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A280864
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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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54
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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
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OFFSET
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1,2
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COMMENTS
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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
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)
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LINKS
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EXAMPLE
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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
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MAPLE
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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:
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MATHEMATICA
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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 *)
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CROSSREFS
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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.
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KEYWORD
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nonn,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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