I have a nice problem for you guys, just a funny problem to stimulate the mind.

 

For each non-negative integer [math]n[/math], let [math]a_n[/math] be the number of digits in the decimal expansion of [math]2^n[/math] that are at least 5. For example [math]a_{16} = 4,since 2^{16}=65536[/math] has four digits that are 5 or higher.

 

What is the sum [math]\sum_{n=0}^\infty \frac{a_n}{2^n}[/math] ?

 

I evaluated machinally the first 299 (see attachment) terms a of the sum. The sum converges to 0.086617 numerically, what is the exact value ?

What is the growth rate of a_n ?

 

Mandrake

a.txt

i would say the growth rate of a_n is about [math]log_{10}2[/math]

 

i still havent studied how to calculate the sum.

Been looking at the problem, but I can't seem to make any inroads into it atm. Have to give it some more thought.

here a_n should be equal to [edit][math][\log_{10}(2^n)][/math][/edit] where [] is denotes the least integer which is greater than

meant least integer greater or equal than

Got a proof?

 

And good luck trying to find the infinite series, looks like a pig, specially with that damned [x] crap lying about.

well, havent got a proof, its rather intiuitive. it works for integers greater than one. must be able to prove by induction somehow.

 

about the series, the most i could do here is to check that the series does indeed converge. about the value, got no idea.

Hmm. I'm just wondering whether an alternative formula for a_n exists.

an alternative form for a_n would be [2^n/10] where [] means the same thing as before

calculated a_n upto n = 33 , found out some stuff on behaviour of a_n (not proven)

 

if n = 0,1,2,3 mod10 then for each grouping of a_n i.e n=0,1,2,3 : 10,11,12,13,...

 

the a_n's are equal. the same happens for n=4,5,6 mod10 and n=7,8,9 mod10.

 

so basically the grouping seems to have a sequence 4,3,3,4,3,3,

 

for example the first few {a_n}

1,1,1,1, 2,2,2, 3,3,3, 4,4,4,4, 5,5,5, 6,6,6, 7,7,7,7, 8,8,8, 9,9,9, 10,10,10,10,

 

gonna try if i can find the value of the series.

oops seems that i have read the question wrong. i thought a_n was the the number of digits in 2^n. it seems that its number of digits in 2^n at least 5

Yeah, that's the hard part

I dont think it is very hard to show that the sum is convergent, since it is bounded by above (clearly a_n is less than the number of digits in 2^n, which seems to grow as O(n) ) and its partial sums are clearly increasing, hence convergent.

 

Mandrake

When I worked it out it converged to ~ 2/9. Merely calculated the sum of the first 2000 terms.

Not really an exact answer for the proof though.

Ja, I know that. Just thought it might give someone else a better idea seeing as MandrakRoot's estimate was 0.086617.

 

Suppose I should have said "converge" instead of leaving out the apostophes.

Here's the Mathematica code if anyone's interested.

 

G[x_] := IntegerDigits[2^x]

A[n_] := Length[select[G[n], #1 > 4 &]]

N[sum[A[n]/(2^n), {n, 0, 2000}], 50]

 

If anyone can see an error please tell me.

The part I have the problem with is trying to find a specific formula for the digits of a_n, if indeed there is one. I've tried analysing the function but there doesn't seem to be a specific pattern.

 

Any guidance would be appreciated