A283832 -id:A283832 - 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

A375024 a(n) is the length of the largest sequence S of distinct integers in the range 1..n such that for any prime number p, any run of consecutive multiples of p in S has length exactly 2, and two consecutive terms in S have some common prime factor.

Original entry on oeis.org

1, 1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 7, 7, 7, 7, 7, 7, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 13, 13, 16, 16, 16, 16, 19, 19, 19, 19, 19, 19

Offset: 1

Rémy Sigrist, Jul 28 2024

Sequences like A280864 can be split into segments of consecutive terms with properties similar to the sequences S that we are considering here.

			Some solutions for small n:
  n   a(n)  Solution S
  --  ----  --------------------------------------------------------------
   1     1  1
   4     2  2,4
   6     3  2,6,3
  10     4  3,6,10,5
  15     7  3,6,10,15,12,14,7
  21    10  3,6,10,15,12,14,21,18,20,5
  33    13  3,6,10,15,12,14,21,18,22,33,24,20,5
  35    16  3,6,10,15,12,14,21,18,20,35,28,22,33,24,26,13
  39    19  3,6,10,15,12,14,21,18,20,35,28,22,33,24,26,39,36,34,17
  45    22  5,10,6,15,20,12,21,14,18,33,22,24,39,26,36,45,40,28,35,30,42,7

Rémy Sigrist, C++ program

Cf. A280864, A283832, A373797.

a(n) <= A373797(n).

a(p) = a(p-1) for any prime number p.

cp's OEIS Frontend

A285190 Records in A283832.

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A285191 Where records occur in A283832.

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A285189 k appears A283832(k+1) times.

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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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A280774 Numbers k such that gcd(A280864(k), A280864(k+1)) = 1.

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A375024 a(n) is the length of the largest sequence S of distinct integers in the range 1..n such that for any prime number p, any run of consecutive multiples of p in S has length exactly 2, and two consecutive terms in S have some common prime factor.

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