u/FormulaDriven

A comment by u/GoldenMuscleGod reminded me of something I saw a while ago. Others can correct me if some of this information is out of date.

Define S(n) to be the sum of 1/p for all primes p less than or equal to n.

S(n) diverges - indeed Euler showed that you can bound it below with an expression that grows with log(log(n+1)) (log is natural log), and we can use log(log(n)) as an estimate of S(n) as n gets large.

That's painfully slow growth. It's not until n = 5,195,977 that S(n) reaches 3. And we have to sum up using all the primes up to around 10^18 to get to 4. But (based on searches I have done), in terms of finding all primes up to n (without gaps), 10^18 or so is roughly as far as computers have taken us - worldwide, that's all we know. So using only the primes we know, S(n) is about 4.

Of course, you will point out that we do know some primes larger than 10^18 , indeed much larger due to people searching for them. But that list is very sparse:

The largest known prime is around 10^41,024,319 and the tenth largest known prime is around 10^11,887,191 . The sum of 1/p for those ten largest primes is a rounding error (basically something less than 10^-11,887,190 ). But between those two primes there must be around 10^41,000,000 primes with a sum of 1/p of around 1, so virtually all that sum is missing from our knowledge.

Indeed S(n) where n is the largest known prime will be around 18. So there's a whole 14 contributed to S(n) by all the primes between 10^18 and the largest known. To get S(n) even up to 5, we would need to find primes up to around 10^50 - that's around 10^48 primes we need to find, and computers have so far only found 10^16 primes, so I don't think that's going to happen.

Overriding principle: play the game the way that brings you the most enjoyment. If reverse rainbows add to the fun for you, no need to downvote me - I'm glad you enjoy that element of the game.

My enjoyment of the game increased long ago when I gave up trying to discern green from yellow, and all that nonsense. I do some pre-solving to avoid red herrings, but I'll submit in whatever order helps me check I'm on the right track. If purple comes last I'll happily sit back and muse before I submit. Anyone else joining me in these sunny uplands of liberation from colour / color?

Otherwise, I guess getting purple first or the full reverse rainbow is important if...

...you like getting the badges and burnishing your stats - but to me they are just a tick of a box that only matters because you think it matters;

...you want to be competitive with other posters on this sub - but I find I satisfy my competitiveness by completing the puzzle with few mistakes and seeing how that compares with others.

...you want to demonstrate your total domination of the game - well... Some skills have value in the wider world - vocabulary, general knowledge and a bit of craftiness for logical leaps and educated guesses - so I get great satisfaction when I succeed with those. But other skills are pretty meaningless beyond Connections - say, mind-reading how Wyna has allocated colours - so I feel no loss ignoring such a task.

Based on their recent comments, this might be of interest to u/Enthooziest or u/Turbo_Fresh 😊.

This question is inspired by a problem on mathriddles set by u/jmarent049 - here for reference, but I give all necessary info below: https://www.reddit.com/r/mathriddles/comments/1tu04oj/what_is_the_longest_binary_string_you_can_make_is/

"Warm-up" question

We will say a natural number n is "2"-free if

 n < 2 
   OR 
 for n ≥ 2, when written as a decimal it does not contain the digit "2" AND floor(n/2) is also "2"-free.

For example, 30 is "2"-free because on repeatedly halving, none of 30, 15, 7, 3, 1 contain "2".

What is the largest "2"-free number?

.

Some observations:

If all n with a ≤ n < b, n is NOT "2"-free

then for all n with 2 a ≤ n < 2 b, n is NOT "2"-free.

That means, if for some r, we know that for

2^r ≤ n < 2^r+1

all n are NOT "2"-free then there cannot be any "2"-free number greater than 2^r+1 (because from that point on we can keep doubling the range and exclude every possible n), so we can conclude 2^r is an upper bound for the greatest "2"-free number.

Applying this we can quickly say all the following n are NOT "2"-free:

2 ≤ n < 3 (obviously)

4 ≤ n < 6 (keep doubling)

8 ≤ n < 12

16 ≤ n < 24 and we extend that to

16 ≤ n < 29 (because 25 to 29 contain "2")

32 ≤ n < 58

64 ≤ n < 116

128 ≤ n < 232 but we can extend that to

128 ≤ n < 299

but that covers 2^7 ≤ n < 2^8

So we know that any "2"-free number must be less than 2^7 .

.

Main question

We will say a natural number n is "10"-free if

 n < 10 
   OR 
 for n ≥ 10, when written as a decimal it does not contain the digit string "10" AND floor(n/2) is also "10"-free.

What is the largest "10"-free number?

.

As with the previous case, if a ≤ n < b is a "10"-free range then 2a ≤ n < 2b is a "10"-free range and if you can show n is "10"-free for all 2^r ≤ n < 2^r+1 then 2^r is an upper bound for all "10"-free numbers. In my reply on the mathriddles post, I used this to show that

2^328 is an upper bound for "10"-free numbers.

But I suspect the largest "10"-free number is a lot lower than that. Is there an efficient way of finding it? (I fear it's just a number-crunching exercise).

.

Follow-up

In general, if for any decimal string "X" we similarly define "X"-free, is it always the case that there is an upper bound for "X"-free numbers? (I suspect yes, but I've not worked out a proof).