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A339055
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Values taken by d(k^2)/d(k) where d(k) is the number of divisors of k and when this ratio is an integer.
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3
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1, 3, 3, 3, 3, 3, 3, 3, 3, 5, 3, 3, 3, 3, 5, 5, 3, 3, 3, 3, 3, 3, 5, 5, 5, 3, 3, 5, 3, 5, 3, 3, 3, 3, 3, 5, 5, 3, 3, 5, 5, 5, 3, 3, 3, 3, 5, 3, 3, 3, 5, 3, 3, 3, 5, 5, 5, 5, 3, 3, 3, 3, 3, 5, 5, 5, 5, 3, 3, 5, 3, 5, 5, 3, 3, 3, 3, 5, 3, 3, 3, 3, 5, 3, 5, 3, 5, 3, 5
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OFFSET
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1,2
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COMMENTS
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This sequence was the subject of the 3rd problem, proposed by Belarus during the 39th International Mathematical Olympiad in 1998 at Taipei (Taiwan) [see the link IMO].
If the prime signature of k is (u_1, u_2, ... , u_q) then d(k^2)/d(k) = Product_{i=1..q} (2*u_i+1)/(u_i+1); now, by a fine induction, we prove that every positive odd integer is a product of fractions of type (2u+1)/(u+1). Hence, the set of possible integer values of the data coincides with the set of all positive odd integers [see Marcin E. Kuczma reference]. The smallest integers k such that d(k^2)/d(k) = n-th odd integer are in A339056.
a(1) = 1 then from a(2) to a(234) the ratio takes only the values 3 and 5.
a(n) = 3 for numbers k whose prime signature is (4, 2) and the smallest such integer is 144 = 2^4 * 3^2 corresponding to a(2) = 3.
a(n) = 5 for numbers k whose prime signature is (4, 2, 2) and the smallest such integer is 3600 = 2^4 * 3^2 * 5^2 corresponding to a(10) = 5.
a(n) = 9 for numbers k whose prime signature is (4, 4, 2, 2) and the smallest such integer is 1587600 = 2^4 * 3^4 * 5^2 * 7^2 corresponding to a(235) = 9.
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REFERENCES
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Marcin E. Kuczma, International Mathematical Olympiads, 1986-1999, The Mathematical Association of America, 2003, pages 134-135.
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LINKS
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The IMO Compendium, Problem 3, 39th IMO 1998.
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FORMULA
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EXAMPLE
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The 4th number k such that d(k^2)/d(k) is an integer is A217584(4) = 400, 400 has 15 divisors and 400^2 = 160000 has 45 divisors, so, a(4) = 45/15 = 3.
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MAPLE
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for n from 1 to 600 do
q:= tau(n^4)/tau(n^2);
if q = floor(q) then print(q); else fi; od:
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MATHEMATICA
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Select[DivisorSigma[0, #^2]/DivisorSigma[0, #] & /@ Range[10^5], IntegerQ] (* Amiram Eldar, Nov 23 2020 *)
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PROG
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(PARI) lista(nn) = {my(q); for (n=1, nn, if (denominator(q=numdiv(n^2)/numdiv(n)) == 1, print1(q, ", ")); ); }
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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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