A187941 Least number with exactly n even divisors.

1, 2, 4, 8, 12, 32, 24, 128, 48, 72, 96, 2048, 120, 8192, 384, 288, 240, 131072, 360, 524288, 480, 1152, 6144, 8388608, 720, 2592, 24576, 1800, 1920, 536870912, 1440, 2147483648, 1680, 18432, 393216, 10368, 2520, 137438953472, 1572864, 73728, 3360, 2199023255552, 5760, 8796093022208

Offset: 0

Author

Juri-Stepan Gerasimov, Mar 16 2011

Keywords

• nonn

Comments

The only odd term in the sequence is 1, having zero even divisors. All larger odd numbers also have zero even divisors.

Conjecture: a(n) = 2^n only if n is prime or if n = 1.

If the prime factorization of a number is 2^k p1^e1...pr^er, then the number of even divisors is k*(e1+1)...(er+1). Hence, to find the least number having n even divisors, factor n and determine k, e1,..., er such that n = k*(e1+1)...(er+1). Then a(n) will have the form 2^k 3^e1 5^e2.... It is obvious that if n is prime, then a(n) = 2^n. Similarly, if n is twice an odd prime p, then a(n) = 2^p * 3. - T. D. Noe, Mar 16 2011

Crossrefs

Cf. A000005, A005179, A038547, A183063.

Programs

Mathematica

evenDivSigma[n_Integer] := Length[Select[Divisors[n], EvenQ]]; Flatten[Table[Take[Select[Range[2, 10^6, 2], evenDivSigma[#] == n &], 1], {n, 20}]] (* Alonso del Arte, Mar 16 2011 *)

Formula

a(n) = 2 * A005179(n) for n > 0.