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Odd perfect numbers

It is not known whether odd perfect numbers exist or not! Mathematicians have been able to prove all sorts of necessary (but not sufficient) requirements for the existence of such numbers without being able to prove either that they do exist or that they don't exist.

Search for odd perfect numbers

For an odd number
 n
to be perfect, we must have
${\displaystyle n=kp^{a},\quad (p^{a},k)=1,}$
where
 k, k > 1,
is an odd factor and
 p a
is an odd prime power factor which is coprime to
 k
, such that
${\displaystyle \sigma (n)=\sigma (k)\,\sigma (p^{a})=\sigma (k)\,\left({\frac {p^{a+1}-1}{p-1}}\right)=2n=2kp^{a},}$

or

${\displaystyle {\frac {\sigma (k)}{2k}}={\frac {p^{a}\,(p-1)}{p^{a+1}-1}},}$
where
 σ (n)
is the sum of divisors of
 n
. One way to search for odd perfect numbers is to consider, for each deficient odd
 k, k > 1,
if some odd prime power
 p a
which is coprime to
 k
, when multiplied by
 k
yields an odd perfect number. (Since all positive multiples of abundant numbers are also abundant, there is no point in considering abundant odd
 k
's.)

One might consider the simpler case

${\displaystyle n=kp,\quad p\nmid k,}$
where
 k, k > 1,
is an odd factor and
 p
is an odd prime factor which does not divide
 k
, such that
${\displaystyle \sigma (n)=\sigma (k)\,\sigma (p)=\sigma (k)\,(p+1)=2n=2kp,}$
where
 σ (n)
is the sum of divisors of
 n
.

Equivalently, we must have

${\displaystyle {\frac {2k}{\sigma (k)}}={\frac {p+1}{p}}=1+{\frac {1}{p}},}$
thus we need to find an odd
 k, k > 1,
such that
${\displaystyle p={\frac {\sigma (k)}{2k-\sigma (k)}}}$
where
 p
happens to be an odd prime.
A008438 Sum of divisors of
 2n  +  1, n   ≥   0
.
{1, 4, 6, 8, 13, 12, 14, 24, 18, 20, 32, 24, 31, 40, 30, 32, 48, 48, 38, 56, 42, 44, 78, 48, 57, 72, 54, 72, 80, 60, 62, 104, 84, 68, 96, 72, 74, 124, 96, 80, 121, 84, 108, 120, ...}

Odd spoof perfect numbers

Main article page: Descartes number

In 1638, Descartes found the following "odd spoof perfect number" (no other has ever been found!):

${\displaystyle 198585576189=3^{2}\cdot 7^{2}\cdot 11^{2}\cdot 13^{2}\cdot 19^{2}\cdot 61,\,}$

that is odd and perfect only if you suppose (incorrectly) that

${\displaystyle 22021=19^{2}\cdot 61\,}$

is a "spoof-prime factor," giving the "spoof prime factorization"

${\displaystyle 198585576189=(3\cdot 7\cdot 11\cdot 13)^{2}\cdot 22021=3003^{2}\cdot 22021=9018009\cdot 22021,\,}$

for which the "freestyle sum of divisors" (i.e. the sum of divisors function where one is free to consider some composite factors as "spoof-prime factors") yields

{\displaystyle {\begin{aligned}\sigma _{\rm {freestyle}}(n)&={\bigg (}{\frac {3^{3}-1}{3-1}}{\bigg )}{\bigg (}{\frac {7^{3}-1}{7-1}}{\bigg )}{\bigg (}{\frac {{11}^{3}-1}{11-1}}{\bigg )}{\bigg (}{\frac {{13}^{3}-1}{13-1}}{\bigg )}(22021+1)={\bigg (}{\frac {26}{2}}{\bigg )}{\bigg (}{\frac {342}{6}}{\bigg )}{\bigg (}{\frac {1330}{10}}{\bigg )}{\bigg (}{\frac {2196}{12}}{\bigg )}(22022)=13\cdot 57\cdot 133\cdot 183\cdot 22022\\&=13\cdot (3\cdot 19)\cdot (7\cdot 19)\cdot (3\cdot 61)\cdot (2\cdot 7\cdot 11^{2}\cdot 13)=2\cdot (3^{2}\cdot 7^{2}\cdot 11^{2}\cdot 13^{2})\cdot ({19}^{2}\cdot 61)=2\cdot (3^{2}\cdot 7^{2}\cdot 11^{2}\cdot 13^{2}\cdot 22021)=2n.\end{aligned}}\,}

References

• Eric W. Weisstein, CRC Encyclopedia of Mathematics, Volume II, CRC Press (2009), p. 2730.