Prime numbers, and their digital roots.

Question

Edit

It is clear that this conjecture is false, in many, many circumstances, and I am grateful to the whole Math Stack Exchange community for helping me to see this.

Thank you!

Let $p \in \mathbb{P},$ where $\mathbb{P}$ is the set of prime numbers, and $S(x)$ be the sum of the digits of $x.$

Also, $S_{n}(x)$ is a function such that:

$S_{2}(x) = S(S(x)),$ and $S_{3}(x) = S(S(S(x))),$ et cetera.

Conjecture

If $S(p) > (10^{l} - 1), l \in \mathbb{Z},$ where $l$ is the smallest solution to $S_{l}(x) \leq 9.$ I believe that $\left(\sum_{i = 1}^{l}{S_{i}(x)}\right) \in \mathbb{P}.$

I have tried it for a few prime numbers, which means that, of course, it may not be true for all prime numbers that satisfy this conjecture, and I am not quite sure if 'conjecture' is the right term for such a statement, but I would like to know if one would be able to prove that this is, indeed, the case, or, whether it is nonsense.

Thank you.

The question would be much easier to understand if you trash the $m$, $f$, and $d$ functions and just write "Let $S(x)$ be the sum of the digits in $x$". — hmakholm left over Monica, Jun 21 '15 at 18:34
@HenningMakholm $m(x), f(n, x),$ and $d(n, x)$ are all functions that are required in order to calculate the $(n + 1)$th digit of $x$. — Taylor, Jun 21 '15 at 18:34
@barakmanos I have replaced it with $p,$ as it was my intention to do so in the beginning. — Taylor, Jun 21 '15 at 18:38
@MohammadAliBaydoun $S_{0}$ would not exist, as far as I know. It would not make sense for it to exist. — Taylor, Jun 21 '15 at 18:45
@Taylor Then why does the summation start from $i = 0$? :p (That's what I was hinting at in my first comment) — user123, Jun 21 '15 at 18:45
@MohammadAliBaydoun I have corrected this error. I am sorry about making so many errors with this. — Taylor, Jun 21 '15 at 18:46
You've edited your question and you dare to down-vote my answer??? Shame on you!!! And by the way, even for $n=1$, it fails as soon as $p$ reaches $89$. Do you want to set the limit on $n=2$ perhaps? — barak manos, Jun 21 '15 at 19:24

barak manos · Answer 1 · 2015-06-22T06:16:04.430

Your conjecture is false in several (probably infinitely many) cases:

p = 13 , n = 0 , t = 4
p = 17 , n = 0 , t = 8
p = 31 , n = 0 , t = 4
p = 53 , n = 0 , t = 8
p = 89 , n = 1 , t = 25

Here is a short Python script to illustrate this:

primes = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97]

def S(x):
    return sum(int(c) for c in str(x))

for p in primes:
    x = S(p)
    n = len(str(x))-1
    t = x
    for i in range(0,n):
        x = S(x)
        t += x
    if t not in primes:
        print 'p =', p, ', n =', n, ', t =', t

hmakholm left over Monica · Answer 2 · 2015-06-21T18:39:33.013

3

Your conjecture is false whenever $n=8, 17, 26,\ldots$.

By a well-known property of digital sums, the numbers $x$, $S(x)$, $S(S(x))$ and so forth are all equivalent modulo 9. Adding 9 such numbers will always yield something that is divisible by $9$.

edited Jun 21 '15 at 18:39

answered Jun 21 '15 at 18:38

hmakholm left over Monica

286,031

@Taylor: Whenever $n$ is one less than a multiple of $9$. In fact every $n$ that is one less than a multiple of $3$ is excluded by the same reasoning. Given this, I don't see any particular reason why it should be true for other $n$s either. – hmakholm left over Monica Jun 21 '15 at 18:40
$29$ is prime, and the sum of its digits is $11.$ The sum of the digits of $11$ is $2.$ $11 + 2 = 13,$ and $13 \in \mathbb{P}.$ – Taylor Jun 21 '15 at 18:41
$37$ is prime, and the sum of its digits is $10.$ The sum of the digits of $10$ is $1.$ $10 + 1 = 11,$ and $11 \in \mathbb{P}.$ – Taylor Jun 21 '15 at 18:42
$59$ is prime, and the sum of its digits is $14.$ The sum of the digits of $14$ is $5.$ $14 + 5 = 19,$ and $19 \in \mathbb{P}.$ – Taylor Jun 21 '15 at 18:44
@Taylor: To make your conjecture even half-way plausible, you will have to test it for all primes up to at least a few thousand. And preferably a few million. My personal feeling is that your conjecture is probably false, so you have to provide a lot of evidence to persuade me. – TonyK Jun 21 '15 at 18:47
Thank you, Henning, for suggesting this, and, because of this, and the post by @barakmanos, I am happy to conclude that my conjecture was, unfortunately, false in many cases.
I am grateful, as always, to the Math Stack Exchange community for helping me. Thank you, all!
– Taylor Jun 21 '15 at 19:45

Prime numbers, and their digital roots.

2 Answers2