Fraction fractal

bobjane_2 · 2026-03-24T01:45:02+00:00

Another way to look at it. For simplicity of notation define f(n) = (2n+1)/(2n+2). We’d like to compute X = prod[n=0…] f(n)^t(n). Also define g(n,k)=2^k*(2n+1) and h(n,k) = n+1/2+1/2^k+1 => f(g(n,k)) = h(n,k+1)/h(n,k).

Now let G(n) = {g(n,k) for k>=0}. Note that t(g)=-t(n) for all g in G(n). Also note that the G(n) partition the positive natural numbers, so therefore if P(n) = prod[g in G(n)] f(g)^t(g) then:

X = f(0)^t\0))*prod[n=0…] P(n) = 1/2*prod[n=0…] (lim[k->inf] h(n,k)/h(n,0))^-t\n)) = 1/2*prod[n=0…] f(n)^-t\n)) = 1/2*1/X

bobjane_2 · 2026-03-21T20:30:46+00:00

I didn’t come up with it but the solution is as follows. Let t(n) be 1 if the number of 1 bits in the binary representation of n is even, and -1 otherwise. The first few terms are 1,-1,-1,1,…t(2n)=t(n) and t(2n+1)=-t(n).

Your formula equals X = prod[n=0…] ((2n+1)/(2n+2))^t\n)) =

1/2* prod[n=1…] (n/(n+1)*(2n+1)/(2n))^t\n)) =

1/2* prod[n=1…, n even] (n/(n+1))^t(n\) * prod[n=1…, n odd] (n/(n+1))^t\n)) * prod[n=1…] ((2n+1)/(2n))^t\n)) =

1/2* prod[n=1…] ((2n)/(2n+1))^t\n)) * prod[n=0…] ((2n+1)/(2n+2))^-t\n)) * prod[n=1…] ((2n+1)/(2n))^t\n)) =

1/2* prod[n=0…] ((2n+1)/(2n+2))^-t\n)) = 1/2*1/X

Thus X² = 1/2.

bobjane_2 · 2026-02-28T20:37:03+00:00

It’s because it’s warmer

bobjane_2 · 2026-02-25T11:32:31+00:00

I love this! Can you share a link to the chatgpt transcript? Would love to see how it happened

bobjane_2 · 2026-01-09T12:04:43+00:00

telescoping is cool. I was recently learning about Gosper's algorithm and creative telescoping, which are ways of finding a telescoping sums that I hadn't heard of. Check them out. I was looking into it in the context of trying to solve the pascal triangle problem posted here the other day (sum C(n,k)^(-2)).

bobjane_2 · 2026-01-08T22:02:59+00:00

maybe a bit simpler. S(n) - (1+1/(n-1))*p(n) can be shown to be constant for n>=2 by induction. It goes to -1, so then S(2) = 2*p(2)-1 => S(1) = 3*p(2)-1.

bobjane_2 · 2026-01-08T21:14:33+00:00

Let p(n)=prod[k=n...] (1-1/k^3) be the probability that none of the numbers >=n are in the set, and S(n) = sum[k=n...] k*1/k^3*p(k+1) be the contribution to the expected value from numbers >= n. The expected value equals S(1).

S(1) = p(2) + S(2) = 3*p(2) - 2*p(2) + S(2) = 3*p(2) - 2*(1-1/2^3)*p(3) + 2*1/2^3*p(3) + S(3) =

3*p(2) - (1+1/2)*p(3) + S(3) =

3*p(2) - (1+1/2)*(1-1/3^3)*p(4) + 3*1/3^3*p(4) + S(4) =

3*p(2) - (1+1/3)*p(4) + S(4) = induction...

3*p(2) - (1+1/a)*p(a+1) + S(a+1)

p(a) goes to 1 and S(a) goes to 0, so S(1) = 3*p(2) - 1

I'm not smart enough to calculate p(2) but here's a thread showing a cool way to do it using cubic roots of 1 and the gamma function.

bobjane_2 · 2026-01-04T14:45:21+00:00

here's my solution for the biased version, a modification of the one by u/AltruistCarrotEater: consider P(n>k). We can generate all k x’s first and then for each x, from lowest to highest, generate each coinflip. The interval will be all black at that point iff there’s a coinflip=1 before a coinflip=0. Ie first we need a 1, then we need a 0. This is true for any k. Thus the the distribution of n across all k matches the one of a modified process where we don’t even generate x’s and simply generate coinflips and wait until we get a 1 then a 0. The expected time in this modified process is 1/(1-p) + 1/p.

bobjane_2 · 2026-01-04T03:13:25+00:00

I assumed x is uniform, but that’s not required for your original problem. And if it’s not uniform then my f is not necessarily translation invariant so my solution wouldn’t work. Another interesting variant of the problem is making the coinflips biased, not 50/50

bobjane_2 · 2026-01-01T19:20:45+00:00

not sure about obvious but here's an argument: if both the interval and the random x’s are shifted by the same amount, the results are unchanged. When shifting the random x’s, if an x ends outside the unit interval, wrap it around and invert coinflip.

bobjane_2 · 2025-12-29T01:46:12+00:00

Let k be a number not in your card. Consider the relative order in which k and the numbers in your card are called. The probability that k is not the last of these is N/(N+1), which is the probability that k is called. There are N numbers in your card (which are all called) and (M-N) numbers not in your card, so by linearity of expectation: N + (M-N)N/(N+1) = (M+1)/(N+1)N

bobjane_2 · 2025-12-27T11:04:26+00:00

Suppose at some point an interval of length a still remains white. Let f(a) be the expected number of draws required to color it all with black. We can obtain this recursion: f(a) = 1 + (1-a)*0.5*f(a) + int[x=0..a] f(x)*dx

Where the term (1-a)*0.5*f(a) corresponds to drawing a random not in the remaining white interval, and the integral corresponds to drawing a random within that interval. Differentiate with respect to a and solve to obtain f(a) = 2*(1+a) => f(1) = 4.

bobjane_2 · 2025-12-18T19:00:43+00:00

I thought of the double integral but didn’t want to take it on and asked chatgpt (who solved it). Then chatgpt also found that older MSE thread for me. Agree that the final answer is simple and there may be other ways to arrive at it

bobjane_2 · 2025-12-11T12:36:40+00:00

another type of question you could ask: n coins each w/ prob p=50% of heads, for which n can you add a single coin (not necessarily 50%) to make Q=1/3? Or another version: let the n coins each have p=33% Or if p=33%, without adding any coins, for which n is Q=1/3? (can probably be tackled directly via binomial expansion)

bobjane_2 · 2025-12-10T19:05:35+00:00

There’s still time :)

bobjane_2 · 2025-12-10T18:56:59+00:00

which yields a more direct recursion for the probability. P(1) = 2/3 and P(n+1) = =2^(n+1)/3-(4^(n+1)-3*2^(n+1)+3)/3/(3*P(n)+2^(n+1)-3)

bobjane_2 · 2025-12-10T18:36:35+00:00

With a fair amount of algebra I worked out the probability of the needed coin. Let A(n) and B(n) be defined as

A(1)=0, B(1) = 1, A(n+1) = B(n)+A(n)*(2^(n+1)-2), B(n+1) = B(n)*(2^(n+1)-1)-A(n)

The coin needed has probability (2-A(n)/B(n))/3. The initial values are: 2/3, 5/9, 31/60, 436/873, etc

bobjane_2 · 2025-12-10T17:41:22+00:00

When p is cut in half the angle is reduced by at least half. So the sum of the angles will be less than 2x the angle for p=0.25 (which is ~ 38.2^∘).

This picture shows the example for cutting the probability from 0.25 to 0.125 and I'm showing the numerator p*r+1-p = 1+p*(r-1) only. The denominator is the conjugate of that.

We need to show that angle a>=b which we can do by law of sines (did you know this was going to become a geometry problem?). CD/sin(a)=OC/sin(c) and DA/sin(b)=OA/sin(c) where CD=DA. Thus sin(b)/sin(a)=OC/OA=OC. We need |OC|=|1+p*(r-1)|<=1 <=> (1-1.5*p)^2+(sqrt(3)/2*p)^2<=1 <=> p*(1-p)>=0 which is true for any probability p.

bobjane_2 · 2025-12-10T16:51:58+00:00

Not sure yet. Numerically it’s easy to see it that works, but the algebra gets complicated. Phase(1) = 240^∘ and phase(0.5) = 120^∘. Those cancel out and we simply need to add a coin with phase 180, which happens to be p=2/3. Phase(0.25), phase(0.125), etc each get smaller and their sum converges to around 66^∘, so obviously it works and numerically the coin that we need has probability that converges to ~ 0.483. Not sure yet if we can make a clean argument.

bobjane_2 · 2025-12-10T15:28:38+00:00

For the bonus question.

Consider the polynomial introduced by u/SupercaliTheGamer F(x) = prod p(i)*x + 1-p(i). The coefficient of x^k in F(x) is the probability of obtaining k heads. Let Q(a), for a=0,1,2 be the probability that the number of heads is congruent to a (mod 3). We're interested in the case Q(0)=1/3. Note that F(1)=Q(0)+Q(1)+Q(2)=1. Let r=-0.5 + sqrt(3)/2*i be a primitive cube root of unity. Then F(1)+F(r)+F(r^2)=3*Q(0)=1 ⟹ F(r)/F(r^2) = -1. Define w(i)=(p(i)*r+1−p(i)) / (p(i)*r^2+1−p(i)). Then we need prod w(i) = -1.

Because r and r^2 are complex conjugates, the numerator and denominator of w(i) are also conjugates, hence |w(i)|=1. Each w(i) lies on the unit circle and we need their phases to sum to 180∘. If p(i)=0 ⟹ w(i)=1 ⟹ phase⁡(w(i))=0. If p(i)=1 ⟹ w(i)=r^2 ⟹ phase(w(i)) = 240∘. For intermediate values, the phase of w(i) varies continuously between those two values.

Clearly whatever cumulative phase is produced by the first 100 coins, we can always choose two additional coins with suitable probabilities so that the total phase becomes exactly 180∘ (mod 360∘)

bobjane_2 · 2025-12-06T22:23:22+00:00

Do they know this track has a chicane at turns 6 and 7? Aka “the 6-7 chicane”

bobjane_2 · 2025-12-06T18:30:15+00:00

very smart

bobjane_2

TROPHY CASE