A097703 - OEIS

%I #15 Aug 31 2024 04:27:33

%S 1,4,7,10,12,13,16,19,22,24,25,28,31,34,37,40,43,46,49,52,55,58,60,61,

%T 62,64,67,70,73,76,79,82,84,85,87,88,91,94,97,100,103,106,109,112,115,

%U 118,121,122,124,127,130,133,136,137,139,142,144,145,148,151,154,157

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

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

%C From _Amiram Eldar_, Aug 31 2024: (Start)

%C 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.

%C 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)

%H Amiram Eldar, <a href="/A097703/b097703.txt">Table of n, a(n) for n = 1..10000</a>

%t 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 *)

%o (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

%Y Complement of A097702.

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

%K nonn

%O 1,2

%A _Ralf Stephan_, Aug 26 2004