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A051521
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Number of ways to represent n as k/d(k), where d(k) = A000005(k) is the number of divisors of k.
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6
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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
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OFFSET
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1,1
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COMMENTS
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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)
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LINKS
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FORMULA
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EXAMPLE
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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.
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MATHEMATICA
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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 *)
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PROG
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(Haskell)
a051521 n = length [k | k <- [1..4*n^2],
let d = a000005 k, divMod k d == (n, 0)]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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