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LegionMammal978

23:01

Not sure how to find a closed-form

This kind of reminds me of nested radical problems :P

Perhaps Math.SE could help

orlp

@LegionMammal978 1 + 1/(2 + 1/(4 + ...?

LegionMammal978

@orlp Yes.

flawr

@LegionMammal978 it is probably difficult

LegionMammal978

As I said, this set is pretty weird

flawr

Is [1;0] a valid element in your set?

orlp

23:07

well

we have

a(n) = 2**n + 1/(a(n+1))

right?

LegionMammal978

@flawr It would expand to [1; 0, 0, 0, ...], which Mathematica tells me is division by 0

@orlp Yes...

flawr

@LegionMammal978 Mathematica != math :D

LegionMammal978

@flawr Wait, no, wrong fraction

flawr

@LegionMammal978 for this particular example you can maybe use the formula for the n-th convergent (idk how to correctly translate)

LegionMammal978

@flawr 1, 3/2, 13/9, 107/74, etc. doesn't make much sense to me

ETHproductions

23:10

@LegionMammal978 Your avatar is driving me insane... :P

Eᴀsᴛᴇʀʟʏ Iʀᴋ

@LegionMammal978 you should put a hat on that blankness, maybe an invisible one?

LegionMammal978

ISC finds nothing for the number

flawr

@LegionMammal978 There is a formula

@LegionMammal978 de.wikipedia.org/wiki/…

(sorry for linking to the german one, but you'll understand it anyway)

LegionMammal978

@flawr Can't find that section on the English version

flawr

let b(n) := 2^n , then the number we're looking at is [b(0); b(1), b(2),...]

LegionMammal978

23:14

Yes...

flawr

and p(n)/q(n) = [b(0); b(1), b(2),...,b(n)]

for p(n) := b(n)*p(n-1)+p(n-2)

and q(n) := b(n)*q(n-1)+q(n-2)

where p(-1) = 1 , p(-2) = 0

LegionMammal978

I see those parts

flawr

and q(-1) =0 , q(-2) = 1

LegionMammal978

The question is, what is the limit of this

flawr

now let us try to find an explicit formula for q(n)

LegionMammal978

23:16

@flawr Mathematica's RSolve refuses

flawr

@LegionMammal978 let's try formal power series

ETHproductions

@LegionMammal978 Same here. I think I've used at least 500 tissues over the last four days...

LegionMammal978

Also, this page briefly mentions my [1; 2] but doesn't give a closed form

> 1.445934640512202668 cf! s= 1/(2^(m-1)+s); 1/s [1;2,4,8,16,32..]

orlp

@LegionMammal978 oeis.org/A061377

the denominators are a(n) = 2^(n-1) a(n-1) + a(n-2)

flawr

Can you solve the functional equation S(x) - p(1) x - p(0) = 2*x*(S(2*x)-p(0)) + x *S(x)

LegionMammal978

23:22

@flawr wat

flawr

@LegionMammal978 the coefficients of x^n in the taylor series of S(x) would be p(n)

LegionMammal978

Umm... how would I give this to Mathematica?

flawr

I don't know mathematica

LegionMammal978

I mean, p(0) = 1 and p(1) = 3

Luis Mendo

@flawr Now only two convolutions

LegionMammal978

23:25

So would it be S(x) - 3x - 1 = 2x(S(2x) - 1) + x S(x)?

flawr

@LegionMammal978 probably, if I didn't mess up=)

But it is probably hopeless anyway

@LuisMendo Now you've got a spare one for free :)

Luis Mendo

@flawr Let's see how soon I can stick it into another answer :)

LegionMammal978

@flawr Mathematica tells me S(x) = (-2x S(2x) - x - 1)/(x - 1)

Doorknob

Well, that confused me for a while.

Luis Mendo

Said the door handle with a fork as hat :P

flawr

23:35

@LuisMendo *spork

@LuisMendo Do you know Sara Niemietz? She's an amazing singer, and she's got an amazing guitarrist: Snuffy Walden youtube.com/watch?v=GRuC4VS5I3w

Luis Mendo

@flawr TIL

flawr

(perhaps you've already seen her in some postmodern-jukebox videos)

(and I can't get this riff outta my head)

orlp

@LegionMammal978 I have insight

Luis Mendo

@flawr He's good, yes! :)

BTW A recently published a song in Soundcloud

Actually I don't play a single instrument in it :)

LegionMammal978

@orlp ?

flawr

23:39

you're singing?

Luis Mendo

Haha no. Listen to it!

flawr

@LuisMendo Oh I like the cover so far=)

Luis Mendo

You may recognize some part of the song. It's entirely made of pieces from other songs

orlp

a(0) = 1; a(1) = 1; a(n) = 2**(n-1) * a(n-1) + a(n-2)
b(0) = 0; b(1) = 1; b(n) = 2**(n-1) * b(n-1) + b(n-2)

Luis Mendo

and movie dialogs

orlp

23:40

@LegionMammal978 1.44... = a(big) / b(big)

Luis Mendo

@flawr The double-slit experiment :-)

LegionMammal978

@orlp It would have to be 2^(n-2)

orlp

@LegionMammal978 no, this is correct

LegionMammal978

Also, don't experiments show that a(0) = b(0) = 1, a(1) = 3, and b(1) = 2?

orlp

@LegionMammal978 that's equivalent

just a different starting point

flawr

23:41

@orlp can't you plug this into mathematica?

@LuisMendo oh now I see=)

LegionMammal978

@flawr As I said, I tried that but RSolve refused

flawr

that's amazing=)

LegionMammal978

@flawr Your avatar is killing my eyes

orlp

@LegionMammal978 wait, did you already have these formulas?

I just made them =/

LegionMammal978

@orlp This onwards

(with b(n) = 2^n)

orlp

23:43

LegionMammal978

But yeah, could you see if there's closed forms for a(n) and b(n)?

A214070 appears to be my constant...

A096641 is my constant minus one, which seems to already have an infinite series representation

orlp

@LegionMammal978 I edited OEIS

awaiting approval

I added a(n) = 2^(n-1) a(n-1) + a(n-2) to oeis.org/A061377

flawr

@LegionMammal978 arxiv.org/pdf/math/0402461 Page 2

orlp

ramanujan is crazy

arda

Was there ever a cowsay code golf?

orlp

23:59

yes

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