A051521 Number of ways to represent n as k/d(k), where d(k) = A000005(k) is the number of divisors of k.

2, 2, 3, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 1, 1, 2, 2, 0, 2, 1, 1, 1, 2, 2, 3, 1, 0, 2, 2, 0, 2, 1, 1, 1, 1, 1, 2, 1, 1, 3, 2, 1, 2, 2, 0, 1, 2, 2, 3, 1, 1, 2, 2, 3, 1, 2, 1, 1, 2, 1, 2, 1, 0, 0, 1, 1, 2, 2, 1, 1, 2, 0, 2, 1, 1, 2, 1, 1, 2, 2, 2, 1, 2, 2, 1, 1, 1

Offset: 1

Author

David W. Wilson

Keywords

• nonn

Comments

From Jianing Song, Nov 25 2018: (Start)

a(9p) = 0 for all primes p. Here is a brief proof: a(18) = a(27) = a(45) = a(63) = 0. Now let p be a prime >= 11.

If there is an x such that d(9p*x) = x, let x = p^a*3^b*y, gcd(p, y) = gcd(3, y) = 1, then p^a*3^b*y = d(p^(a+1)*3^(b+2)*y) = (a + 2)*(b + 3)*d(y). Since y >= d(y), we must have (a + 2)*(b + 3) >= p^a*3^b >= 11^a*3^b. If a >= 1, then 3 >= (b + 3)/3^b >= 11^a/(a + 2) >= 11/3, a contradiction. So a = 0. 3^b/(b + 3) <= 2, so b = 0, 1, 2.

Case (i): b = 0, then y = 6*d(y), which has a unique solution y = 72. But gcd(3, 72) != 1, a contradiction,

Case (ii): b = 1, then y = (8/3)*d(y), which has no solution.

Case (iii): b = 2, then y = (10/9)*d(y), which has no solution.

Similarly, it can be proved that a(81p) = 0 for all primes p. (End)

Examples

			There are a(1) = 2 numbers k for which k/d(k) = 1, namely k = 1 and k = 2.
There are a(2) = 2 numbers k for which k/d(k) = 2, namely k = 8 and k = 12.
There are a(3) = 3 numbers k for which k/d(k) = 3, namely k = 9, 18 and 24.
		

Links

• T. D. Noe, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A000005 (number of divisors), A033950, A036762, A036763 (indices of 0s), A036764, A051278 (indices of 1s), A051279 (indices of 2s).

Programs

Haskell

a051521 n = length [k | k <- [1..4*n^2],
                        let d = a000005 k, divMod k d == (n,0)]
-- Reinhard Zumkeller, Dec 28 2011

Mathematica

a[n_] := Count[Table[n == k/DivisorSigma[0, k], {k, 1, 4*n^2}], True]; Table[a[n], {n, 1, 100}]  (* Jean-François Alcover, Oct 22 2012 *)

Formula

a(A036763(n)) = 0; a(A051278(n)) = 1; a(A051279(n)) = 2. - Reinhard Zumkeller, Dec 28 2011