%I #51 Jun 18 2019 07:23:59
%S 6,24,28,126,496,8128,5594428,33550336,8589869056,17589794838,
%T 35439846824,49380301744
%N Numbers k such that k = sum of nonabundant proper divisors of k (A294888).
%C Naturally, all the terms of A000396, including 137438691328, are in this sequence. - _Antti Karttunen_, Dec 01 2017
%C Thus, if there are infinitely many Mersenne primes, then this sequence is also, by definition of even perfect numbers, infinite. - _Iain Fox_, Dec 02 2017
%C All non-perfect terms are abundant. Proof: Assume d is a deficient number in this sequence. Because multiples of abundant numbers are abundant, d cannot have an abundant divisor, thus all its divisors are nonabundant. Since d is in this sequence, the sum of its proper divisors, which are all nonabundant, must equal d. However, if this were true, then d would be perfect. Therefore, this sequence contains no deficient numbers. - _Iain Fox_, Dec 07 2017
%C Questions from _Iain Fox_, Dec 07 2017: (Start)
%C Are there an infinite number of abundant terms?
%C Are all abundant terms in this sequence even?
%C (End)
%C No other terms up to 10^10. - _Iain Fox_, Dec 07 2017
%C a(13) > 6*10^10. - _Giovanni Resta_, Dec 11 2017
%C In comparison, the numbers which are the sum of their abundant proper divisors seems to be scarcer: up to 6*10^10 only 19514300 and 16333377500 have this property. - _Giovanni Resta_, Dec 11 2017
%C From _Iain Fox_, Dec 11 2017: (Start)
%C The first abundant term without a perfect divisor is 35439846824.
%C This term and any other abundant terms without perfect divisors are also terms in A125310.
%C (End)
%H <a href="/index/O#opnseqs">Index entries for sequences where any odd perfect numbers must occur</a>
%o (PARI) isok(n) = sumdiv(n, d, if ((d<n) && (sigma(d)<=(2*d)), d)) == n; \\ _Michel Marcus_, Nov 17 2017
%o (PARI) normalize(f)=f=select(v->v[2],f~)~;if(vecmax(matsize(f)),f,factor(1));
%o is(n,f=factor(n))=
%o {
%o my(p=Mat(f[,1]),g,s);
%o forvec(v=apply(k->[0,k],f[,2]~),
%o g=normalize(concat(p,v~));
%o if(sigma(g,-1)<=2,
%o s+=factorback(g)
%o );
%o );
%o s==if(sigma(f,-1)>2,n,2*n);
%o }
%o forfactored(n=6,10^9, if(is(n[1],n[2]), print1(n[1]", "))) \\ _Charles R Greathouse IV_, Dec 08 2017
%Y Fixed points of A294888.
%Y Subsequence of A005835; A000396 is a subsequence.
%Y Cf. A125310.
%K hard,nonn,more
%O 1,1
%A _Antti Karttunen_, Nov 14 2017
%E a(9) from _Iain Fox_, Dec 07 2017
%E a(10)-a(12) from _Giovanni Resta_, Dec 11 2017
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