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Lexicographically earliest sequence of distinct primes such that the sequence of squarefree numbers that are coprime to these primes has an asymptotic density 1/2.
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%I #9 Oct 16 2023 09:30:18

%S 5,79,1831,1856917,517136788981,25309896984298197131551,

%T 9737146484866113825954170751740726870607451

%N Lexicographically earliest sequence of distinct primes such that the sequence of squarefree numbers that are coprime to these primes has an asymptotic density 1/2.

%C The corresponding sequence of squarefree numbers is A366641.

%C Equivalently, lexicographically earliest sequence of distinct primes such that Product_{n>=1} (1 + 1/a(n)) = 12/Pi^2.

%C The next term has 85 digits and is too large to be included in the data section.

%H Amiram Eldar, <a href="/A366640/b366640.txt">Table of n, a(n) for n = 1..11</a>

%e The asymptotic density of the squarefree numbers is 6/Pi^2 = 0.607... (A059956). Without the even numbers, the density of the odd squarefree numbers (A056911) is 4/Pi^2 = 0.405... (A185199), which is smaller than 1/2. Without the multiples of 3, the density of the squarefree numbers that are not divisible by 3 (A261034) is 9/(2*Pi^2) = 0.455... (A088245), which is also smaller than 1/2. Without the multiples of 5, the density of the squarefree numbers that are not divisible by 5 (A274546) is 5/Pi^2 = 0.506..., which is larger than 1/2. Therefore, a(1) = 5.

%e The asymptotic density of the squarefree numbers that are coprime to the primes a(1)..a(n), for n=1..8, is:

%e n a(n) density

%e - ---------------- ------------------------------------------------------

%e 1 5 5/Pi^2 = 0.506605...

%e 2 79 79/(16*Pi^2) = 0.500273...

%e 3 1831 144649/(29312*Pi^2) = 0.500000269...

%e 4 1856917 268601187133/(54429980416*Pi^2) = 0.500000000000966...

%e 5 517136788981 1/2 + 1.975... * 10^(-23)

%e 6 2.530... * 10^22 1/2 + 5.134... * 10^(-44)

%e 7 9.737... * 10^42 1/2 + 3.775... * 10^(-85)

%e 8 1.324... * 10^84 1/2 + 2.993... * 10^(-167)

%t seq[len_] := Module[{s = {}, r = 12/Pi^2, p}, Do[p = NextPrime[1/(r - 1)]; r *= (1/(1 + 1/p)); AppendTo[s, p], {len}]; s]; seq[8]

%Y Cf. A005117, A056911, A261034, A274546, A366641.

%Y Cf. A059956, A088245, A185199.

%K nonn

%O 1,1

%A _Amiram Eldar_, Oct 15 2023