Clément Vovard - cp's OEIS frontend

User: Clément Vovard

Clément Vovard has authored 3 sequences.

A358338 a(n) = abs(a(n-1) - count(a(n-1))) where count(a(n-1)) is the number of times a(n-1) has appeared so far in the sequence, a(1)=0.

0, 1, 0, 2, 1, 1, 2, 0, 3, 2, 1, 3, 1, 4, 3, 0, 4, 2, 2, 3, 1, 5, 4, 1, 6, 5, 3, 2, 4, 0, 5, 2, 5, 1, 7, 6, 4, 1, 8, 7, 5, 0, 6, 3, 3, 4, 2, 6, 2, 7, 4, 3, 5, 1, 9, 8, 6, 1, 10, 9, 7, 3, 6, 0, 7, 2, 8, 5, 2, 9, 6, 1, 11, 10, 8, 4, 4, 5, 3, 7, 1, 12, 11, 9, 5, 4

Offset: 1

Clément Vovard, Nov 10 2022

This sequence is related to the inventory sequence (A342585) as it uses the number of times a number has occurred so far in the sequence.
The following comments are only empirical observations:
ceiling(sqrt(2n)) is an excellent envelope of a(n) with no exceptions found in the first 50000 terms.
When x > 3 appears for the first time, it seems to always be preceded by a 1 and followed by x-1. Also, x-1 will already have occurred earlier in the sequence (new highest terms grow by 1).
The number of times x > 0 appears in the first k terms seems to approximately equal sqrt(2k)-x-1. Therefore, 1 appears approximately sqrt(2k) times. The highest term that has appeared in k terms is then approximately sqrt(2k), which also makes sense considering the number of times 1 appears and the fact that a new number is preceded by 1. The only exception is 0, which appears approximately sqrt(2k)/2 times.

			For n=2, a(2-1)=0 and 0 has occurred 1 time so far so a(2)=abs(0-1)=1.
For n=12, a(12-1)=1 and 1 has occurred 4 times so far so a(12)=abs(1-4)=3.

Clément Vovard, Table of n, a(n) for n = 1..10000

Cf. A337835, A340488, A342585.

from collections import Counter
def aupton(terms):
    alst, inventory = [0], Counter([0])
    for n in range(2, terms+1):
        c = abs(alst[-1] - inventory[alst[-1]])
        alst.append(c); inventory[c] += 1
    return alst
print(aupton(85)) # Michael S. Branicky, Nov 10 2022

A343299 a(n) = n + A000120(a(n-1)) - a(n-1), with n > 1, a(1) = 1, where A000120(x) is the binary weight of x.

Original entry on oeis.org

1, 2, 2, 3, 4, 3, 6, 4, 6, 6, 7, 8, 6, 10, 7, 12, 7, 14, 8, 13, 11, 14, 12, 14, 14, 15, 16, 13, 19, 14, 20, 14, 22, 15, 24, 14, 26, 15, 28, 15, 30, 16, 28, 19, 29, 21, 29, 23, 30, 24, 29, 27, 30, 28, 30, 30, 31, 32

Offset: 1

Clément Vovard, Apr 11 2021

The 'crossings' that appear in the graph seem to occur when a(n) is a power of 2.

			a(2) = 2 + A000120(1) - 1 = 2 + 1 - 1 = 2.
a(6) = 6 + A000120(4) - 4 = 6 + 1 - 4 = 3.

Clément Vovard, Table of n, a(n) for n = 1..10000
Michael De Vlieger, Log-log scatterplot of a(n) for n = 1..2^16.
Michael De Vlieger, Log-log scatterplot of a(n) for n = 1..2^10 with color function registering binary weight of a(n-1).
Michael De Vlieger, Log-log scatterplot of a(n) for n = 1..2^10 with even n in red and odd n in blue.

Cf. A000120, A011371

a[1] = 1; a[n_] := a[n] = n + DigitCount[a[n - 1], 2, 1] - a[n - 1]; Array[a, 100] (* Amiram Eldar, Apr 12 2021 *)

lista(nn) = {my(va = vector(nn)); va[1] = 1; for (n=2, nn, va[n] = n + hammingweight(va[n-1]) - va[n-1];); va;} \\ Michel Marcus, Apr 12 2021

a(n) = n - A011371(a(n-1)).

A339141 a(n) = reverse(10*n - a(n-1)), with n>1, a(1) = 1.

Original entry on oeis.org

1, 91, -16, 65, -51, 111, -14, 49, 14, 68, 24, 69, 16, 421, -172, 233, -36, 612, -224, 424, -412, 236, -6, 642, -293, 355, -58, 833, -345, 546, -632, 259, 17, 323, 72, 882, -215, 595, -502, 209, 102, 813, -383, 328, 221, 932, -264, 447, 34, 664, -451

Offset: 1

Clément Vovard, Nov 25 2020

Note that for x<0, reverse(x) is defined by -1*reverse(-x).

Starting the sequence with other numbers also gives similar-looking graphs.

			For n = 2, 10*n = 10*2 = 20, 20 - a(n-1) = 20 - 1 = 19, reverse(19) = 91.
For n = 3, 10*n = 10*3 = 30, 30 - a(3-1) = 30 - 91 = -61, reverse(-61) = -16.

Clément Vovard, Table of n, a(n) for n = 1..10000

Cf. A004086, A166525.

a:= proc(n) option remember; `if`(n<2, n, (p-> signum(p)* (f->
      parse(cat(f[-i]$i=1..length(f))))(""||(abs(p))))(10*n-a(n-1)))
    end:
seq(a(n), n=1..60);  # Alois P. Heinz, Jan 06 2021

nmax=51; a[1]=1; a[n_]:=Sign[10n-a[n-1]]IntegerReverse[10n-a[n-1]]; Table[a[n],{n,nmax}] (* Stefano Spezia, Dec 05 2020 *)

rev(n) = sign(n)*fromdigits(Vecrev(digits(n)));
a(n) = if (n==1, 1, rev(10*n-a(n-1))); \\ Michel Marcus, Dec 05 2020

a(n) = reverse(10*n - a(n-1)) where reverse means reverse the order of the digits.

cp's OEIS Frontend

User: Clément Vovard

A358338 a(n) = abs(a(n-1) - count(a(n-1))) where count(a(n-1)) is the number of times a(n-1) has appeared so far in the sequence, a(1)=0.

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A343299 a(n) = n + A000120(a(n-1)) - a(n-1), with n > 1, a(1) = 1, where A000120(x) is the binary weight of x.

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A339141 a(n) = reverse(10*n - a(n-1)), with n>1, a(1) = 1.

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