%I #13 Feb 25 2022 11:24:52

%S 120,672,963,1036,264768,523776,459818240,1476304896

%N Numbers k such that A342925(k) = k + 3*A003415(k).

%C 2^33 < a(9) <= A005820(6) = 51001180160.

%t Select[Range[6*10^5], #3 == #1 + 3 #2 & @@ Prepend[Map[If[# < 2, 0, # Total[#2/#1 & @@@ FactorInteger[#]]] &, {#, DivisorSigma[1, #]}], #] &] (* _Michael De Vlieger_, Feb 25 2022 *)

%o (PARI)

%o A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));

%o A342925(n) = A003415(sigma(n));

%o isA342923(n) = (A342925(n)==(n+(3*A003415(n))));

%Y Cf. A003415, A342922, A342925.

%Y Cf. A005820 (subsequence).

%K nonn,more

%O 1,1

%A _Antti Karttunen_, Apr 08 2021

%E a(8) from _Antti Karttunen_, Feb 25 2022