A097703 Numbers j such that m = 216*j + 108 satisfies sigma(m) != 2*usigma(m).

1, 4, 7, 10, 12, 13, 16, 19, 22, 24, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58, 60, 61, 62, 64, 67, 70, 73, 76, 79, 82, 84, 85, 87, 88, 91, 94, 97, 100, 103, 106, 109, 112, 115, 118, 121, 122, 124, 127, 130, 133, 136, 137, 139, 142, 144, 145, 148, 151, 154, 157

Offset: 1

Author

Ralf Stephan, Aug 26 2004

Keywords

• nonn

Comments

Conjecture: all numbers of form 3k + 1 are here. Other terms are listed in A097704.

From Amiram Eldar, Aug 31 2024: (Start)

The conjecture is true. If j = 3*k+1, then m = 324*(2*k+1). Let e = A007949(2*k+1) >= 0, so 2*k+1 = 3^e * i and i coprime to 6. Then sigma(m)/(2 * usigma(m)) = (7/20) * (3^(e+5)-1)/(3^(e+4)+1) * sigma(i)/usigma(i) >= 847/820 > 1, because sigma(i)/usigma(i) >= 1 for all i.

If m = 216*j + 108 = 108*(2*j+1) then sigma(m) = 2*usigma(m) if and only if 2*j+1 is a squarefree number coprime to 3 (see A097702), i.e., 2*j+1 is a term of A276378. Therefore this sequence consists of numbers j such that 2*j+1 is either a multiple of 3 or nonsquarefree (or both). (End)

Links

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

Crossrefs

Complement of A097702.

Cf. A000203, A007949, A013929, A034448, A063880, A097704, A276378.

Programs

Mathematica

usigma[n_] := Block[{d = Divisors[n]}, Plus @@ Select[d, GCD[ #, n/# ] == 1 &]]; Complement[ Range[157], (Select[ Range[37000], DivisorSigma[1, # ] == 2usigma[ # ] &] - 108)/216] (* Robert G. Wilson v, Aug 28 2004 *)

PARI

is(k) = {my(f = factor(216*k + 108)); sigma(f) != 2 * prod(i = 1, #f~, 1 + f[i,1]^f[i,2]);} \\ Amiram Eldar, Aug 31 2024