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In aliquot sequences, am I right in saying that all perfect numbers when you add up the digits the result always add up to $10$ ( apart from the perfect no $6$?) if so, is this just a coincidence?

Example: "$496\mapsto 4+9+6=19\mapsto 1+9=10$".

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    Hi, Jonas, I did not explain myself very well at all. 4+9+6=19 1+9= 10 it seems to be the same with them all, unless I have made an error. Thank you so much for looking at this. Donna – 2017-01-17
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    Noticing something special about the second, third, and fourth number in the sequence is not enough to make a conjecture about the sequence as a whole. Here are a few more terms in the sequence: $6,28,496,8128,33550336,8589869056,137438691328,\dots$ Perhaps you should try to reword your statement further. Maybe you mean to suggest that perfect numbers are congruent to $1\pmod{9}$ – 2017-01-17
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    @JMoravitz: I guess it would be stronger than just congruence to $1\mod 9$, because it would imply you can't get to $100$ for example. – 2017-01-17
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    Googling `digit sums of perfect numbers` led to https://primes.utm.edu/notes/proofs/Theorem3.html – 2017-01-17
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    Perfect numbers are, as @JMoravitz suggests, congruent to $1$ modulo $9$, but I would be mildly surprised if they all passed through $10$ in their successive digital sums. – 2017-01-17
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    It is worth pointing out that the proofs talked about so far have been specifically for **even** perfect numbers and are not valid for odd perfect numbers (*if any actually exist, it remains an open problem to prove the existence or nonexistence of odd perfect numbers*). – 2017-01-17
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    @BrianTung: As long as the sum of digits exceeds $10$, all numbers equivalent to $1 \bmod 9$ will pass through some power of $10$ because those are the only numbers with sum of digits $1$. As OP seems not to have tried numbers large enough to have a sum of digits exceeding $100$ they will all go through $10$. – 2017-01-17
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    I do not see evidence OP didn't try larger numbers. It's true for the first 20 perfect numbers, and at that point the digit sums exceed 10,000, but this is still not good evidence. I don't have an idea how to try to prove or disprove other than finding a counterexample Mathematica can handle if there is one. – 2017-01-17
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    Plus, I don't have a calculator and added them up in my head ( which now hurts ) so I did get a bit tired by the 15th. Sadly, I am a lawyer, not a mathematician, and I think that you chaps are great. I have just always been fascinated by number patterns. Donna – 2017-01-17
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    @DonnaSkelly: You are on the internet, so you have calculators! How about Wolfram Alpha for example? – 2017-01-17
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    Will do, Jonas. This iPad is ancient and the signal is rubbish in the wilds of Scotland. I have to waive it out the window. – 2017-01-17
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    @JonasMeyer How interesting such a simple observation is so much harder to prove than just congruence to 1 mod 9. I wonder what else can be said about certain sets of numbers congruent modulo $c$, and whether they have a seemingly constant second-to-last digit sum as well. With a little research and programming a separate question might be in order. I'll look into this later – 2017-01-18
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    Checking all known perfect numbers in Mathematica, they all come to 10. E.g., the digit sum for the largest known perfect number is 201071323. Not surprising for such a small number of examples, but if there are infinitely many Mersenne primes I would (naively) be surprised if this always happens. – 2017-01-19
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    @JonasMeyer Wow, Jonas, isn't nature brilliant. Maybe we have found something here. I'm always looking out for patterns. Prime Numbers drive me nuts. Again, I'm sorry I'm a lawyer and not a mathematician. Thank you so much for all of your help. Keep me updated if anyone else finds anything. Donna Skelly. – 2017-01-19

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For any odd $k$, $(2^k-1)2^{k-1}$ is congruent to $1 \bmod 9$, and all even perfect numbers are of this form (although in one case, $6$, $k$ is even). Unless it happens that, in forming the ultimate digit sum for some perfect number, an intermediate total hits on a higher power of $10$, say $100$, it's likely that $10$ will be reached in the digit-sum sequence.

Since the order of $2\bmod 9$ is $6$, you have essentially demonstrated the truth of the above proposition by testing $k=3,5,7$ already.

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    It's speculated that there are an infinite, if widely-spaced, number of even perfect numbers. If so, then it seems inevitable that one such will in fact reach some higher power $10$ in the digit sum sequence and render the OP claim false. – 2017-01-17
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It is certainly true that even perfect numbers $>6$ are all congruent to $1\pmod 9$. Thus if you keep on adding the digits together until you get a single digit, that digit will always be $1$.

To see this, note that, as per Euclid, even perfect numbers have the form $n_p=2^{p-1}(2^p-1)$ for some prime $p$ such that $2^p-1$ is also prime.

Excluding $p=2$ (which gives us $6$) we may assume $p$ is odd.

We note that $n_3=28$ which is indeed $1\pmod 9$. Form now on we'll assume that $p>3$.

It is easy to verify that $2$ has order $6 \pmod 9$. $p$ odd and $>3$ implies that $\equiv \pm 1\pmod 6$ and it is now a straightforward exercise to confirm that in both cases $n_p\equiv 1 \pmod 9$.

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    @JonasMeyer's comment above is important here... the OP seems to be making a stronger claim than just congruence to $1 \pmod 9$... instead, the digit sum must ALWAYS reach $10$, and not $100$ or $1000$ for example – 2017-01-17
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    You are all brilliant. Thank you! Donna – 2017-01-17
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    @BrevanEllefsen Ah, didn't see that. I would be surprised were that true, but I don't know. – 2017-01-17
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    Euclid observed that if $2^p-1$ is prime, then $2^{p-1}(2^p-1)$ is perfect, but it was Euler who proved that even perfect numbers always have this form. – 2017-01-18
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An even number is perfect if and only if it is of the form $N=2^n\cdot(2^{n+1}-1)$ and $2^{n+1}-1$ is prime (Euler).

This implies that $n+1$ is also prime (although this is not a sufficient condition for $2^{n+1}-1$ to be prime).

On the other hand, if $S(n)$ is the sum of the digits of $n$, then $S(n)\equiv n\pmod 9$. So we need to see if $2^n\cdot (2^{n+1}-1)\equiv 1\pmod 9$.

We have that $2$ is a primitive root modulo $9$, that is, the first power of $2$ that is $1$ in $\Bbb Z_9$ is $2^6$. Now write $n=6q+r$. Since $n+1$ must be prime, if we assume that $n+1$ is not $2$ or $3$, $r+1$ must be $1$ or $5$, so $r$ is $0$ or $4$. Putting all together we get $$N\equiv2^r(2^{r+1}-1)=2^{2r+1}-2^r\equiv1\pmod 9$$ in both cases.