A327831 -id:A327831 - cp's OEIS frontend

A327830 Numbers m such that the geometric mean of tau(m) and sigma(m) is an integer.

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

Offset: 1

Bernard Schott, Sep 27 2019

nonn

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.

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

Cf. A000005 (tau), A000203 (sigma), A064840 (tau*sigma).
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).
Subsequences: A066436, A256152.

[k:k in [1..1150]| IsSquare(#Divisors(k)*DivisorSigma(1,k))]; // Marius A. Burtea, Sep 27 2019

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

Select[Range[1000], IntegerQ @ Sqrt[DivisorSigma[0, #] * DivisorSigma[1, #]] &] (* Amiram Eldar, Sep 27 2019 *)

isok(m) = issquare(numdiv(m)*sigma(m)); \\ Michel Marcus, Sep 27 2019

A341940 Numbers m such that phi(m)*tau(m) is a square but phi(m)/tau(m) is not the square of an integer.

Original entry on oeis.org

54, 1026, 1280, 2187, 2304, 3840, 4352, 6750, 8802, 9072, 9900, 12500, 13056, 13718, 17496, 18700, 21870, 25856, 36900, 37500, 41154, 41553, 47682, 50432, 56100, 57078, 65792, 69700, 77568, 78786, 79200, 84240, 100000, 102656, 103586, 111100, 117666, 125712

Offset: 1

Bernard Schott, Feb 24 2021

nonn

If phi(m)/tau(m) is a square of an integer (m is in A341939) then phi(m)*tau(m) is also a square (m is in A341938), but the converse is false. This sequence consists of these counterexamples (see the Examples section).

			phi(54) = 18, tau(54) = 8, phi(54)*tau(54) = 18*8 = 144 = 12^2 but phi(54)/tau(54) = 9/4 = (3/2)^2 is not the square of an integer, hence 54 is a term.
phi(1026) = 324, tau(1026) = 16, phi(1026)*tau(1026) = 324*16 = 5184 = 72^2 but phi(1026)/tau(1026) = 324/16 = 81/4 = (9/2)^2 is not the square of an integer, hence 1026 is another term.

Similar for: A327624 (phi(n) and sigma(n)), A327831 (sigma(n) and tau(n)).
Equals A341938 \ A341939.
Cf. A000005 (phi), A000010 (tau).

with(numtheory): filter:= r -> phi(r)/tau(r) <> floor(phi(r)/tau(r)) and issqr(phi(r)*tau(r)) : select(filter, [$1..50000]);

Select[Range[10^5], IntegerQ /@ Sqrt[{(e = EulerPhi[#])*(d = DivisorSigma[0, #]), e/d}] == {True, False} &] (* Amiram Eldar, Feb 24 2021 *)

isok(m) = my(x=eulerphi(m), y = numdiv(m)); issquare(x*y) && (denominator(x/y) != 1); \\ Michel Marcus, Feb 24 2021

cp's OEIS Frontend

A327830 Numbers m such that the geometric mean of tau(m) and sigma(m) is an integer.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Magma

Maple

Mathematica

PARI