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A327830
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Numbers m such that the geometric mean of tau(m) and sigma(m) is an integer.
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2
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1, 7, 17, 22, 30, 31, 71, 94, 97, 115, 119, 127, 138, 154, 164, 165, 199, 210, 214, 217, 232, 241, 260, 265, 318, 337, 343, 374, 382, 449, 497, 510, 513, 517, 527, 577, 647, 658, 668, 679, 682, 705, 745, 759, 805, 862, 881, 889, 894, 930, 966, 967, 996, 1102, 1125
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OFFSET
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1,2
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COMMENTS
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The first 20 terms of this sequence are also the first 20 terms of A144695: m such that sigma(m)/tau(m) is a square. Indeed, if sigma(m)/tau(m) is a square then sigma(m)*tau(m) is also a square, but the converse is false. These counterexamples are in A327831; the first one is a(21) = 232.
The primes p of the form 2*k^2 - 1: 7, 17, 31, 71, ... (A066436) form a subsequence because sigma(p) * tau(p) = (2*k)^2.
Another subsequence consists of the terms m such that sigma(m) and tau(m) are both squares; this occurs when m is the product of two distinct primes p*q, p < q where sigma(m) = (p+1)*(q+1) is a square and tau(m) = 4. The first few terms are 22, 94, 115, 119, 214, ... They are in A256152.
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LINKS
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EXAMPLE
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sigma(30) = 72 and tau(30) = 8, sigma(30)*tau(30) = 576 = 24^2, hence 30 is a term.
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MAPLE
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filter:= s -> issqr(sigma(s)*tau(s)) : select(filter, [$1..2500]);
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MATHEMATICA
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Select[Range[1000], IntegerQ @ Sqrt[DivisorSigma[0, #] * DivisorSigma[1, #]] &] (* Amiram Eldar, Sep 27 2019 *)
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PROG
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(Magma) [k:k in [1..1150]| IsSquare(#Divisors(k)*DivisorSigma(1, k))]; // Marius A. Burtea, Sep 27 2019
(PARI) isok(m) = issquare(numdiv(m)*sigma(m)); \\ Michel Marcus, Sep 27 2019
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CROSSREFS
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Cf. A011257 (similar, with phi(m) and sigma(m)), A144695 (sigma(m)/tau(m) is a square), A327831 (sigma(m) * tau(m) is a square but sigma(m)/tau(m) is not an integer).
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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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