login
a(1) = 1, a(2) = 2. For n > 2, let i = a(n-2), j = a(n-1). Then a(n) is least novel k such that A007947(i*j*k) is the smallest possible primorial number, subject to no more than two consecutive terms having the same number of distinct prime divisors.
1

%I #10 Jul 08 2024 08:51:43

%S 1,2,3,6,4,8,12,9,16,18,24,27,32,36,48,64,54,72,81,96,108,128,144,162,

%T 243,192,216,256,288,324,512,384,432,729,486,576,1024,648

%N a(1) = 1, a(2) = 2. For n > 2, let i = a(n-2), j = a(n-1). Then a(n) is least novel k such that A007947(i*j*k) is the smallest possible primorial number, subject to no more than two consecutive terms having the same number of distinct prime divisors.

%C For all consecutive i,j,k rad(i*j*k) = A002110(2) = 6.

%C If A001221(i,j) = omega(i,j) = {1,1}, omega(k) is constrained to 2, if omega(i,j) = {2,2}, omega(k) must = 1. If omega(i,j) = {1,2} or {2,1} then omega(k) is 1 or 2, depending on smallest missing number conforming to the definition.

%C The sequence is a greedy permutation of the 3-smooth numbers (A003586). A power of 2 or 3 follows any pair of consecutive terms each having two distinct prime factors, but the converse is not true (8,12 --> 9 and 12,9 -->16) for example.

%e a(1,2) = 1,2 with omega(1,2) = 0,1 so a(3) = 3 since rad(1*2*3) = 6, the omega condition is satisfied (omega(3) = 1) and 3 is least such term.

%e a(2,3) = 2,3, omega(2,3) = 1,1 so a(4) cannot be 4 or 5 since both have omega = 1, even though rad(2*3*4) and rad(2*3*5) are both primorial. Therefore a(4) = 6, omega(6) = 2, and rad(2*3*6) = 6.

%e a(5) = 4 because rad(3*6*4) = 6, omega(3,6,4) = 1,2,1 and 4 is least such term.

%Y Cf. A001221, A002110, A007947.

%K nonn,more

%O 1,2

%A _David James Sycamore_, Jun 01 2024