FormulaDriven

I think we can reasonably surmise that when Tim Curry died in late August, Wyna (the setter of Connections) decided to design a puzzle that referenced his career, and on Wednesday we got that puzzle.

Wyna likes to throw in a rainbow herring, a sometimes playful theme which you might or might not spot, but which raises a smile if you do. Typically the words for these will appear in the top two rows, or even across the top row, giving a big hint that they are a rainbow herring. A recent example was when PURPLE was placed top left, with CORVETTE, BERET and DOVES nearby, a little bit of fun for Prince fans.

So Wednesday's puzzle had, sitting either end of the top two rows, four of the most ce

Answer to today's Plusword

WARNING: replies on this thread might not use spoiler tags, so read with caution!!

Across

1 >!GRIST!<

6 >!LANKY!<

7 >!END UP!<

8 >!AGILE!<

9 >!MEALS!<

Down

1 >!GLEAM!<

2 >!RANGE!<

3 >!INDIA!<

4 >!SKULL!<

5 >!TYPES!<

PW

!SAUCE!<

Answer to today's Plusword

WARNING: replies on this thread might not use spoiler tags, so read with caution!!

Across

1 >!OHBOY!<

6 >!SAUNA!<

7 >!ATTIC!<

8 >!KETCH!<

9 >!ASYET!<

Down

1 >!OSAKA!<

2 >!HATES!<

3 >!BUTTY!<

4 >!ONICE!<

5 >!YACHT!<

PW

!PESKY!<

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 Others can correct me if some of this information is out of date.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 f

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 bad

Answer to today's Plusword

WARNING: replies on this thread might not use spoiler tags, so read with caution!!

Across

1 >!RIOJA!<

6 >!ANNUL!<

7 >!STAID!<

8 >!PRICE!<

9 >!SORER!<

Down

1 >!RASPS!<

2 >!INTRO!<

3 >!ONAIR!<

4 >!JUICE!<

5 >!ALDER!<

PW

!DELTA!<

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.

Answer to today's Plusword

WARNING: replies on this thread might not use spoiler tags, so read with caution!!

Across

1 >!SAGES!<

6 >!EXEAT!<

7 >!PITTA!<

8 >!INTER!<

9 >!AGONE!<

Down

1 >!SEPIA!<

2 >!AXING!<

3 >!GETTO!<

4 >!EATEN!<

5 >!STARE!<

PW

!PRONG!<

Answer to today's Plusword

WARNING: replies on this thread might not use spoiler tags, so read with caution!!

Across

1 >!SAGAS!<

6 >!CREST!<

7 >!RETIE!<

8 >!ANODE!<

9 >!PANEL!<

Down

1 >!SCRAP!<

2 >!ARENA!<

3 >!GETON!<

4 >!ASIDE!<

5 >!STEEL!<

PW

!CHIRP!<

Answer to today's Plusword

WARNING: replies on this thread might not use spoiler tags, so read with caution!!

Across

1 >!INSTA!<

6 >!DATED!<

7 >!IDAHO!<

8 >!OILER!<

9 >!TAKEN!<

Down

1 >!IDIOT!<

2 >!NADIA!<

3 >!STALK!<

4 >!TEHEE!<

5 >!ADORN!<

PW

!OCEAN!<

Answer to today's Plusword

WARNING: replies on this thread might not use spoiler tags, so read with caution!!

Across

1 >!SCOTS!<

6 >!LATHE!<

7 >!OCHRE!<

8 >!SHEER!<

9 >!HERES!<

Down

1 >!SLOSH!<

2 >!CACHE!<

3 >!OTHER!<

4 >!THREE!<

5 >!SEERS!<

PW

!LICIT!<

I know from a quick search that the sum for n = 0 to infinity of

(-1)^n / (n^2 + k^(2))

is

1/(2k^(2)) + (𝜋 / k) / (e^k𝜋 - e^(-k𝜋))

(you can use the function cosech(x) = 2 / (e^x - e^-x ) if you want to be fancy).

I think you can prove it using partial fractions over complex numbers and gamma functions and the like, but I'm wondering if that's the most direct way? How would you prove it - or at least what are the key steps / results you would use to prove it?

Answer to today's Plusword

WARNING: replies on this thread might not use spoiler tags, so read with caution!!

Across

1 >!QUICK!<

6 >!UNMAN!<

7 >!AMPLE!<

8 >!DELVE!<

9 >!STYES!<

Down

1 >!QUADS!<

2 >!UNMET!<

3 >!IMPLY!<

4 >!CALVE!<

5 >!KNEES!<

PW

!ALPHA!<

There's a lot of money changing hands.

some shady activity