For the HW exercise at 9:39 i will show my answer: We want to prove that ⌊√5(3m+⌊m√5⌋)⌋ = 5m+3⌊m√5⌋, This is the same as proving : √5(3m+⌊m√5⌋) ≥ 5m+3⌊m√5⌋ and √5(3m+⌊m√5⌋)
[LaTeX notation for math] Once you reach the 2-recursive expression x_{n+2} = 6 x_{n+1} - 4 x_n, it would be easier now to turn to the characteristic equation t^2 - 6t + 4 = 0 (which you need to solve anyway), solve it to get the roots r_0 and r_1, then pose x_n = A (r_0)^n + B (r_1)^n and compute the coefficients A and B from x_0 = 1 and x_1 = 5.
I didn’t see this comment. See my comment above. I was wondering the same thing. Because I’m in computer science, recurrences are like second nature, so I was so excited to see this question.
9:26 Let A=sqrt(5)(3m+floor(m*sqrt(5)) and B=5m+3floor(m*sqrt(5)) Note that B is an integer. To show floor(A)=B, we need to show A>=B and B+1>A, or equivalently, 0
24:43 Almost the 100K subs. It’s so great to see this channel growing so fast! Anyway, homework.... Show that in the arithmetic progression with first term 1 and common difference 729, there are infinitely many powers of 10.
a(n+1)=1+3⁶(n) So if a(n+1)=10^k we have 10^k-1=3⁶(n) so we have to show 3⁶ divide infinitly many numbers 10^k-1 By LTE we have V3(10^k-1)=V3(10-1)+V3(k)=2+V3(k) so if k=3⁴*a then 10^k-1 is divisible by 3^6 so for every 10^(3⁴*a) there exist number in this sequece
Using generating functions to solve for the general form of a sequence always gives me differential equations vibes, i.e. solving a differential equation with the Laplace transform.
There is also Simple way to get closed form of recursion xn=6x(n-1)-4x(n-2) let's name M,N roots of the polynomial x²=6x-4. Then we say that x(n)=A*M^n+B*N^n where A and B are some constans that satisfy A+B=x(0) and AM+BN=x(1) and we get the same result
Nice solution and explanation, as usual. Using the first hint, I feel we can come up with an easier way to find the answer. First we notice that the nth Fibonacci number is (1/sqrt(5))(phi^n-phib^n) where phi is the golden ratio and phib its conjugate, (1-sqrt(5))/2. When n is big, since |phib|
9:00 proof ==> let m be a positive integer and epsilon = m * sqrt(5) - floor( m * sqrt(5)), which is between 0 and 1 exclusive. We know we can take integers from floor functions directly. Distribute sqrt(5) in the lhs and we have lhs = floor (3 * sqrt(5) * m + sqrt(5) * (sqrt(5) * m - epsilon)) = floor( 3 * floor(sqrt(5) * m ) + 3*epsilon + 5 m - sqrt(5) * epsilon) = 5 * m + floor( 3 * floor(sqrt(5) * m )) + (3-sqrt(5)) * epsilon ) = 5 * m + floor( 3 * floor(sqrt(5) * m ))) + floor( (3-sqrt(5)) * epsilon ) notice that floor( 3 * floor(sqrt(5) * m )) = 3 * floor(sqrt(5) * m ) (3-sqrt(5)) * epsilon is a product of two numbers less than one so floor( (3-sqrt(5)) * epsilon) = 0 Therefore we have the claimed identity.
9:31 you can prove it by induction : . x0 = 1 is a natural number . suppose xn is a natural number xn+1 = xn + Ent(xn*sqrt(5)) = sum of 2 natural numbers then xn+1 is a natural number .for all n >= 0, xn is a natural number
You really should have provided some explanation for how you obtained the two-step recurrence relation. That was the key step and the rest was just standard procedure.
I love this channel and everything about it, the solutions are always so neat, your voice is good too listen to and the problems are challenging! My only problem is the sheer amount of ads lol, but that's on UA-cam's side... Kinda insane though, never expected something as "unpopular" as maths to get consistent double ad midrolls every 7 minutes lol
Wow, you did the 5.12 when there is a 5.8 trade route right next door... great if you're up to it I guess. :) But (as others point out too) once you have the second order recursion isn't it much easier to go to an auxiliary equation to get x_n = A (3 + srt(5) )^n + B (3 - sqrt(5) )^n then fit A and B to x_0 and x_1? You get nice neat answers in a few lines that way.
Hi Michael, I love ur videos, a friend from numebr theory found a shorter way (the only problem is that it uses university methods) Let K=\Q(\sqrt{5}) and S its Galois automorphism Let a = 3+\sqrt{5}, b = s(a)=3-\sqrt{5} and they r both algebric integers A=\frac{\sqrt{5}+2}{2\sqrt{5}}, B=s(A)=\frac{\sqrt{5}-2}{2\sqrt{5}} Define y_n = Aa^n + Bb^n. Now s(y_n)=y_n so all r rationals and later we will show that they r all algebric integers and also (we chose the right A and B): y_0=A+B=1=x_0, y_1=Aa+Bb=5=x_1 Let's assume by induction that y_n=x_n and it's also an integer then: y_{n+1} = a y_n - Bb^n(a-b) = 3x_n + (\sqrt{5}x_n - Bb^n(a-b)) Now, the 2 numbers y_{n+1},3x_n r integers and b^n eopnentially decays so (\sqrt{5}x_n - Bb^n(a-b)) = \lfloor\sqrt{5}x_n floor
Could you please show a solution to ANY combinatorics problem? Until i started watching your videos i didnt get number theory and somehow you fixed it, please consider making some combinatorics videos
I found an error in minute 21:44, you forgot a minus sign, beta-alpha = -√5/2, therefore, A & B are different. A=(-1+√5)/(8√5), B=(1+√5)/(8√5) I already checked them rejoining partial fractions.
not saying I would do this, but when a math competition says to write this in closed form, would one receive points for doing 2017 manipulations on 1? lol
Maybe I have a proof for 9:24 I don't know if this is the easiest proof, because it is done with just highschool maths. First we can say that the left hand of the equation (the part inside the floor function) is greater than/equal to some natural n and less than n+1. Then we try to isolate floor(m*sqrt5) and we get: -m*(3 - sqrt5) + 3/sqrt5 *floor(m*sqrt5)
@@elkincampos3804 Also, the difference you just described is greater than 0. Btw, if I'm not wrong you proved it in less than 3! lines!! That IS a nice and short proof.
Not sure what I am missing: It appears that for the generating function, F(1) = 0. Isn't F(1) just the sum of all x_n from zero to infinity. This seems odd since all x_n are positive, right?
A shorter way without recursion, after we guessed the answer use the explicit formula for Fibonacci numbers: fib(n)=1/S*(a^n-b^n) with S=sqrt(5) and a=(1+S)/2, b=(1-S)/2 then with induction assume that x(n)=2^(n-1)*fib(2*n+3), so we need x(n+1)=2^n*fib(2*n+5). Use the definition of x: x(n+1)=3*x(n)+floor(S*x(n))=floor((3+S)*x(n))= =floor((3+S)*2^(n-1)/S*a^(2*n+3)-(3+S)*2^(n-1)/S*b^(2*n+3)) notice that 3+S=2*a^2, so we can write: x(n+1)=floor(2^n/S*a^(2*n+5)-(3+S)*2^(n-1)/S*b^(2*n+3)) use the formula for fib(2*n+5), so subtract then add the conjectured 2^n*fib(2*n+5), we got: x(n+1)=2^n*fib(2*n+5)+floor(-(3+S)*2^(n-1)/S*b^(2*n+3)^n+1/S*2^n*b^(2*n+5))= =2^n*fib(2*n+5)+floor(1/S*(2*b^2)^n*b^3*(b^2-(3+S)/2)) here 0
I would have used continued fractions of sqrt(5) = 2+1/(2+1/... to get something like floor( xn sqrt(5) ) = floor ( xn an/bn) = usual quotient, for an/bn a sufficiently accurate aporoximant of the continued fraction expansion. But this solution is waaaaay clever :) how the hell you guessed the linear recurrence??
Hi Michael. Your solutions are fantastic, but often I see you doing complicated generating function analysis rather than just using characteristic equation stuff. Your solutions would be far neater. W
Homework: I think this is it. To prove: floor(sqrt(5) * (3m + floor(sqrt*5)*m)) = 5m+3*floor(m*sqrt(5)) where m is any natural number. Let m*sqrt(5) = x + c where x is a natural number and 0
we have ⌊√5(3m+⌊m√5⌋)⌋ = 5m+3⌊m√5⌋ then ⌊√5(3m+⌊m√5⌋)⌋ - 5m+3⌊m√5⌋ = 0 5m+3⌊m√5⌋ is an integer number so we can move it into the floor ⌊√5(3m+⌊m√5⌋) - 5m+3⌊m√5⌋] = 0 to be true we need to have √5(3m+⌊m√5⌋) - 5m+3⌊m√5⌋ < 1 with simple calculation become (3 - √5)(m√5 - ⌊m√5⌋) < 1 both factors are less the 1 so the product is less than 1
I wonder why you didn’t use characteristic equations to solve the recurrence: x(n) = 6x(n-1) - 4x(n-2) Assume the solution is r^n So we have: r^2 - 6r + 4 = 0 r = 6 +/- sqrt (36 - 16) ----------- 2 r = 3 +/- sqrt (5) So the solution is x(n) = A (3+sqrt 5)^n + B (3-sqrt 5)^n x(0) = 1 = A+B x(1) = 5 = A(3+sqrt5) + B (3-sqrt 5) Solve this and obtain the solution. A = 1/2 + 1/sqrt 5, B= 1/2 - 1/sqrt 5 So x(n) = (1/2 + 1/sqrt5) (3+sqrt5)^n + (1/2- 1/sqrt5)(3-sqrt5)^n I have no idea whether my answer is the same as the Professor’s or not.
Yes it will be the same if there are no slips on either side. I think the reason he choses a diffferent method is that he likes to take a pure mathematical approach rather than say the appproach of an engineer or appied mathematician. On the other hand the real difficulty in this problem was the start which did seem to require a bit of tial and error.
For the HW exercise at 9:39 i will show my answer:
We want to prove that ⌊√5(3m+⌊m√5⌋)⌋ = 5m+3⌊m√5⌋,
This is the same as proving :
√5(3m+⌊m√5⌋) ≥ 5m+3⌊m√5⌋ and
√5(3m+⌊m√5⌋)
Michael Penn: "So now we are gonna jump into the solution."
[An ad starts rolling.]
Me: 😑
Every four minutes on his videos I get an ad. I hope he's making a lot of money
[LaTeX notation for math] Once you reach the 2-recursive expression x_{n+2} = 6 x_{n+1} - 4 x_n, it would be easier now to turn to the characteristic equation t^2 - 6t + 4 = 0 (which you need to solve anyway), solve it to get the roots r_0 and r_1, then pose x_n = A (r_0)^n + B (r_1)^n and compute the coefficients A and B from x_0 = 1 and x_1 = 5.
I didn’t see this comment. See my comment above. I was wondering the same thing. Because I’m in computer science, recurrences are like second nature, so I was so excited to see this question.
HW 9:27
set sqrt5*m = k + p
where k is natural number, and 0
WOW!!! The BEST solution by far
oh no it happened! the proof is trivial and left as an exercise to the reader.
9:26 Let A=sqrt(5)(3m+floor(m*sqrt(5)) and B=5m+3floor(m*sqrt(5))
Note that B is an integer. To show floor(A)=B, we need to show A>=B and B+1>A, or equivalently, 0
24:43
Almost the 100K subs. It’s so great to see this channel growing so fast! Anyway, homework....
Show that in the arithmetic progression with first term 1 and common difference 729, there are infinitely many powers of 10.
by 'ratio' do you mean the common difference?
@@divyanshaggarwal6243 Had the same doubt, I believe it should be the common difference
729 = 9^3
1 + a*9^3 = 10^ n
a*9^3 = 10^n - 1
= 9*( 10^n-1 + 10^n-2 … + 1)
a*9^2 = 10^n-1 + 10^n-2 … + 1
a*81 = 10^n-1 + 10^n-2 … + 1
a*81 = 0 (mod 9)
10^n-1 + 10^n-2 … + 1 = 1^n-1 + 1^n-2 … + 1 (mod 9)
0 = n (mod 9)
Choose n = 9x, where x is a positive integer to get a solution and so there are infinite solutions.
(I hope this is right)
a(n+1)=1+3⁶(n)
So if a(n+1)=10^k we have
10^k-1=3⁶(n) so we have to show 3⁶ divide infinitly many numbers 10^k-1
By LTE we have V3(10^k-1)=V3(10-1)+V3(k)=2+V3(k) so if k=3⁴*a then 10^k-1 is divisible by 3^6 so for every 10^(3⁴*a) there exist number in this sequece
@@divyanshaggarwal6243 Fixed
Using generating functions to solve for the general form of a sequence always gives me differential equations vibes, i.e. solving a differential equation with the Laplace transform.
this one felt a little bit hand-wavy, but still fun nonetheless
@@angelmendez-rivera351 Agreed, but the 6 and 4 did seem like it was on his belt next to his light saber.
If x1+x2+...+xn=0 and x1²+x2²+...+xn²=1, what is the maximum possible value of x1³+x2³+...+xn³?
From Brazilian Math Olympiad 2018
If you want more hard maths loving questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun .
The answer according to me is most probably 0. If you want the solution then you may write (comment) down.
@@abidahasnain2638 gimme solution pls
I believe 22:10 should be (1-sqrt(5))/4 and not (-1-sqrt(5))/4. Thanks for the content!
There is also Simple way to get closed form of recursion xn=6x(n-1)-4x(n-2) let's name M,N roots of the polynomial x²=6x-4. Then we say that x(n)=A*M^n+B*N^n where A and B are some constans that satisfy A+B=x(0) and AM+BN=x(1) and we get the same result
It works because M²=6M-4 and N²=6n-4
Nice solution and explanation, as usual.
Using the first hint, I feel we can come up with an easier way to find the answer.
First we notice that the nth Fibonacci number is (1/sqrt(5))(phi^n-phib^n) where phi is the golden ratio and phib its conjugate, (1-sqrt(5))/2. When n is big, since |phib|
My blue (1+...+n)^2 Michael Penn hoodie has just arrived. Very nice!
9:00 proof ==>
let m be a positive integer and epsilon = m * sqrt(5) - floor( m * sqrt(5)), which is between 0 and 1 exclusive. We know we can take integers from floor functions directly.
Distribute sqrt(5) in the lhs
and we have lhs = floor (3 * sqrt(5) * m + sqrt(5) * (sqrt(5) * m - epsilon)) = floor( 3 * floor(sqrt(5) * m ) + 3*epsilon + 5 m - sqrt(5) * epsilon)
= 5 * m + floor( 3 * floor(sqrt(5) * m )) + (3-sqrt(5)) * epsilon )
= 5 * m + floor( 3 * floor(sqrt(5) * m ))) + floor( (3-sqrt(5)) * epsilon )
notice that floor( 3 * floor(sqrt(5) * m )) = 3 * floor(sqrt(5) * m )
(3-sqrt(5)) * epsilon is a product of two numbers less than one so floor( (3-sqrt(5)) * epsilon) = 0
Therefore we have the claimed identity.
9:31
you can prove it by induction :
. x0 = 1 is a natural number
. suppose xn is a natural number
xn+1 = xn + Ent(xn*sqrt(5)) = sum of 2 natural numbers
then xn+1 is a natural number
.for all n >= 0, xn is a natural number
You really should have provided some explanation for how you obtained the two-step recurrence relation. That was the key step and the rest was just standard procedure.
@@angelmendez-rivera351 I always find math so unreadable when it is written digitally
Try it on x_2, 3 (which you called manually) and you have a 2d system of linear eqs
For A and B as he calls them
If you know the closed form of the Fibonacci sequence, then that sqrt(5) should point you at it.
I love this channel and everything about it, the solutions are always so neat, your voice is good too listen to and the problems are challenging!
My only problem is the sheer amount of ads lol, but that's on UA-cam's side... Kinda insane though, never expected something as "unpopular" as maths to get consistent double ad midrolls every 7 minutes lol
Very nice indeed
Wow, you did the 5.12 when there is a 5.8 trade route right next door... great if you're up to it I guess. :) But (as others point out too) once you have the second order recursion isn't it much easier to go to an auxiliary equation to get x_n = A (3 + srt(5) )^n + B (3 - sqrt(5) )^n then fit A and B to x_0 and x_1? You get nice neat answers in a few lines that way.
Hi Michael, I love ur videos, a friend from numebr theory found a shorter way (the only problem is that it uses university methods)
Let
K=\Q(\sqrt{5}) and S its Galois automorphism
Let
a = 3+\sqrt{5}, b = s(a)=3-\sqrt{5} and they r both algebric integers
A=\frac{\sqrt{5}+2}{2\sqrt{5}}, B=s(A)=\frac{\sqrt{5}-2}{2\sqrt{5}}
Define
y_n = Aa^n + Bb^n.
Now
s(y_n)=y_n so all r rationals and later we will show that they r all algebric integers and also (we chose the right A and B):
y_0=A+B=1=x_0, y_1=Aa+Bb=5=x_1
Let's assume by induction that
y_n=x_n and it's also an integer then:
y_{n+1} = a y_n - Bb^n(a-b) = 3x_n + (\sqrt{5}x_n - Bb^n(a-b))
Now, the 2 numbers
y_{n+1},3x_n r integers and
b^n eopnentially decays so
(\sqrt{5}x_n - Bb^n(a-b)) = \lfloor\sqrt{5}x_n
floor
Could you please show a solution to ANY combinatorics problem? Until i started watching your videos i didnt get number theory and somehow you fixed it, please consider making some combinatorics videos
Yea i agree, we want to see some combinatory problems
I hear you guys! My semester is wrapping up and I will be able to do some new types of problems soon. I'll make sure and do some combinatorics.
And he has done it
@@tomatrix7525 And it didnt disappoint
There is a extra - sign.
22:10 it should be (1 - sqrt5)/4 instead of
( -1 - sqrt5)/4
your persistence in heavy calculations is praiseworthy
@@madhukushwaha4578 congratulations, your suggestion just showed how your IQ is close to your age.
@Sanjay Surya well if you need this level of education feel free to go learn there :)
I found an error in minute 21:44, you forgot a minus sign, beta-alpha = -√5/2, therefore, A & B are different.
A=(-1+√5)/(8√5), B=(1+√5)/(8√5)
I already checked them rejoining partial fractions.
@@madhukushwaha4578 yo stop spamming
Please look at the functional equation from 1988 IMO.
21:44 Wouldn't that be negative sqrt(5)/2 ?
Also, 11:08 Wouldn't that be (-2-sqrt(5))/4?
not saying I would do this, but when a math competition says to write this in closed form, would one receive points for doing 2017 manipulations on 1? lol
If you want more hard maths loving questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun .
Theoretically you would receive credit, but that would give you terms on the order of 5^2007 which is far too large to do computations with by hand.
Outline of the homework:
1.- Write [a] = b as a-1 < b
How do we get the Xn in the very last step from the equation of F(t)
Maybe I have a proof for 9:24
I don't know if this is the easiest proof, because it is done with just highschool maths.
First we can say that the left hand of the equation (the part inside the floor function) is greater than/equal to some natural n and less than n+1. Then we try to isolate floor(m*sqrt5) and we get: -m*(3 - sqrt5) + 3/sqrt5 *floor(m*sqrt5)
Just realized prof. Penn asked for a "nice solution". This is absolutely NOT a nice solution XD
Thus, assume m>0 then 1
@@elkincampos3804 Also, the difference you just described is greater than 0. Btw, if I'm not wrong you proved it in less than 3! lines!! That IS a nice and short proof.
Since that √5 is irrational . Set {a} is dense in interval (0,1). Then the difference is 0
7:23 you forgot to close the normal bracket
Not sure what I am missing:
It appears that for the generating function, F(1) = 0. Isn't F(1) just the sum of all x_n from zero to infinity. This seems odd since all x_n are positive, right?
A shorter way without recursion, after we guessed the answer use the explicit formula for Fibonacci numbers:
fib(n)=1/S*(a^n-b^n) with S=sqrt(5) and a=(1+S)/2, b=(1-S)/2
then with induction assume that x(n)=2^(n-1)*fib(2*n+3), so
we need x(n+1)=2^n*fib(2*n+5).
Use the definition of x:
x(n+1)=3*x(n)+floor(S*x(n))=floor((3+S)*x(n))=
=floor((3+S)*2^(n-1)/S*a^(2*n+3)-(3+S)*2^(n-1)/S*b^(2*n+3))
notice that 3+S=2*a^2, so we can write:
x(n+1)=floor(2^n/S*a^(2*n+5)-(3+S)*2^(n-1)/S*b^(2*n+3))
use the formula for fib(2*n+5), so subtract then add the conjectured 2^n*fib(2*n+5), we got:
x(n+1)=2^n*fib(2*n+5)+floor(-(3+S)*2^(n-1)/S*b^(2*n+3)^n+1/S*2^n*b^(2*n+5))=
=2^n*fib(2*n+5)+floor(1/S*(2*b^2)^n*b^3*(b^2-(3+S)/2))
here 0
I have a unique solution to the 2019 Putnam B2 problem. I can send it you and could you share it on your vid?
I would have used continued fractions of sqrt(5) = 2+1/(2+1/... to get something like floor( xn sqrt(5) ) = floor ( xn an/bn) = usual quotient, for an/bn a sufficiently accurate aporoximant of the continued fraction expansion.
But this solution is waaaaay clever :) how the hell you guessed the linear recurrence??
It was so great.thank u.
This is Sloane A018903, by the way.
Hmm, feels like you already knew the answer before you started solving the problem. Not sure if anyone in the contest would take this approach.
Hi Michael. Your solutions are fantastic, but often I see you doing complicated generating function analysis rather than just using characteristic equation stuff. Your solutions would be far neater. W
les valeurs dans le tableau sont ils exactes?
ok ils sont exactes mes excuses
He left the only tricky part as homework. All the rest is just standard methods.
Homework: I think this is it.
To prove:
floor(sqrt(5) * (3m + floor(sqrt*5)*m)) = 5m+3*floor(m*sqrt(5)) where m is any natural number.
Let m*sqrt(5) = x + c where x is a natural number and 0
Totally proven
just solve the entire thing by hand. At 1:40 ish theres the table for xn, with n from 0 to 6. just keep writing that out until you hit n=2007 bro
I feel the only hard part here (the key) was to notice that double step recursion. T
we have ⌊√5(3m+⌊m√5⌋)⌋ = 5m+3⌊m√5⌋ then
⌊√5(3m+⌊m√5⌋)⌋ - 5m+3⌊m√5⌋ = 0
5m+3⌊m√5⌋ is an integer number so we can move it into the floor
⌊√5(3m+⌊m√5⌋) - 5m+3⌊m√5⌋] = 0
to be true we need to have
√5(3m+⌊m√5⌋) - 5m+3⌊m√5⌋ < 1
with simple calculation become
(3 - √5)(m√5 - ⌊m√5⌋) < 1
both factors are less the 1 so the product is less than 1
I wonder why you didn’t use characteristic equations to solve the recurrence:
x(n) = 6x(n-1) - 4x(n-2)
Assume the solution is r^n
So we have:
r^2 - 6r + 4 = 0
r = 6 +/- sqrt (36 - 16)
-----------
2
r = 3 +/- sqrt (5)
So the solution is x(n) = A (3+sqrt 5)^n + B (3-sqrt 5)^n
x(0) = 1 = A+B
x(1) = 5 = A(3+sqrt5) + B (3-sqrt 5)
Solve this and obtain the solution.
A = 1/2 + 1/sqrt 5, B= 1/2 - 1/sqrt 5
So x(n) = (1/2 + 1/sqrt5) (3+sqrt5)^n + (1/2- 1/sqrt5)(3-sqrt5)^n
I have no idea whether my answer is the same as the Professor’s or not.
Yes it will be the same if there are no slips on either side. I think the reason he choses a diffferent method is that he likes to take a pure mathematical approach rather than say the appproach of an engineer or appied mathematician. On the other hand the real difficulty in this problem was the start which did seem to require a bit of tial and error.
Wish I could clean my blackboard that fast
22:15 Shouldn’t that -1 be a +1?
Uffff, very tricky problem
I respect Mr. Penn's videos a lot but this one felt like a complete cheat, I don't like the approach
Edit: β-α= half negative root of 5
And 1 - α should be (1 - sqrt(5)) / 4 i guess :D
Professor Penn really likes those floors!
Hard af
I need to watch these videos during my spanish class and ap world history class lol
Try to place a comment once again...
HW; A=3√5m+√5⌊√5m⌋=3(⌊√5m⌋+{√5m})+√5(√5m - {√5m}) = 3⌊√5m⌋+5m+{√5m}(3-√5) => ⌊A⌋=3⌊√5m⌋+5m
I wonder what AI would be like at solving this kind of problem without a priori knowledge of the question. Totally useless I would imagine.
cool