Jan Ritsema van Eck - cp's OEIS frontend

User: Jan Ritsema van Eck

Jan Ritsema van Eck has authored 6 sequences.

A332028 Savannah problem: number of distinct possible populations after n weeks, not allowing new populations after the empty set.

3, 5, 7, 11, 14, 17, 22, 26, 30, 34, 40, 45, 50, 55, 60, 67, 73, 79, 85, 91, 97, 105, 112, 119, 126, 133, 140, 147, 156, 164, 172, 180, 188, 196, 204, 212, 222, 231, 240, 249, 258, 267, 276, 285, 294, 305, 315, 325, 335, 345

Offset: 1

Jan Ritsema van Eck and Ali Sada, Feb 05 2020

nonn

The Savannah math problem (Ali Sada, 26 Dec 2019 email to Seqfan list) is about a savannah ecosystem consisting of zebras, fed lions and hungry lions. Assume we start with an empty savannah. Each week, the following things happen in this order:
1. All hungry lions (if any) die.
2. All fed lions (if any) become hungry.
3. One animal enters the savannah. This can be a zebra, a fed lion or a hungry lion.
4. Let m = min(number of zebras, number of hungry lions); then m hungry lions eat m zebras and become m fed lions.
The Savannah math problem is to determine how many distinct populations are possible after n weeks. There are two versions. This sequence gives the number of possible populations if continuation from the empty set is not allowed (See A332027 for the other version).
Since all populations with one fed lion remain stationary as long as one zebra enters the savannah each week, and all populations with more than one lion follow directly or indirectly from those with one fed lion, we know for all population with one or more lions (except one hungry lion) that if they are possible in week n, they will also be possible in week n+1. The population with one hungry lion can only be reached via the empty set, so it is not possible after week 1. The same goes for the population of 1 zebra without lion. The population of k zebras without lion can only be reached from k-1 zebras without lion. So it is only possible in week k. This means that in week n, there are n populations that were possible in previous weeks but not any more: 1 hungry lion, and 1 through (n-1) zebras without lion. Therefore, the total number of possible populations in week n is equal to the number if continuation of the empty set is allowed (A332027), minus n (except in week 1, when it is 3).

			After one week, there are 3 possible populations, depending on which animal entered the savannah: one zebra (Z), one fed lion (F), one hungry lion (H). After two weeks, we have from Z: 2Z, ZF, and (ZH->) F; and from F (which becomes H in the second step): (ZH->) F, FH and 2H. Population which follow from H (which becomes the empty set in the first step), are not allowed. Overall, there are 5 distinct possible populations after the second week: 2Z, ZF, FH, F and 2H.

Cf. A060432, A002024, A332026, A332027.

from math import isqrt
def A332028(n): return (k:=(r:=isqrt(m:=n+1<<1))+int((m<<2)>(r<<2)*(r+1)+1)-1)*(6*n-2-k*(k+3))//6+(isqrt(n<<3)+1>>1)+n if n>1 else 3 # Chai Wah Wu, Jun 07 2025

a(n) = A060432(n) + A002024(n), for n>1.

A332027 Savannah problem: number of distinct possible populations after n weeks, allowing populations after the empty set.

Original entry on oeis.org

3, 7, 10, 15, 19, 23, 29, 34, 39, 44, 51, 57, 63, 69, 75, 83, 90, 97, 104, 111, 118, 127, 135, 143, 151, 159, 167, 175, 185, 194, 203, 212, 221, 230, 239, 248, 259, 269, 279, 289, 299, 309, 319, 329, 339, 351, 362, 373, 384

Offset: 1

Jan Ritsema van Eck and Ali Sada, Feb 05 2020

nonn

The Savannah math problem (Ali Sada, 26 Dec 2019 email to Seqfan list) is about a savannah ecosystem consisting of zebras, fed lions and hungry lions. Assume we start with an empty savannah. Each week, the following things happen in this order:
1. All hungry lions (if any) die.
2. All fed lions (if any) become hungry.
3. One animal enters the savannah. This can be a zebra, a fed lion or a hungry lion.
4. Let m = min (number of zebras, number of hungry lions); then m hungry lions eat m zebras and become m fed lions.
The Savannah math problem is to determine how many distinct populations are possible after n weeks. There are two versions. This sequence gives the number of possible populations if continuation from the empty set is allowed (see A332028 for the other version).
Since the three 1-animal populations (1 zebra, 1 fed lion and 1 hungry lion) can be reached directly from 1 hungry lion (via the empty set), they will be possible every week. Since all other possible populations are reached via one of these, any population that is possible in week n is also possible in week n+1. Therefore, the total number of possible populations in week n is the sum of all new populations in weeks 1 through n: A(n) is the partial sum of A332026.

			After one week, there are 3 possible populations, depending on which animal entered the savannah: one zebra (Z), one fed lion (F), one hungry lion (H). After two weeks, we have from Z: 2Z, ZF, and (ZH->) F; from F (which becomes H in the second step): (ZH->) F, FH and 2H; and from H (which becomes the empty set in the first step): Z, F and H. Overall, there are 7 distinct possible populations after the second week: 2Z, ZF, Z, FH, F, 2H and H.

Cf. A002024, A332026, A060432, A332028.

from math import isqrt
def A332027(n): return (k:=(r:=isqrt(m:=n+1<<1))+int((m<<2)>(r<<2)*(r+1)+1)-1)*(6*n-2-k*(k+3))//6+(isqrt(n<<3)+1>>1)+(n<<1) # Chai Wah Wu, Jun 07 2025

a(n) = A060432(n) + A002024(n) + n.

A332026 Savannah problem: number of new possibilities after n weeks.

Original entry on oeis.org

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

Offset: 1

Jan Ritsema van Eck and Ali Sada, Feb 05 2020

nonn

The Savannah math problem (Ali Sada, 26 Dec 2019 email to Seqfan list) is about a savannah ecosystem consisting of zebras, fed lions and hungry lions. Assume we start with an empty savannah. Each week, the following things happen in this order:
1. All hungry lions (if any) die.
2. All fed lions (if any) become hungry.
3. One animal enters the savannah. This can be a zebra, a fed lion or a hungry lion.
4. Let m = min(number of zebras, number of hungry lions); then m hungry lions eat m zebras and become m fed lions.
The Savannah math problem is to determine how many distinct populations are possible after n weeks.
After step 4, there are either zero zebras or zero hungry lions (or both). Therefore, all possible populations can be positioned in a two-dimensional table, with the number of zebras on the positive part of one axis, the number of hungry lions on the negative part of the same axis and the total number of lions on the other axis (thus, a hungry lion is a lion with a deficit of one zebra). Tabulating the minimum number of weeks in which each population can occur, we get:
zebras
   8 |   8    9   11   14   18
   7 |   7    8   10   13   17
   6 |   6    7    9   12   16
   5 |   5    6    8   11   15
   4 |   4    5    7   10   14
   3 |   3    4    6    9   13
   2 |   2    3    5    8   12
   1 |   1    2    4    7   11
   0 |   0    1    3    6   10
  -1 |        1    2    5    9
  -2 |             2    4    8
  -3 |                  4    7
  -4 |                       7
     +------------------------------
         0    1    2    3    4 lions
A(n) is the number of n's in a sufficiently large version of this table.
The row for Z=0 is equal to the sequence of triangular numbers (A000217).
The number of columns with new entries in week n is the integer inverse of triangular number (A002024) plus 1. The number of new entries in week n is usually the number of columns, except when n is a triangular number plus 1; then a new column is started and the number of new entries is the number of columns plus 1.

			After one week, there are 3 possible populations, depending on which animal entered the savannah: one zebra (Z), one fed lion (F), one hungry lion (H). After two weeks, from Z we get: 2Z, ZF, and (ZH->) F; from F (which becomes H in the second step) we get: (ZH->) F, FH and 2H. From H (which becomes the empty set in the first step): Z, F and H. Overall, there are 4 new possible populations that were not possible after the first week: 2Z, ZF, FH, and 2H.

Cf. A002024, A010054.

See A332027 and A332028 for the total number of possibilities.

from math import isqrt
from sympy.ntheory.primetest import is_square
def A332026(n): return (isqrt(m:=n<<3)+1>>1)+is_square(m-7)+1 # Chai Wah Wu, Jun 07 2025

a(n) = A002024(n) + A010054(n) + 1.

A309814 Where records occur in A171898.

Original entry on oeis.org

1, 2, 3, 9, 27, 30, 41, 56, 75, 92, 152, 187, 212, 276, 309, 442, 561, 578, 632, 746, 893, 1276, 1834, 1894, 1929, 1938, 2216, 2562, 2714, 2807, 3573, 3767, 4124, 4534, 9371, 9581, 9875

Offset: 1

Jan Ritsema van Eck, Aug 18 2019

Records in A171898 (the forward van Eck transform of A181391) are a subset of A171866 (records in A181391).

Cf. A309813, A171866, A171867, A171898, A181391.

a(i) + A309813(i) + 1 gives the position of the corresponding record in A181391.

A309813 Records in A171898.

Original entry on oeis.org

1, 2, 6, 42, 56, 131, 170, 307, 650, 1302, 2021, 2028, 2238, 2944, 3151, 4466, 7365, 7936, 9344, 17384, 23470, 28328, 30318, 31806, 36385, 37283, 40245, 48934, 89509, 108985, 113677, 123092, 173614, 295701, 300255, 311221, 363758

Offset: 1

Jan Ritsema van Eck, Aug 18 2019

nonn

This is a subset of A171866 (records in A181391). Proof: each positive term m in A171866 also occurs in A181391, and the other way around, with a shift of m+1 positions. Consider a record r in A171866, which occurs in that sequence at position i; it occurs in A181391 at position i+r+1. Any larger term s>r must occur in A171866 at a position j>i (or r would not be a record), so its position in A181391, j+s+1, is larger than i+r+1. Therefore, r is also a record in A181391.

Cf. A309814 (positions), A171898, A181391, A171866.

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

cp's OEIS Frontend

User: Jan Ritsema van Eck

A332028 Savannah problem: number of distinct possible populations after n weeks, not allowing new populations after the empty set.

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A332027 Savannah problem: number of distinct possible populations after n weeks, allowing populations after the empty set.

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A332026 Savannah problem: number of new possibilities after n weeks.

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A309814 Where records occur in A171898.

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A309813 Records in A171898.

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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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