A064180 Composite numbers k such that the sum of the proper divisors of k not including 1, (Chowla's function, A048050) and their product (A007956) are both perfect squares.

117, 208, 292, 320, 475, 539, 549, 567, 873, 964, 1737, 2107, 2692, 2997, 3573, 3904, 4477, 4802, 5275, 5284, 5968, 6057, 7267, 7488, 7492, 9189, 9457, 9475, 10084, 10377, 11072, 11728, 11737, 12717, 13769, 14373, 14692, 16219, 16399, 17397

Offset: 1

Author

Robert G. Wilson v, Oct 14 2001

Keywords

• easy
• nonn

Examples

			117 is in the sequence because the divisors of 117 are 1, 3, 9, 13, 39 and 117. Being squarefree itself, the product of divisors is a perfect square. The sum of the divisors in question, 3+9+13+39 = 64 and it is a perfect square.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000290, A048050, A007956.

Programs

Magma

[k:k in [1..18000]| not IsPrime(k) and IsSquare((&+Divisors(k))-1-k) and IsSquare((&*Divisors(k))/k) ]; // Marius A. Burtea, Jul 03 2019

Mathematica

Select[ Range[2, 25000], IntegerQ[ Sqrt[ Apply[ Plus, Delete[ Divisors[ # ], -1]] - 1]] && IntegerQ[ Sqrt[ Apply[ Times, Delete[ Divisors[ # ], -1]]]] && ! PrimeQ[ # ] & ]
aQ[n_] := CompositeQ[n] && IntegerQ[Sqrt[n^(DivisorSigma[0, n]/2 - 1)]] && IntegerQ[Sqrt[DivisorSigma[1, n] - 1 - n]]; Select[Range[18000], aQ] (* Amiram Eldar, Jul 03 2019 *)