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Numbers m such that the geometric mean of tau(m) and sigma(m) is an integer.
2

%I #37 Sep 08 2022 08:46:24

%S 1,7,17,22,30,31,71,94,97,115,119,127,138,154,164,165,199,210,214,217,

%T 232,241,260,265,318,337,343,374,382,449,497,510,513,517,527,577,647,

%U 658,668,679,682,705,745,759,805,862,881,889,894,930,966,967,996,1102,1125

%N Numbers m such that the geometric mean of tau(m) and sigma(m) is an integer.

%C 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.

%C 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.

%C 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.

%e sigma(30) = 72 and tau(30) = 8, sigma(30)*tau(30) = 576 = 24^2, hence 30 is a term.

%p filter:= s -> issqr(sigma(s)*tau(s)) : select(filter, [$1..2500]);

%t Select[Range[1000], IntegerQ @ Sqrt[DivisorSigma[0, #] * DivisorSigma[1, #]] &] (* _Amiram Eldar_, Sep 27 2019 *)

%o (Magma) [k:k in [1..1150]| IsSquare(#Divisors(k)*DivisorSigma(1,k))]; // _Marius A. Burtea_, Sep 27 2019

%o (PARI) isok(m) = issquare(numdiv(m)*sigma(m)); \\ _Michel Marcus_, Sep 27 2019

%Y Cf. A000005 (tau), A000203 (sigma), A064840 (tau*sigma).

%Y 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).

%Y Subsequences: A066436, A256152.

%K nonn

%O 1,2

%A _Bernard Schott_, Sep 27 2019