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

0,2

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

Links

Table of n, a(n) for n=0..43.

Formula

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

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 *)

Crossrefs

Cf. A000005, A005179, A038547, A183063.

Sequence in context: A082906 A364768 A204088 * A085083 A076745 A007374

Adjacent sequences: A187938 A187939 A187940 * A187942 A187943 A187944

Keyword

nonn

Author

Juri-Stepan Gerasimov, Mar 16 2011

Status

approved