A283312 a(n) = smallest missing positive number, unless a(n-1) was a prime in which case a(n) = 2*a(n-1).

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

Offset

1,2

Comments

Comments from N. J. A. Sloane, Nov 02 2020: (Start)

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)

A toy model of A280864, A280985, and A127202.

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

N. J. A. Sloane, Table of n, a(n) for n = 1..75000

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)

a(2*n-1 + A369610(n)) = prime(n). - David James Sycamore, Jan 27 2024

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

Cf. A127202, A280864, A280985.

See A283313 for smallest missing number, A338362 for inverse, A338361 for indices of primes, A338357 for first differences.

For records see A338356 and A001747.

Sequence in context: A082560 A191598 A338221 * A280985 A127202 A179869

Adjacent sequences: A283309 A283310 A283311 * A283313 A283314 A283315

Keyword

nonn

Author

N. J. A. Sloane, Mar 08 2017

Extensions

Entry revised by N. J. A. Sloane, Nov 03 2020

Status

approved