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

1, 3, 12, 60, 420, 3780, 41580, 540540, 8648640, 147026880, 2793510720, 64250746560, 1606268664000, 46581791256000, 1444035528936000, 53429314570632000, 2190601897395912000, 94195881588024216000, 4427206434637138152000, 216933115297219769448000, 11497455110752647780744000

Offset: 1

Author

Juri-Stepan Gerasimov, Feb 06 2025

Keywords

• nonn

Comments

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]

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

Examples

			1 is in the sequence because tau(1*2) = tau(2) = 2 = 2^1;
3 is in the sequence because tau(3*2) = tau(6) = 4 = 2^2;
12 is in the sequence because tau(12*2) = tau(24) = 8 = 2^3;
		

Crossrefs

Cf. A000005, A000028, A005843, A037992, A099777.

Formula

a(n) = A037992(n)/2. - Amiram Eldar, Feb 06 2025