A276457 -id:A276457 - cp's OEIS frontend

A276463 Position at which n first appears in A276457, with a(0)=0.

0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 120, 160, 158, 140, 337, 171, 189, 207, 165, 248, 230, 263, 122, 124, 126, 128, 130, 132, 134, 136, 138, 244, 757, 142, 144, 146, 148, 150, 152, 154, 156, 374, 725, 399, 279, 291, 556, 295, 423, 167, 169, 502, 429, 173, 175, 177, 179, 181, 183, 185, 187, 496, 508, 191, 193, 195, 197

Offset: 0

Yuriy Sibirmovsky, Sep 04 2016

The conjecture is that every number will appear in A276457, so this sequence should be infinite.
1187 doesn't appear in the first 100000 terms of A276457, thus the b-file ends with 1186.
The first 11 terms are successive even numbers (A005843).

Yuriy Sibirmovsky, Table of n, a(n) for n = 0..1186

Cf. A276457, A005843.

Nm=1000;
Aa=Table[0,{j,1,Nm}];
Aa[[3]]=1;
Do[B=Table[IntegerDigits[Aa[[l]]],{l,1,j-1}];
Aa[[j+1]]=SequenceCount[Flatten[B],IntegerDigits[Aa[[j]]]],{j,3,Nm-1}];
Table[Position[Aa,k][[1]]-1,{k,0,50}]

A181391 Van Eck's sequence: For n >= 1, if there exists an m < n such that a(m) = a(n), take the largest such m and set a(n+1) = n-m; otherwise a(n+1) = 0. Start with a(1)=0.

Original entry on oeis.org

0, 0, 1, 0, 2, 0, 2, 2, 1, 6, 0, 5, 0, 2, 6, 5, 4, 0, 5, 3, 0, 3, 2, 9, 0, 4, 9, 3, 6, 14, 0, 6, 3, 5, 15, 0, 5, 3, 5, 2, 17, 0, 6, 11, 0, 3, 8, 0, 3, 3, 1, 42, 0, 5, 15, 20, 0, 4, 32, 0, 3, 11, 18, 0, 4, 7, 0, 3, 7, 3, 2, 31, 0, 6, 31, 3, 6, 3, 2, 8, 33, 0, 9, 56, 0, 3, 8, 7, 19, 0, 5, 37, 0, 3, 8, 8, 1

Offset: 1

Jan Ritsema van Eck, Oct 17 2010, Oct 19 2010

The name "Van Eck's sequence" is due to N. J. A. Sloane, not the author!
Note that it is obvious from the definition that a(n) < n for all n. - N. J. A. Sloane, Jun 20 2019. Even stronger, a(n)+a(n+1) < n for all n, since the a(n+1) implies that a(n-a(n+1)) = a(n). - Jan Ritsema van Eck, Jun 30 2019
Starting with a number different from 0, for instance with 1, gives a different but similar sequence. See A171911-A171918 for examples.
Examination of the first 10^6 terms suggests that lim sup a(n)/n = 1. Cf. A171866/A171867. - David Applegate and N. J. A. Sloane, Oct 18 2010
From Jan Ritsema van Eck, Oct 25 2010: (Start)
Theorem: There are infinitely many zeros.
Proof: Suppose not. Then from a certain point on, no new terms appear, so the sequence is bounded and nonzero. Let M be the maximal term. Any block of length M determines the rest of the sequence.
But there are only M^M different blocks of length M containing the numbers 1 through M.
So a block must repeat, and so the sequence eventually becomes periodic. The periodic part does not contain any zeros.
Suppose the period has length p, and starts at term r, with a(r)=x, ..., a(r+p-1)=z, a(r+p)=x, ... There is another z after q <= p steps, which is immediately followed by q.
But this q implies that a(r-1)=z. Therefore the periodic part really began at step r-1.
Repeating this shows that the periodic part starts at a(1). But a(1)=0, and the periodic part cannot contain a 0. Contradiction. (End)
An alternative wording of the definition: For n>=1, if there exists an m < n such that a(m) = a(n), take the largest such m, otherwise take m = n; set a(n+1) = n-m. Start with a(1)=0. - Arie Bos, Dec 10 2010
Conjectures: (i) lim sup a(n)/n = 1; (ii) gaps between 0's are about log_10 n; (iii) every number eventually appears. - N. J. A. Sloane, in lecture "The OEIS: The Major Problems", Conference for 50th Anniversary of the OEIS, Rutgers University, Oct 10 2014. (Added Jun 16 2019.)
Conjecture: every pair of nonnegative integers (x,y) other than (1,1) and (x,x+1) for x>0 appear as consecutive entries (that is, a(i) = x, a(i+1) = y, for some i). - László Kozma, Aug 09 2016. Correction from Tomas Rokicki, Jun 19 2019: The pair (x,x+1) only occurs at (0,1), as it would imply distinct values x positions previously.
As mentioned in an earlier comment, for any k >= 0 there is a "van Eck" sequence E(k) starting with k and extended using the same rule (cf. A171911-A171918). The initial E(k)(1) = k is followed by at least k initial terms from this sequence A181391 = E(0): E(k)(n+1) = E(0)(n) for all n <= k and beyond, as long as E(0)(n) != k. - M. F. Hasler, Jun 11 2019
Comment from Jordan Linus, circa Jun 16 2019, from an online comment added to the Haran-Sloane video: (Start)
Theorem: limsup a(n)/sqrt(n) >= 1.
Proof: Whenever a(n)=0, either there have been sqrt(n) zeros in the sequence, thus sqrt(n) new distinct numbers (and at least one bigger than sqrt(n)), or there have been fewer than sqrt(n) zeros in the sequence, and thus there is a gap of at least sqrt(n) between two zeros (and the term after the second zero is at least sqrt(n)). So either way there is a term >= sqrt(n). (End)
The long-term behavior of the sequence E(k) appears to be the same for all k, although the individual numbers differ. Empirically modeled up to 2^25 terms of E(k) for k between 0 and 9. - Po-chia Chen, Jun 18 2019
After searching the first 318 billion entries, the smallest number not appearing is 645315850; the smallest pair not of the form (1,1), (0,a), or (x,x+1) not appearing is (268,5). - Tomas Rokicki, Jun 19 2019
Theorem: i-a(i) is unique for all i, a(i)>0. Put differently: i-a(i)<>j-a(j) for all i,j, a(i)>0 and j>i. Proof: if a(i-1)=a(j-1)=x, a(j) can be at most j-i because there is by definition not another x between a(j-1) and a(j-a(j)-1). If a(i-1)<>a(j-1), it follows that a(i-a(i)-1)<>a(j-a(j)-1). Either way i-a(i)<>j-a(j). A special case of this is that pairs (x,x+1) cannot occur for x>0 (as remarked above); similarly, triples (x,y,x+2) cannot occur and so on. Also, since a(i-a(i+1))=a(i) by definition, i-a(i+1)-a(i) is unique for all i, a(i+1) and a(i)>0. A simple example is that triples (x,y,x+1) cannot occur for y>0. Many other "impossible patterns" can be derived. - Jan Ritsema van Eck, Jul 22 2019
After 10^12 terms, the smallest number not appearing is 1732029957; the smallest pair not of the form (1,1), (0,a), or (x,x+1) not appearing is (528,5); there have been 90689534032 zeros. - Benjamin Chaffin, Sep 11 2019
Similar to E(k) above, the sequence E(k,l,...,m) could be defined as the sequence starting with k,l,...,m and continuing using Van Eck's rule. For example, E(1,1) is 1,1,1,... and has period 1. The smallest possible period greater than 1 is 42, attained by E(37, 42, 7, 42, 2, 5, 22, 42, 4, 11, 42, 3, 21, 42, 3, 3, 1, 25, 38, 42, 6, 25, 4, 14, 42, 5, 20, 42, 3, 13, 42, 3, 3, 1, 17, 36, 42, 6, 17, 4, 17, 2). More info: https://redd.it/dbdhpj - Michiel De Muynck, Sep 30 2019
If the rule a(n+1)=0 (when a(n) has not been seen before) is replaced by a(n+1)=a(a(n)) and a(0) is set to 0, then the resulting sequence is A025480. - David James Sycamore, Nov 01 2019

			We start with a(1) = 0. 0 has not appeared before, so the rule says a(2) = 0. Now 0 HAS occurred before, at a(1), which is 1 term before, so a(3) = 1. 1 has not occurred before, so a(4) = 0. 0 appeared most recently at term a(2), which is 2 terms earlier, so a(5) = 2. 2 has not occurred before, so a(6) = 0. And so on.

N. J. A. Sloane, Table of n, a(n) for n = 1..100000
Benjamin Chaffin, Frequency of values 0..500 over the first 10^12 terms
Benjamin Chaffin, Frequency of values 0..1000000 over the first 10^12 terms
Benjamin Chaffin, Average distance between 0s over the first 10^12 terms
Po-chia Chen, Additional comments on A181391, Jun 19 2019 [These comments have not yet been reviewed. - _N. J. A. Sloane_, Jun 20 2019]
Po-chia Chen, Further comments on A181391, Jun 20 2019 [These comments have also not yet been reviewed. - _N. J. A. Sloane_, Jul 05 2019]
Code Golf Stack Exchange, Nth term of Van Eck Sequence
Brady Haran and N. J. A. Sloane, Don't Know (the Van Eck Sequence), Numberphile video (2019).
N. J. A. Sloane, Fortran program

Cf. A171862, A171863, A171864, A171865, A171866, A171867, A171887, A171888, A171889, A171896, A171897 (numbers in order of appearance), A171898, A171899.
Cf. also A171911-A171918 (starting with other numbers than 0), A171951-A171956, A171957, A171958, A175041, A175100, A268755, A274425, A309363 (using 2 instead of 0 to mark a new value).
A276457 and A337980 are of the same type.
Cf. A025480.

import Data.List (findIndex, unfoldr)
import Data.Maybe (fromMaybe)
a181391 n = a181391_list !! (n-1)
a181391_list = 0 : (unfoldr g [0]) where
   g xs = Just (m, m : xs) where
        m = 1 + fromMaybe (-1) (findIndex (== head xs) $ tail xs)
-- Reinhard Zumkeller, Oct 31 2011

NB. see www.Jsoftware.com
(,  # <:@- }:  i: {:)^:({.`}.) 100 0 NB. Arie Bos, Dec 10 2010

function A181391(len)
    L = [0, 0]
    for n in 2:len
        k = findlast(m -> L[n] == L[m], 1:n-1)
        push!(L, k == nothing ? 0 : n - k)
    end
L end
println(A181391(96)) # Peter Luschny, May 19 2019

M:=10000;
a:=Array(1..M);
last:=Array(0..M,-1);
a[1]:=0;
a[2]:=0;
last[0]:=2;
nxt:=1;
for n from 3 to M do
hist:=last[nxt];
a[n]:=nxt;
last[nxt]:=n;
nxt:=0;
if hist>0 then nxt:=n-hist; fi;
od:
[seq(a[n],n=1..M)];
# N. J. A. Sloane, Oct 18 2010

m = 100; ClearAll[a, last]; a[] = 0; last[] = -1; last[0] = 2; nxt = 1; Do[ hist = last[nxt]; a[n] = nxt; last[nxt] = n; nxt = 0; If[ hist > 0 , nxt = n - hist], {n, 3, m}]; Table[a[n], {n, 1, m}] (* Jean-François Alcover, Dec 01 2011, after Maple program by N. J. A. Sloane *)
A181391L = Nest[# /. {{Longest[p___], a_, q___, a_} :> {p, a, q, a, Length[{a, q}]}, {a___} :> {a, 0}} &, {}, #] &; A181391L[97] (* JungHwan Min, Jan 14 2017 *)

A181391_vec(N,a=0,i=Map())={vector(N,n,a=if(n>1,iferr(n-mapget(i,a),E,0)+mapput(i,a,n)))} \\ M. F. Hasler, Jun 11 2019

A181391 = [0,0]
for n in range(1,10**4):
    for m in range(n-1,-1,-1):
        if A181391[m] == A181391[n]:
            A181391.append(n-m)
            break
    else:
        A181391.append(0) # Chai Wah Wu, Aug 14 2014

A181391 = [0]
last_pos = {}
for i in range(10**4):
    new_value = i - last_pos.get(A181391[i], i)
    A181391.append(new_value)
    last_pos[A181391[i]] = i
# Ehsan Kia, Jun 12 2019

vaneckw <-function(howmany = 100){
howmany = round(howmany[1])
ve = c(0,0)
for (jj in 2:(howmany)) {
thefind <- which(ve[1:(jj-1)] == ve[jj])
if (length(thefind)){
ve <- c(ve,jj-thefind[length(thefind)])
} else ve <- c(ve,0)
}
return(invisible(ve))
} #Carl Witthoft, Jun 14 2019

A278288 a(n) is the number of ways to represent a(n-1) as a sum of two distinct terms from a(0) to a(n-2). a(0) = 0. a(1) = a(2) = 1.

Original entry on oeis.org

0, 1, 1, 1, 2, 3, 3, 4, 6, 2, 4, 8, 3, 8, 4, 12, 6, 10, 10, 11, 12, 15, 9, 12, 19, 7, 15, 17, 10, 18, 22, 17, 12, 22, 21, 22, 25, 25, 26, 22, 26, 26, 27, 32, 25, 30, 27, 35, 21, 23, 31, 31, 32, 37, 37, 38, 37, 39, 37, 40, 40, 41, 45, 28, 37, 42, 38, 50, 33, 43, 58, 34

Offset: 0

Yuriy Sibirmovsky, Nov 16 2016

nonn

a(n) is the number of pairs (j, k) such that a(j) + a(k) = a(n-1), with 0 <= j, k < n-1 and j != k.
For large n, a(n) on average follows linear law a(n) ~ 0.7 n with linear spread (see the plot).
Unlike sequences such as A248034 or A276457, this sequence is base-independent.

			a(2) = a(1) + a(0), so a(3) = 1.
a(3) = a(1) + a(0) = a(2) + a(0), so a(4) = 2.
a(4) = a(3) + a(2) = a(3) + a(1) = a(2) + a(1), so a(5) = 3.

Yuriy Sibirmovsky, Table of n, a(n) for n = 0..2999
Yuriy Sibirmovsky, The plot of a(n) for n=0..2999 (with linear regression)

Cf. A248034, A276457.

Nm=100;
A=Table[1,{n,1,Nm}];
A[[1]]=0;
Do[Nc=0;
Do[If[A[[j]]+A[[k]]==A[[n]] && k!=j,Nc++],{j,1,n-1},{k,1,j}];
A[[n+1]]=Nc,{n,3,Nm-1}];
A

A309342 a(1) = 0, and for n >= 1, a(n+1) is the number of times the binary representation of a(n) appears in the concatenation of the binary representations of a(1), ..., a(n-1).

Original entry on oeis.org

0, 0, 1, 0, 2, 1, 2, 2, 3, 1, 7, 1, 11, 2, 5, 5, 7, 7, 9, 0, 12, 1, 30, 2, 12, 2, 14, 4, 3, 26, 4, 4, 5, 15, 13, 7, 24, 0, 34, 0, 39, 2, 27, 3, 45, 3, 49, 0, 49, 1, 89, 2, 36, 2, 39, 3, 60, 3, 66, 1, 112, 2, 45, 4, 19, 6, 26, 6, 28, 6, 30, 12, 12, 13, 20, 4

Offset: 1

Rémy Sigrist, Jul 24 2019

This sequence is a binary variant of A276457.

This sequence is necessarily unbounded.

			The first terms, alongside the binary representations of a(n) and of a(1), ..., a(n-1) with marks in front of occurrences of a(n), are:
  n   a(n)  bin(a(n))  bin(a(1)...a(n-1))
  --  ----  ---------  --------------------
   1     0          0
   2     0          0                    [0
   3     1          1                    00
   4     0          0                 [0[01
   5     2         10                 00[10
   6     1          1              00[10[10
   7     2         10             00[10[101
   8     2         10          00[10[101[10
   9     3         11          001010[11010
  10     1          1  00[10[10[1[10[10[1[1
  11     7        111       00101011010[111

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Rémy Sigrist, Logarithmic scatterplot of the first 250000 terms
Rémy Sigrist, Perl program for A309342

Cf. A276457.

Perl
```
See Links section.
```

A335497 a(1) = 1, and for any n > 0, a(n+1) is the number of times the decimal representation of a(n) appears in the concatenation of the first n terms, possibly with overlap.

Original entry on oeis.org

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

Offset: 1

Rémy Sigrist, Jun 14 2020

This sequence is a variant of A276457.
This sequence is unbounded.
It seems that lim sup a(n)/(n*log(n)) = 0.03 approximately. - Ya-Ping Lu, Dec 16 2021

			The first terms, alongside their concatenations with a star in front of each occurrence of a(n), are:
  n   a(n)  cat(a(1)...a(n))
  --  ----  ---------------------------------
   1     1  *1
   2     1  *1*1
   3     2  11*2
   4     1  *1*12*1
   5     3  1121*3
   6     1  *1*12*13*1
   7     4  112131*4
   8     1  *1*12*13*14*1
   9     5  11213141*5
  ...
  17     9  1121314151617181*9
  18     1  *1*12*13*14*15*16*17*18*19*1
  19    10  112131415161718191*10
  20     1  *1*12*13*14*15*16*17*18*19*1*10*1
  21    12  1*12131415161718191101*12
  22     2  11*21314151617181911011*2*2

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Rémy Sigrist, Logarithmic scatterplot of the first 1000000 terms
Rémy Sigrist, Perl program for A335497

Cf. A179801, A276457.

Perl
```
See Links section.
```

a1 = 1; print(a1, end =', '); S = str(a1)
for n in range(2, 100): ct = S.count(str(a1)); S += str(ct); print(ct, end = ', '); a1 = ct # Ya-Ping Lu, Dec 16 2021

A360257 a(1) = 1; for n > 1, a(n) is the number of preceding terms having the same sum of divisors as a(n-1).

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9, 1, 10, 1, 11, 2, 2, 3, 2, 4, 2, 5, 2, 6, 3, 3, 4, 3, 5, 3, 6, 4, 4, 5, 4, 6, 5, 5, 6, 6, 7, 2, 7, 3, 7, 4, 7, 5, 7, 6, 8, 2, 8, 3, 8, 4, 8, 5, 8, 6, 9, 2, 9, 3, 9, 4, 9, 5, 9, 6, 10, 2, 10, 3, 10, 4, 10, 5, 10, 6, 11, 12, 1, 12, 2, 11, 13, 1

Offset: 1

Scott R. Shannon, Jan 31 2023

nonn

The most common sum of divisor count of all previous terms changes as n increases; these values, up to 2 million terms, are 1, 12, 24, 12, 24, 72, 720, 72, 720, 72, 720, 72, 720. The value 72 holds the record from a(6998) = 71 to a(1271035) = 563. After a(1285242) = 264 the divisor sum 720 becomes the most common sum to well beyond 10 million terms. It is likely the record becomes arbitrarily large as n increases.

			a(22) = 2 as a(21) = 11 and 11 has a divisor sum of A000203(11) = 12. However, A000203(6) also equals 12, and as a(11) = 6 there are two previous terms with a divisor sum of 12.

Scott R. Shannon, Image of the first 2000000 terms.

Cf. A000203, A276457, A356348.

lista(nn) = my(va = vector(nn), vs=vector(nn)); va[1] = 1; vs[1] = 1; for (n=2, nn, va[n] = #select(x->(x==vs[n-1]), vs); vs[n] = sigma(va[n]);); va; \\ Michel Marcus, Jan 31 2023

cp's OEIS Frontend

A276463 Position at which n first appears in A276457, with a(0)=0.

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A181391 Van Eck's sequence: For n >= 1, if there exists an m < n such that a(m) = a(n), take the largest such m and set a(n+1) = n-m; otherwise a(n+1) = 0. Start with a(1)=0.

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A278288 a(n) is the number of ways to represent a(n-1) as a sum of two distinct terms from a(0) to a(n-2). a(0) = 0. a(1) = a(2) = 1.

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A309342 a(1) = 0, and for n >= 1, a(n+1) is the number of times the binary representation of a(n) appears in the concatenation of the binary representations of a(1), ..., a(n-1).

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Perl

A335497 a(1) = 1, and for any n > 0, a(n+1) is the number of times the decimal representation of a(n) appears in the concatenation of the first n terms, possibly with overlap.

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Python

A360257 a(1) = 1; for n > 1, a(n) is the number of preceding terms having the same sum of divisors as a(n-1).

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PARI