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A340816 Numbers k for which sigma(k^2)/k^2 reaches a new record, where sigma = A000203.

%I A340816 #14 Sep 02 2023 17:35:52
%S A340816 1,2,4,6,12,24,30,60,120,180,210,360,420,840,1260,2520,4620,9240,
%T A340816 13860,27720,55440,60060,120120,180180,360360,720720,1441440,1801800,
%U A340816 2042040,3063060,6126120,12252240,24504480,30630600,36756720,38798760,58198140,116396280
%N A340816 Numbers k for which sigma(k^2)/k^2 reaches a new record, where sigma = A000203.
%C A340816 Appears to be almost identical to A308471.
%H A340816 MathOverflow, <a href="https://mathoverflow.net/questions/381843/on-superabundant-like-numbers">On superabundant-like numbers</a>
%e A340816 a(1) = 1 with sigma(1^2)/1^2 = 1.
%e A340816 a(2) = 2 with sigma(2^2)/2^2 = 7/4 > 1.
%e A340816 3 is not in the sequence because sigma(3^2)/3^2 = 13/9 < 7/4.
%e A340816 a(3) = 4 with sigma(4^2)/4^2 = 31/16 > 7/4.
%p A340816 wmax:= 0: R:= NULL:
%p A340816 for n from 1 to 10^6 do
%p A340816   w:= numtheory:-sigma(n^2)/n^2;
%p A340816   if w > wmax then
%p A340816     wmax:= w; R:= R, n;
%p A340816   fi;
%p A340816 od:
%p A340816 R;
%t A340816 DeleteDuplicates[Table[{k,DivisorSigma[1,k^2]/k^2},{k,31*10^5}],GreaterEqual[#1[[2]],#2[[2]]]&][[;;,1]] (* The program generates the first 30 terms of the sequence. To generate more increase the k constant (now set at 31*10^5) but the program may take a long time to run. *) (* _Harvey P. Dale_, Sep 02 2023 *)
%Y A340816 Cf. A000203, A004394, A308471.
%K A340816 nonn
%O A340816 1,2
%A A340816 _Robert Israel_, Jan 22 2021