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A332993 a(1) = 1, for n > 1, a(n) = n + a(A032742(n)).

%I A332993 #25 Apr 05 2020 18:49:49
%S A332993 1,3,4,7,6,10,8,15,13,16,12,22,14,22,21,31,18,31,20,36,29,34,24,46,31,
%T A332993 40,40,50,30,51,32,63,45,52,43,67,38,58,53,76,42,71,44,78,66,70,48,94,
%U A332993 57,81,69,92,54,94,67,106,77,88,60,111,62,94,92,127,79,111,68,120,93,113,72,139,74,112,106,134,89,131,80,156,121
%N A332993 a(1) = 1, for n > 1, a(n) = n + a(A032742(n)).
%C A332993 Sum of those divisors of n that can be obtained by repeatedly taking the largest proper divisor (of previous such divisor, starting from n, which is included in the sum), up to and including the terminal 1.
%H A332993 Antti Karttunen, <a href="/A332993/b332993.txt">Table of n, a(n) for n = 1..16384</a>
%H A332993 Antti Karttunen, <a href="/A332993/a332993.txt">Data supplement: n, a(n) computed for n = 1..65537</a>
%F A332993 a(1) = 1; and for n > 1, a(n) = n + a(A032742(n)).
%F A332993 a(n) = n + A006022(n).
%F A332993 a(n) = A332994(n) + A333791(n).
%F A332993 a(n) = A000203(n) - A333783(n).
%F A332993 It seems that for all n >= 1, a(n) <= A073934(n) <= A333794(n).
%e A332993 a(18) = 18 + 18/2 + 9/3 + 3/3 = 18 + 9 + 3 + 1 = 31.
%o A332993 (PARI) A332993(n) = if(1==n,n,n + A332993(n/vecmin(factor(n)[,1])));
%Y A332993 Cf. A000203, A006022, A032742, A064097, A073934, A332994, A333783, A333791, A333794.
%K A332993 nonn
%O A332993 1,2
%A A332993 _Antti Karttunen_, Apr 04 2020