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A294900
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Numbers k such that k = sum of nonabundant proper divisors of k (A294888).
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3
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6, 24, 28, 126, 496, 8128, 5594428, 33550336, 8589869056, 17589794838, 35439846824, 49380301744
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OFFSET
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1,1
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COMMENTS
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Naturally, all the terms of A000396, including 137438691328, are in this sequence. - Antti Karttunen, Dec 01 2017
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
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
Questions from Iain Fox, Dec 07 2017: (Start)
Are there an infinite number of abundant terms?
Are all abundant terms in this sequence even?
(End)
No other terms up to 10^10. - Iain Fox, Dec 07 2017
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
The first abundant term without a perfect divisor is 35439846824.
This term and any other abundant terms without perfect divisors are also terms in A125310.
(End)
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LINKS
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PROG
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(PARI) isok(n) = sumdiv(n, d, if ((d<n) && (sigma(d)<=(2*d)), d)) == n; \\ Michel Marcus, Nov 17 2017
(PARI) normalize(f)=f=select(v->v[2], f~)~; if(vecmax(matsize(f)), f, factor(1));
is(n, f=factor(n))=
{
my(p=Mat(f[, 1]), g, s);
forvec(v=apply(k->[0, k], f[, 2]~),
g=normalize(concat(p, v~));
if(sigma(g, -1)<=2,
s+=factorback(g)
);
);
s==if(sigma(f, -1)>2, n, 2*n);
}
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CROSSREFS
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KEYWORD
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hard,nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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