A375029 -id:A375029 - cp's OEIS frontend

A280864 Lexicographically earliest infinite sequence of distinct positive terms such that, for any prime p, any run of consecutive multiples of p has length exactly 2.

1, 2, 4, 3, 6, 8, 5, 10, 12, 9, 7, 14, 16, 11, 22, 18, 15, 20, 24, 21, 28, 26, 13, 17, 34, 30, 45, 19, 38, 32, 23, 46, 36, 27, 25, 35, 42, 48, 29, 58, 40, 55, 33, 39, 52, 44, 77, 49, 31, 62, 50, 65, 78, 54, 37, 74, 56, 63, 51, 68, 60, 75, 41, 82, 64, 43, 86

Offset: 1

Rémy Sigrist, Jan 09 2017

In other words, each multiple of a prime p has exactly one neighbor that is also a multiple of p.
This sequence is similar to A280866; the first difference occurs at n=42: a(42)=55 whereas A280866(42)=50.
Conjectured to be a permutation of the positive integers.
Sometimes referred to as the "cup of coffee" sequence, since it feels as if just one more cup of coffee is all it would take to prove that this is indeed a permutation of the positive integers. - N. J. A. Sloane, Nov 04 2020
There are several short cycles, and apparently at least two infinite cycles. For a list see the attached file "Properties of A280864". - N. J. A. Sloane, Feb 03 2017
Properties (For proofs, see the attached file "Properties of A280864")
Theorem 1: This sequence contains every prime and every even number. (Added by N. J. A. Sloane, Jan 15 2017)
Theorem 2: The sequence contains infinitely many odd composite numbers. (Added by N. J. A. Sloane, Feb 14 2017)
Theorem 3: If p is an odd prime, the sequence contains infinitely many odd multiples of p. (Added by N. J. A. Sloane, Mar 12 2017, with corrected proof Apr 03 2017)
There are two types of primes in this sequence: Type I, the first time a term a(n) is divisible by p is when a(n)=p for some n; Type II, the first time a term a(n) is divisible by p is when a(n)=k*p for some n and some k>1 (the Type II primes are listed in A280745).
Conjecture 4: If a prime p divides a(n) then p <= n. - N. J. A. Sloane, Apr 07 2017 and Apr 16 2017
Theorem 5: The sequence is a permutation of the natural numbers iff it contains every square. - N. J. A. Sloane, Apr 14 2017
From Bob Selcoe, Apr 03 2017: (Start)
Define the "radical class" C_R to be the set of numbers which have the same radical R (or the same largest squarefree divisor - i.e., the same product of their prime factors). These are the columns in A284311. So for example C_10 is the set of numbers with radical 10 or prime factors {2,5}: {10, 20, 40, 50, 80, 100, 160, ...}.
If the sequence contains any members of C_R, then those members must appear in order; so for example, if 160 has appeared, {10, 20, 40, 50, 80} will have already appeared, in that order. Naturally, this holds for prime powers; for example, C_5: if 3125 has appeared, {5, 25, 125, 625} will have appeared earlier, in that order.
After seeing a(n), let S be smallest missing number (A280740) and let prime(G) be largest prime already appearing in the sequence. Conjecture: Prime(G) < S <= prime(G+1), and a(35) = 25 = S is the only nonprime S term (following a(31) = 23, preceding a(39) = 29). (End)

			The first terms, alongside their required and forbidden prime factors are:
n   a(n)  Required  Forbidden
--  ----  --------  ---------
1      1  none      none
2      2  none      none
3      4  2         none
4      3  none      2
5      6  3         none
6      8  2         3
7      5  none      2
8     10  5         none
9     12  2         5
10     9  3         2
11     7  none      3
12    14  7         none
13    16  2         7
14    11  none      2
15    22  11        none
16    18  2         11
17    15  3         2
18    20  5         3
19    24  2         5
20    21  3         2
21    28  7         3
22    26  2         7
23    13  13        2
24    17  none      13
25    34  17        none
26    30  2         17
27    45  3, 5      2
28    19  none      3, 5
29    38  19        none
30    32  2         19
31    23  none      2
32    46  23        none
33    36  2         23
34    27  3         2
35    25  none      3
36    35  5         none
37    42  7         5
38    48  2, 3      7
39    29  none      2, 3
40    58  29        none
41    40  2         29
42    55  5         2

N. J. A. Sloane, Table of n, a(n) for n = 1..100000 (First 10000 terms from Rémy Sigrist)
Dana G. Korssjoen, Biyao Li, Stefan Steinerberger, Raghavendra Tripathi, and Ruimin Zhang, Finding structure in sequences of real numbers via graph theory: a problem list, arXiv:2012.04625, Dec 08, 2020
Rémy Sigrist, PARI program for A280864
N. J. A. Sloane, Properties of A280864 [Revised, Apr 25 2017]
N. J. A. Sloane, Table of n, a(n) for n = 1..1000000, computed using Sigrist's PARI program.
N. J. A. Sloane, Confessions of a Sequence Addict (AofA2017), slides of invited talk given at AofA 2017, Jun 19 2017, Princeton. Mentions this sequence.

See A280738, A280740, A280741 (inverse), A280742, A280743, A280744, A280745, A280746, A280755, A280770, A280771, A280773, A280774, A283832, A284724, A284725, A284726, A284785, A285181 for various subsidiary sequences.
A280754 gives fixed points.
Cf. A280866.
In the same spirit as A064413 and A098550.
A338338, A338444, and A375029 are variants.
A373797 is a finite version.

N:= 1000: # to get all terms until the first term > N
A[1]:= 1:
A[2]:= 2:
G:= {}:
Avail:= [$3..N]:
found:= true:
lastn:= 2:
for n from 3 while found and nops(Avail)>0 do
  found:= false;
  H:= G;
  G:= numtheory:-factorset(A[n-1]);
  r:= convert(G minus H,`*`);
  s:= convert(G intersect H, `*`);
  for j from 1 to nops(Avail) do
    if Avail[j] mod r = 0 and igcd(Avail[j],s) = 1 then
      found:= true;
      A[n]:= Avail[j];
      Avail:= subsop(j=NULL,Avail);
      lastn:= n;
      break
    fi
  od;
od:
seq(A[i],i=1..lastn); # Robert Israel, Mar 22 2017

terms = 100;
rad[n_] := Times @@ FactorInteger[n][[All, 1]];
A280864 = Reap[present = 0; p = 1; pp = 1; Do[forbidden = GCD[p, pp]; mandatory = p/forbidden; a = mandatory; While[BitGet[present, a] > 0 || GCD[forbidden, a] > 1, a += mandatory]; Sow[a]; present += 2^a; pp = p; p = rad[a], terms]][[2, 1]] (* Jean-François Alcover, Nov 23 2017, translated from Rémy Sigrist's PARI program *)

Added "infinite" to definition. - N. J. A. Sloane, Sep 28 2019

A373797 a(n) = maximal k such that there exists a sequence S = [s_1, s_2, ..., s_k] with s_i distinct and in the range 1 <= s_i <= n such that if any s_i is divisible by a prime p, then p also divides exactly one of s_{i-1} and s_{i+1}.

Original entry on oeis.org

1, 1, 1, 3, 3, 4, 4, 6, 6, 8, 8, 10, 10, 11, 11, 13, 13, 14, 14, 16, 16, 17, 17, 19, 19, 21, 22, 24, 24, 25, 25

Offset: 1

N. J. A. Sloane, Jul 26 2024

This is a finite version of the "cup of coffee" sequence A280864.
An obvious upper bound is a(n) <= n-C, where C is the number of primes p <= n such that 2*p > n. For example a(4) <= 3, since we cannot make use of the prime 3.
Note that for a prime p <= n/2, the number of terms in S that are divisible by p must be even. So if floor(n/p) is odd for any p <= n/2, we have a(n) <= n-C-1.
When going from n=p-1 to n=p (where p is a prime), the new p cannot be used, so a(p) = a(p-1) for p prime.
We can obtain a sequence of lower bounds by taking S to consist of the first A280774(m) terms of A280864, for m = 1,2,3,...  This gives
   a(m) >= max_{1 <= i <= A280774(m)} A280864(i).... (**)
 For example, with m=4, we can take the first A280774(4) = 10 terms of A280864, that is, the sequence S = 1,2,4,3,6,8,5,10,12,9, to get a(12) >= 10.  In fact equality holds: a(12) = 10 (see EXAMPLES below).
It was possible that equality would always hold in (**). (**) gives a(4) >= 3, a(8) >= 6, a(12) >= 10, a(16) >= 13 (equality holds up to this point), then a(28) >= 23, a(45) >= 27, ... However, Rémy Sigrist has shown that a(28) = 24.
a(24) = 19 so a(25) >= 19. Here is an argument that shows that a(25) = 20 is impossible. We can't use the big primes 13 17 19 23. There are 5 multiples of 5 we could use, 5 10 15 20 25, but we can only use 4 of them.  There are 3 multiples of 7 we could use, namely 7 14 21, but we can only use 2 of them. These two lists are disjoint. So a(25) >= 25 - 4 - 2 = 19. - N. J. A. Sloane, Jul 27 2024

			Solutions for small n (the solutions are a long way from being unique, but see A375030):
  n   a(n)    Solution S
  1    1        1
  4    3        1,2,4
  6    4        1,2,6,3
  8    6        1,2,4,3,6,8
  10   8        1,2,4,3,9,5,10,8
  12   10       1,2,4,3,6,8,5,10,12,9
  14   11       1,2,4,3,6,10,5,7,14,12,9
  16   13       1,2,4,3,6,8,5,10,12,9,7,14,16
As an example, let us verify that the prime-divisibility condition holds for the n=14 solution (we write Y to indicate divisibility):
  S = 1,2,4,3,6,10,5,7,14,12,9
  2?....Y.Y...Y..Y......Y..Y..
  3?........Y.Y............Y.Y
  5?.............Y.Y..........
  7?.................Y..Y.....
The Y's occur in disjoint pairs, as required.
Also, a(18) = 14, from S = 1,8,16,3,6,14,7,5,15,18,4,9,12,2. (We cannot use 11, 13, and 17, and there are an odd number of multiples of 2 and of 5, so we must lose at least one more term - we can take care of this by sacrificing 10 - so 18-4 = 14 is optimal. This implies a(19) = 14 also.)

Rémy Sigrist, C++ program

Cf. A280864, A280774, A375024, A375029, A375030.

See A373800 for the right-hand side of (**).

from itertools import permutations
from sympy import primefactors, primepi
def A373797(n):
    c = [set()]+[set(primefactors(i)) for i in range(1,n+1)]+[set()]
    for k in range(n-primepi(n)+primepi(n>>1),0,-1):
        for a in permutations(range(1,n+1),k):
            alist = [0]+list(a)+[n+1]
            if all((p in c[alist[i-1]])^(p in c[alist[i+1]]) for i, d in enumerate(alist[1:-1],1) for p in c[d]):
                return k # Chai Wah Wu, Jul 26 2024

# Given a list S = [s_1, s_2, ..., s_k], this function checks if the terms s_i are such that if any s_i is divisible by a prime p, then p also divides exactly one of s_{i-1} and s_{i+1}.
from sympy import primefactors
def isSolution(S: list[int]) -> bool:
    return all(not any((S[i-1] % p == 0) == (S[i+1] % p == 0)
           for p in primefactors(S[i])) for i in range(1, len(S)-1))
# Peter Luschny, Jul 27 2024
(C++) // See Links section.

a(20)-a(24) from Rémy Sigrist, Jul 27 2024
a(25) from N. J. A. Sloane, Jul 27 2024
a(26)-a(31) from Rémy Sigrist, Jul 28 2024

A374191 Triangle read by rows: T(n) is a permutation of [0, 1, 2, ..., n] subject to an extended Sigrist condition (A280864).

Original entry on oeis.org

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

Offset: 0

Peter Luschny, Jul 31 2024

In a series of submissions (A280864, A280866, A375029, A375030) Rémy Sigrist studies sequences whose terms are locally connected via their prime factors, i.e., where neighbors influence each other in their divisibility.
Sigrist chooses IN = {1, 2,...} as domain. The triangle discussed here is based on Sigrist's idea but chooses IN = {0, 1, 2,...} as the domain. We recall that all numbers divide 0, but 0 only divides 0. Neither 0 nor 1 have prime factors.
For a row of the triangle T(n) = [T(n, k) | k=0..n] and for a term t in this row, let 't be its predecessor and t' its successor. The terms of a row are subject to the following two conditions:
(1) For all k in 1..n-1 and t = T(n, k), t has no prime factor, or it is not the case that there is a prime factor of t such that (p | 't) <=> (p | t'). Expressed more succinctly (in pseudo-Python):
  all(not any((p | 't) <=> (p | t') for p in primefactors(t)) for k = 1..n-1).
(2) If t = T(n, n), then t has no prime factor or for all prime factors of t, (p | t) => (p | t').
The row itself must also satisfy two conditions:
(3) T(n) is a permutation of {0, 1, 2, ..., n}.
(4) T(n) is the lexicographically earliest list among all lists whose terms satisfy conditions (1) and (2).
From Peter Luschny, Aug 05 2024: (Start)
On StackExchange (see link), user Bubbler found by exhaustive analysis for n < 29 that only n <= 14 and n = 16 have a solution. Bubbler also states that "since the 4th Ramanujan prime is 29, there are at least four primes that are greater than n/2 (i.e., prime factors that appear only once) when n >= 29 but there are only 3 positions that such primes can go (both sides of 0 and the first element), which proves that there is no solution when n >= 29."
We set all terms of row 15 equal to 0 by convention to make the sequence finite and full. (End)

			Triangle starts:
  [ 0] (0)
  [ 1] (0, 1)
  [ 2] (0, 2, 1)
  [ 3] (1, 2, 0, 3)
  [ 4] (0, 3, 1, 2, 4)
  [ 5] (1, 2, 4, 3, 0, 5)
  [ 6] (1, 2, 0, 5, 3, 6,  4)
  [ 7] (2, 1, 3, 6, 4, 5,  0,  7)
  [ 8] (1, 2, 4, 3, 6, 8,  5,  0,  7)
  [ 9] (3, 1, 2, 4, 5, 0,  7,  8,  6,  9)
  [10] (1, 2, 4, 3, 9, 5, 10,  8,  7,  0,  6)
  [11] (6, 1, 2, 4, 3, 9,  5, 10,  8,  7,  0, 11)
  [12] (1, 2, 4, 3, 6, 8,  5, 10, 12,  9,  7,  0, 11)
  [13] (7, 1, 2, 4, 3, 6,  8,  5, 10, 12,  9, 11,  0, 13)
  [14] (2, 1, 3, 6, 4, 5, 10,  8,  7, 14, 12,  9, 11,  0, 13)         (*)
  [15] (0, 0, 0, 0, 0, 0,  0,  0,  0,  0,  0,  0,  0,  0,  0, 0)
  [16](11, 1, 2, 4, 3, 6,  8,  5, 10, 12,  9,  7, 14, 16, 13, 0, 15)  (*)
  (*) Found by Bubbler (see link).
.
The terms of T(11, k) alongside their prime factors are:
    k   T(11,k)   prime factors
  --  -------  ---------------
   0     6      2  3
   1     1
   2     2      2
   3     4      2
   4     3         3
   5     9         3
   6     5            5
   7    10      2     5
   8     8      2
   9     7              7
  10     0
  11    11                11

User Bubbler, Comments and solution to challenge: Lexicographically earliest permutation of the initial segment of nonnegative integers subject to divisibility constraints, code golf, StackExchange.

Cf. A280864, A280866, A375029, A375030.

from sympy import primefactors
from itertools import permutations
def test(a: int, b: int, p: int) -> bool:
    return (a % p == 0) == (b % p == 0)
def isSolution(S: tuple[int,...]) -> bool:
    if len(S) == 1: return True
    if not all(test(S[-2], S[-1], p)
           for p in primefactors(S[-1])):
        return False
    return all(not any(test(S[i-1], S[i+1], p)
           for p in primefactors(S[i]))
           for i in range(1, len(S) - 1))
def Trow(r: int) -> tuple[int,...] | None:
    C = list(range(r + 1))
    for P in permutations(C):
        if isSolution(P): return P
for n in range(9): print(Trow(n))

cp's OEIS Frontend

A280864 Lexicographically earliest infinite sequence of distinct positive terms such that, for any prime p, any run of consecutive multiples of p has length exactly 2.

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A373797 a(n) = maximal k such that there exists a sequence S = [s_1, s_2, ..., s_k] with s_i distinct and in the range 1 <= s_i <= n such that if any s_i is divisible by a prime p, then p also divides exactly one of s_{i-1} and s_{i+1}.

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A374191 Triangle read by rows: T(n) is a permutation of [0, 1, 2, ..., n] subject to an extended Sigrist condition (A280864).

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