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

Author

N. J. A. Sloane, Mar 08 2017

Keywords

• nonn

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

Examples

			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.
		

Links

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

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.

Programs

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 *)

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

Extensions

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