|
| |
|
|
A283312
|
|
a(n) = smallest missing positive number, unless a(n-1) was a prime in which case a(n) = 2*a(n-1).
|
|
10
|
|
|
|
1, 2, 4, 3, 6, 5, 10, 7, 14, 8, 9, 11, 22, 12, 13, 26, 15, 16, 17, 34, 18, 19, 38, 20, 21, 23, 46, 24, 25, 27, 28, 29, 58, 30, 31, 62, 32, 33, 35, 36, 37, 74, 39, 40, 41, 82, 42, 43, 86, 44, 45, 47, 94, 48, 49, 50, 51, 52, 53, 106, 54, 55, 56, 57, 59, 118, 60, 61, 122, 63, 64, 65, 66, 67, 134, 68, 69
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,2
|
|
|
COMMENTS
|
Alternatively, this is the lexicographically earliest infinite sequence of distinct positive numbers such that every prime is followed by its double.
Theorem: This is a permutation of the positive integers.
Proof. Sequence is clearly infinite, so for any k there is a number N_0(k) such that n >= N_0(k) implies a(n) > k.
Suppose m is missing. Consider a(n) for n = N_0(m). Then a(n) must be a prime p (otherwise it would have been m, which is missing), a(n+1) = 2*p, and a(n+2) = m, a contradiction. QED.
(End)
Alternative definition: a(1,2) = 1,2. Let P(k) = rad(a(1)*a(2)*...*a(k)), then for n > 2, a(n) = P(n)/P(n-1), where rad is A007947. - David James Sycamore, Jan 27 2024
|
|
|
LINKS
|
|
|
|
FORMULA
|
There is an explicit formula for the n-th term of the inverse permutation: see A338362.
The graph: Numbers appear in the sequence in their natural order, except when interrupted by the appearance of primes. Suppose a(n)=x, where x is neither a prime nor twice a prime. Then if p is a prime in the range x/2 < p < x, 2p appears in the sequence between p and p+1. Therefore we have the identity
n = x + pi(x) - pi(x/2). ... (1)
If a(n) = x = a prime, then (1) is replaced by
n = x + pi(x) - pi(x/2) - 1. ... (2)
If a(n) = x = twice a prime then
n = x/2 + pi(x/2) - pi(x/4). ... (3)
These equations imply that the lower line in the graph of the sequence is
x approx= n(1 - 1/(2*log n)) ... (4)
while the upper line is
x approx= 2n(1 - 1/(2*log n)). ... (5)
|
|
|
EXAMPLE
|
The offset is 1. What is a(1)? It is the smallest missing positive number, which is 1. Similarly, a(2)=2.
What is a(3)? Since the previous term was the prime 2, a(3) = 4.
And so on.
|
|
|
MAPLE
|
a:=[1];
H:=Array(1..1000, 0); MMM:=1000;
H[1]:=1; smn:=2; t:=2;
for n from 2 to 100 do
if t=smn then a:=[op(a), t]; H[t]:=1;
if isprime(t) then a:=[op(a), 2*t]; H[2*t]:=1; fi;
t:=t+1;
# update smallest missing number smn
for i from smn+1 to MMM do if H[i]=0 then smn:=i; break; fi; od;
else t:=t+1;
fi;
od:
a;
|
|
|
MATHEMATICA
|
Module[{nmax = 100, smn = 1}, Nest[Append[#, If[PrimeQ[Last[#]], 2*Last[#], While[MemberQ[#, ++smn]]; smn]]&, {1}, nmax-1]] (* Paolo Xausa, Feb 12 2024 *)
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
