defaultdict
def A363764_gen():
    x = 1
    freq = defaultdict(int)
    freq[x] = f0 = 1
    freqfreq = defaultdict(int)
    freqfreq[1] = 1
    for n in count(1):
        yield x
        x = n-f0*freqfreq[f0]
        freq[x] = f0 = freq[x]+1
        if f0 != 1: freqfreq[f0-1] -= 1
        freqfreq[f0] += 1
def A363764_list(nmax):
    return list(islice(A363764_gen(), nmax)) # Pontus von Brömssen, Jul 01 2023 (after an idea by Kevin Ryde)

A363301 a(1)=1, and thereafter, a(n) is the number of terms in the sequence thus far that appear with a frequency equal to or less than that of a(n-1).

Original entry on oeis.org

1, 1, 2, 1, 4, 2, 3, 2, 8, 3, 4, 5, 2, 13, 3, 11, 4, 13, 5, 6, 3, 21, 4, 23, 5, 13, 14, 6, 7, 6, 18, 7, 8, 9, 6, 35, 7, 21, 10, 7, 40, 8, 22, 9, 12, 9, 26, 10, 13, 49, 10, 27, 11, 14, 15, 10, 56, 11, 30, 12, 17, 12, 34, 13, 64, 14, 37, 15, 18, 19, 14, 66, 15, 40, 20, 15, 71, 16, 17, 24, 17, 44, 18, 46, 19, 24

Offset: 1

Neal Gersh Tolunsky, May 26 2023

nonn

			a(3)=2 because a(2)=1 occurs with a frequency of 2 and there are two terms in the sequence thus far that appear with a frequency of 2 or less (1, 1).
a(6)=3 because a(5)=2 occurs with a frequency of 2 and there are three terms that appear with a frequency of 2 or less (2, 2, and 4).

Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10000
Neal Gersh Tolunsky, Graph of the first 100000 terms

Cf. A350768.

from itertools import islice
from collections import Counter
def agen(): # generator of terms
    an, c, freqs = 1, Counter(), Counter()
    while True:
        if an in c:
            freqs[c[an]] -= 1
        c[an] += 1
        freqs[c[an]] += 1
        yield an
        an = sum(v*freqs[v] for v in range(1, c[an]+1))
print(list(islice(agen(), 86))) # Michael S. Branicky, May 26 2023

A372596 a(1)=-1; thereafter a(n) is (the number of terms that appear with a different frequency from that of a(n-1)) minus (the number of terms that appear with the same frequency).

Original entry on oeis.org

-1, -1, -2, 1, 0, -1, 0, 3, 2, 1, 2, -1, 4, 7, 6, 5, 4, 1, 12, 7, 4, 9, 10, 9, 8, 11, 10, 7, 10, 5, 14, 17, 16, 15, 14, 15, 12, 9, 8, 11, 8, 5, 0, -5, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 36, 19, 34, 21, 32, 29, 26, 23, 20, 17, 16, 19

Offset: 1

Neal Gersh Tolunsky, May 05 2024

sign

			Let t'(n) be the number of terms that appear with a different frequency from that of a(n-1) and t(n) be the number of remaining terms.
The resulting a(n) = t'(n) - t(n) are then:
   n  t'(n)  t(n) a(n)
  --  ----  ----  ----
   1     *     *    -1
   2     0     1    -1
   3     0     2    -2
   4     2     1     1
   5     2     2     0
   6     2     3    -1
   7     3     3     0
   8     5     2     3
   9     5     3     2
  10     5     4     1

Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10000
Rémy Sigrist, PARI program.
Neal Gersh Tolunsky, Graph of first 200000 terms.

Cf. A372598, A350768, A363764.

PARI
```
\\ See Links section.
```

A372598 a(1)=-2; thereafter a(n) is (the number of terms thus far that appear with a different frequency from that of a(n-1)) minus (the number of terms that appear with the same frequency).

Original entry on oeis.org

-2, -1, -2, -1, -4, 3, 2, 1, 0, -1, 4, -1, 4, 5, 2, 3, 0, -3, 10, 9, 8, 7, 6, 5, 0, 19, 8, 3, 16, 11, 10, 7, 4, 15, 16, 7, 12, 19, 10, 9, 12, 9, 6, 11, 8, 3, 30, 37, 36, 35, 34, 33, 32, 31, 30, 19, 14, 33, 22, 35, 20, 37, 18, 39, 38, 37, 18, 23, 40, 39, 22, 19

Offset: 1

Neal Gersh Tolunsky, May 06 2024

sign

			Let t'(n) be the number of terms that appear with a different frequency from that of a(n-1) and t(n) be the number of remaining terms.
  The resulting a(n) = t'(n) - t(n) are then:
   n  t'(n)  t(n) a(n)
  --  ----  ----  ----
   1     *     *    -2
   2     0     1    -1
   3     0     2    -2
   4     1     2    -1
   5     0     4    -4
   6     4     1     3
   7     4     2     2
   8     4     3     1
   9     4     4     0
  10     4     5    -1

Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10000
Rémy Sigrist, PARI program.
Neal Gersh Tolunsky, Graph of first 200000 terms.

Cf. A372596, A350768, A363764.

PARI
```
\\ See Links section.
```

A359010 Variant of the inventory sequence: Record the number of terms whose value occurs once thus far in the sequence, then the number of terms whose value occurs twice thus far, and so on; a row ends when a 0 that would repeat infinitely is reached.

Original entry on oeis.org

0, 1, 0, 1, 4, 0, 1, 0, 3, 4, 0, 1, 2, 0, 4, 0, 2, 2, 6, 4, 0, 2, 0, 0, 12, 0, 3, 2, 0, 8, 5, 0, 4, 2, 0, 4, 0, 12, 0, 3, 2, 3, 8, 0, 6, 7, 0, 2, 6, 3, 4, 5, 0, 7, 8, 0, 0, 6, 3, 8, 0, 6, 7, 8, 0, 0, 4, 3, 4, 10, 0, 7, 8, 9, 0, 2, 4, 0, 8, 5, 0, 14, 0, 9, 10, 0

Offset: 1

Neal Gersh Tolunsky, Dec 11 2022

Note that we are counting terms with repetition. For example, to find a(5)=4, we are looking for the number of terms that appear twice. 0 and 1 each occur twice, which is 2+2=4 (not 1+1=2). This means that each column contains only multiples of the number of occurrences it is counting.

A row ends when a 0 is reached as the k-th term in a row and the only value left occurring greater than or equal to k times is 0. - Neal Gersh Tolunsky, Feb 08 2025

			First few rows of irregular triangle:
  0;
  1,  0;
  1,  4,  0;
  1,  0,  3,  4,  0;
  1,  2,  0,  4,  0;
  2,  2,  6,  4,  0;
  2,  0,  0, 12,  0;
  3,  2,  0,  8,  5,  0;
  4,  2,  0,  4,  0, 12,  0;
  3,  2,  3,  8,  0,  6,  7,  0;
  2,  6,  3,  4,  5,  0,  7,  8,  0;

Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10073 (first 136 rows)

Cf. A342585, A350768.

from collections import Counter
from itertools import count, islice
def end_cond(I, k): # the only value left occurring >= k times is 0
    return I[0] >= k and not any(I[i] >= k for i in I if i > 0)
def agen(): # generator of terms
    I = Counter()
    while True:
        for i in count(1):
            c = sum(v for v in I.values() if v==i)
            yield c
            I[c] += 1
            if c == 0 and end_cond(I, i):
                break
print(list(islice(agen(), 86))) # Michael S. Branicky, Jan 28 2025

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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(n-1).

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A363301 a(1)=1, and thereafter, a(n) is the number of terms in the sequence thus far that appear with a frequency equal to or less than that of a(n-1).

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A372596 a(1)=-1; thereafter a(n) is (the number of terms that appear with a different frequency from that of a(n-1)) minus (the number of terms that appear with the same frequency).

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A372598 a(1)=-2; thereafter a(n) is (the number of terms thus far that appear with a different frequency from that of a(n-1)) minus (the number of terms that appear with the same frequency).

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A359010 Variant of the inventory sequence: Record the number of terms whose value occurs once thus far in the sequence, then the number of terms whose value occurs twice thus far, and so on; a row ends when a 0 that would repeat infinitely is reached.

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Python