%I #26 Aug 16 2022 15:30:28
%S 1,3,8,12,23,26,32,66,68,78,83,106,116,169,181,201,210,216,234,273,
%T 282,296,427,436,501,504,513,538,547,583,655,688,711,738,751,851,866,
%U 947,1065,1088,1155,1274,1277,1285,1350,1369,1389,1456,1594,1615,1702,1734
%N a(n) is the first value to occur consecutively in the sequence b_n defined by p_2k(b_n(k)) = p_k(n)^2, k=1,2,3,..., where p_k(n) is the n-th k-almost prime.
%C A k-almost prime has exactly k prime factors, repetitions included.
%C Conjecture: Each sequence b_n repeats indefinitely. (Example: for n=3, b_n = 9, 8, 8, 8, 8, 8, .... It looks like b_3(k) is 8 for all k > 1.)
%C The conjecture follows from the formula that uses A078843 below (and the strict monotonicity of A078843). However the first repeated value is not for every n the value that repeats indefinitely. For example a(8) = b_8(2) = b_8(3) = 66, but b_8(k) = 64 for k >= 4. - _Peter Munn_, Aug 05 2019
%F From _Peter Munn_, Aug 05 2019: (Start)
%F b_n(k) = A058933(A078840(k,n)^2).
%F a(n) = b_n(min {k : b_n(k) = b_n(k+1)}).
%F If n < A078843(k+1) and b_n(k) < A078843(2k+1) then b_n(i) = b_n(k) for i >= k.
%F (End)
%e for k = 1, 2, 3, 4, 5, 6, ...:
%e p_k(3) = 5, 9, 18, 36, 72, 144, ... (the 3rd k-almost prime);
%e p_k(3)^2 = 25, 81, 324, 1296, 5184, 20736, ...;
%e b_3(k) = 9, 8, 8, 8, 8, 8, ... (index in the 2k-almost primes);
%e so since b_3(3) = b_3(2) = 8, a(3) = 8.
%o (PARI) get_p(m,k) = {local(i,n);i=0;n=1;while(i<k,n=n+1;if(bigomega(n)==m,i=i+1));n}
%o get_k(m,p) = {local(i,n);i=0;n=1;while(n<p,n=n+1;if(bigomega(n)==m,i=i+1));i};
%o A209934(n) = {local(m,k,k_old);m=3;k_old=get_k(2,get_p(1,n)^2);k=get_k(4,get_p(2,n)^2);while(k<k_old,k_old=k;k=get_k(2*m,get_p(m,n)^2);m=m+1);k} \\ _Michael B. Porter_, Mar 20 2012
%Y Cf. A058933, A078840, A078843, A091538.
%K nonn
%O 1,2
%A _Daniel Tisdale_, Mar 15 2012
%E Edited, correcting the subscripting, by _Peter Munn_, Aug 04 2019