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A380836 a(n) is the smallest k such that tau(2*k) is equal to 2^n, where tau = A000005.

%I A380836 #43 Feb 16 2025 22:30:38
%S A380836 1,3,12,60,420,3780,41580,540540,8648640,147026880,2793510720,
%T A380836 64250746560,1606268664000,46581791256000,1444035528936000,
%U A380836 53429314570632000,2190601897395912000,94195881588024216000,4427206434637138152000,216933115297219769448000,11497455110752647780744000
%N A380836 a(n) is the smallest k such that tau(2*k) is equal to 2^n, where tau = A000005.
%C A380836 a(n) = A000028(n+1) | a(n+1) ? [I interpret this as saying "Is it true that if a(n) = A000028(n+1) then a(n) divides a(n+1)?" - _N. J. A. Sloane_, Feb 15 2025]
%C A380836 For n > 2, a(n) is a Zumkeller number (A083207). Proof: Let p_r be the maximum prime in the prime factorization of a(n). First, in the prime factorization p_r must be to the power of one (otherwise we could build a smaller term with the same number of divisors, which would contradict the definition). Second, p_1 must be 2 and floor(log_2(p_r)) <= e_1, where e_1 is the exponent of p_1 (same reason as above). Therefore, there exists a power e <= e_1 such that p_1^e*p_r is a primitive Zumkeller number (see A180332). Then p_1^e1*p is a Zumkeller number (see Theorem 4.13 in Mahanta et al. JNT paper at A083207). Then a(n) = p_1*e_1*p_2^e_2*...*p_r is a Zumkeller number (see Corollary 5 in Rao/Peng JNT paper at A083207). - _Ivan N. Ianakiev_, Feb 15 2025
%F A380836 a(n) = A037992(n)/2. - _Amiram Eldar_, Feb 06 2025
%e A380836 1 is in the sequence because tau(1*2) = tau(2) = 2 = 2^1;
%e A380836 3 is in the sequence because tau(3*2) = tau(6) = 4 = 2^2;
%e A380836 12 is in the sequence because tau(12*2) = tau(24) = 8 = 2^3;
%Y A380836 Cf. A000005, A000028, A005843, A037992, A099777.
%K A380836 nonn
%O A380836 1,2
%A A380836 _Juri-Stepan Gerasimov_, Feb 06 2025