A051521 - OEIS

%I #41 Apr 18 2022 04:13:10

%S 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,

%T 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,

%U 1,1,2,0,2,1,1,2,1,1,2,2,2,1,2,2,1,1,1

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

%C From _Jianing Song_, Nov 25 2018: (Start)

%C 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.

%C 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.

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

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

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

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

%H T. D. Noe, <a href="/A051521/b051521.txt">Table of n, a(n) for n = 1..1000</a>

%F a(A036763(n)) = 0; a(A051278(n)) = 1; a(A051279(n)) = 2. - _Reinhard Zumkeller_, Dec 28 2011

%e There are a(1) = 2 numbers k for which k/d(k) = 1, namely k = 1 and k = 2.

%e There are a(2) = 2 numbers k for which k/d(k) = 2, namely k = 8 and k = 12.

%e There are a(3) = 3 numbers k for which k/d(k) = 3, namely k = 9, 18 and 24.

%t 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 *)

%o (Haskell)

%o a051521 n = length [k | k <- [1..4*n^2],

%o                         let d = a000005 k, divMod k d == (n,0)]

%o -- _Reinhard Zumkeller_, Dec 28 2011

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

%K nonn

%O 1,1

%A _David W. Wilson_