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A275816
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Least number k such that the number of its divisors is n times, with n>1, the number of its prime factors, counted without multiplicity.
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2
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2, 4, 8, 16, 32, 64, 128, 256, 432, 1024, 864, 4096, 1728, 2592, 3456, 65536, 6912, 262144, 10368, 14400, 27648, 4194304, 21600, 32400, 110592, 50400, 43200, 268435456, 64800, 1073741824, 86400, 230400, 1769472, 129600, 151200, 68719476736, 7077888, 921600
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OFFSET
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2,1
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COMMENTS
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The number of divisors of p^(n-1) is n times the number of prime factors of p^(n-1), where p is prime. It means that a solution exists for every n>1.
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LINKS
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FORMULA
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EXAMPLE
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a(10) = 432 because the number of divisors of 432 is 20, the number of different prime factors of 432 is 2 (2, 3), and 20 = 10 * 2.
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MAPLE
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with(numtheory): P:=proc(q) local k, n; for n from 2 to q do for k from 1 to 2^(n-1) do
if tau(k)=n*nops(factorset(k)) then print(k); break; fi; od; od; end: P(10^9);
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MATHEMATICA
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a[n_] := Block[{k = 2}, If[PrimeQ[n], 2^n/2, While[ DivisorSigma[0, k]/ PrimeNu[k] != n, k++]; k]]; a /@ Range[2, 25] (* Giovanni Resta, Nov 16 2016 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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