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A332878 a(1)=1; a(n+1) is the smallest number not yet seen in the sequence which is divisible by 1 + Omega(a(n)). 1

%I #33 Nov 25 2021 12:41:02

%S 1,2,4,3,6,9,12,8,16,5,10,15,18,20,24,25,21,27,28,32,30,36,35,33,39,

%T 42,40,45,44,48,54,50,52,56,55,51,57,60,65,63,64,7,14,66,68,72,78,76,

%U 80,84,70,88,75,92,96,49,69,81,85,87,90,95,93,99,100,105,104

%N a(1)=1; a(n+1) is the smallest number not yet seen in the sequence which is divisible by 1 + Omega(a(n)).

%C Here Omega = A001222, the number of prime factors counted with multiplicity.

%C Conjecture: This is a permutation of the natural numbers in which the primes appear in their natural order. Prime p > 2 arises as a(k) if and only if a(k-1) = 2^(p-1), in which case a(k+1) = 2*p. The sequence of numbers k such that a(k) is prime starts 2, 4, 10, 42, ... How does it continue?

%C a(636) = 11, a(2530) = 13, a(39731) = 17. It appears that the prime p occurs roughly at index 2^(p-2)*(1 + O(1/log p)). It is followed by 2p and then a multiple of 3. The graph of the sequence has several "branches" which can be labeled by odd primes: Most numbers occur on the main (p=3) branch which has an initial slope of about 1.61 increasing to 1.65 in the range 1e4 .. 4e4. A smaller fraction of the numbers lie on a second (p=5) and third (p=7) branch with slope of roughly 1.25 resp. 1.11 around n ~ 4e4, and a very small fraction lies on the branches with even lower slope (about 0.15 for the p=11 and 0.035 for the p=13 branch). - _M. F. Hasler_, Mar 04 2020

%H M. F. Hasler, <a href="/A332878/b332878.txt">Table of n, a(n) for n = 1..10000</a>

%H M. F. Hasler, <a href="/A332878/a332878.gif">Scatterplot of a(1..40000)</a>.

%e a(1) = 1 => 1 + Omega(a(1)) = 1, so a(2) must be 2.

%e Then 1 + Omega(a(2)) = 2, so a(3) = 4.

%e Since 1 + Omega(4) = 3, a(4) = 3; then 1 + Omega(3) = 2 and a(5) = 6, etc.

%t a[1]=1; a[n_] := a[n] = Block[{s = Array[a, n-1], k, o = 1 + PrimeOmega@ a[n-1]}, k = o; While[ MemberQ[s, k], k += o]; k]; Array[a, 56] (* _Giovanni Resta_, Mar 04 2020 *)

%o (PARI) {A332878_vec(N,a=1,u=1)=vector(N,n, a=n=bigomega(a)+1; while(bittest(u,a), a+=n); u+=1<<a; a)} \\ _M. F. Hasler_, Mar 04 2020

%Y Cf. A000040, A001222, A061286.

%K nonn

%O 1,2

%A _David James Sycamore_, Feb 28 2020

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