%I #12 Nov 11 2019 09:27:07
%S 6,68,828,3444,6237,10755,14595,15687,16056,18837,27495,42228,44408,
%T 46548,50436,55750,68103,88730,108560,129105,132924,161490,168730,
%U 168756,181935,189112,249543,260540,273195,299115,304425,313677,345975,369472,424215,430402
%N Let s denote the sum of the deficient numbers in the aliquot parts of x. Sequence lists numbers x such that sigma(s) = usigma(x), where usigma(x) is the sum of the unitary divisors of x (A034448).
%H Amiram Eldar, <a href="/A258134/b258134.txt">Table of n, a(n) for n = 1..1500</a>
%e Aliquot parts of 6 are 1, 2, 3, which are all deficient numbers. Then sigma(1+2+3) = sigma(6) = 12 = usigma(6).
%e Aliquot parts of 68 are 1, 2, 4, 17, 34, twhich are all deficient numbers. Then sigma(1+2+4+17+34) = sigma(58) = 90 = usigma(68).
%e Aliquot parts of 828 are 1, 2, 3, 4, 6, 9, 12, 18, 23, 36, 46, 69, 92, 138, 207, 276, 414. Deficient numbers are 1, 2, 3, 4, 9, 23, 46, 69, 92 and 207. Then sigma(1+2+3+4+9+23+46+69+92+207) = sigma(456) = 1200 = usigma(828).
%p with(numtheory); P:=proc(q) local a,b,d,k,n; for n from 1 to q do
%p a:=sort([op(divisors(n))]); b:=0; d:=0;
%p for k from 1 to nops(a)-1 do if sigma(a[k])<2*a[k] then b:=b+a[k]; fi; od;
%p for k from 1 to nops(a) do if gcd(a[k],n/a[k])=1 then d:=d+a[k]; fi; od;
%p if sigma(b)=d then print(n); fi; od; end: P(10^9);
%Y Cf. A000203, A034448.
%K nonn
%O 1,1
%A _Paolo P. Lava_, May 21 2015