How to summon a Turtle...

[Turtle]

6

28

496

8128

33550336

8589869056

137438691328

2305843008139952128

2658455991569831744654692615953842176

19156194260823610729479337808430363813099732154816 9216

13164036458569648337239753460458722910223472318386 943117783728128

…

 

It’s just , [math]2^{n-1} \cdot (2^n-1)[/math], where [math]2^n-1[/math] is prime. A thinly veiled list of the Mersenne primes.

Ah contraire mon ami; an unveiled list of the Perfect numbers hiding in plain view. ;)

 

Speaking of which, I must return to hiding in plane view myself; adieu.

 

Anything of a math nature asserted by BD or C, requires multiple reviews. BD holds that the numbers listed interests us; so they do. However, C holds that the list, perfect they may be, are a genesis(?) of Mersenne primes wheras as T is going to hold that Mersenne primes are a genesis of the perfect numbers and the perfect numbers are a genesis of the figurate/polygonal numbers.

 

Have we got a game?

[TheBigDog]

Anything of a math nature asserted by BD or C, requires multiple reviews. BD holds that the numbers listed interests us; so they do. However, C holds that the list, perfect they may be, are a genesis(?) of Mersenne primes wheras as T is going to hold that Mersenne primes are a genesis of the perfect numbers and the perfect numbers are a genesis of the figurate/polygonal numbers.

 

Have we got a game? :lol:

Behold the Turtle! Lets play!

 

Bill

[Turtle]

Anything of a math nature asserted by BD or C, requires multiple reviews. BD holds that the numbers listed interests us; so they do. However, C holds that the list, perfect they may be, are a genesis(?) of Mersenne primes wheras as T is going to hold that Mersenne primes are a genesis of the perfect numbers and the perfect numbers are a genesis of the figurate/polygonal numbers.

 

Have we got a game?

Behold the Turtle! Lets play!

 

Bill

 

Roger. The crux of my position is that perfect numbers, that is the concept or very nature-of perfect numbers, requires no symbols whatsoever for their explanation or demonstration. If you have a pile of beans, you can make another pile of any number, and procede to rearrange them into all the possible arrangements of equal piles. Say a pile of 28 beans. You can have 2 piles of 14, or 14 piles of two, or 4 piles of seven, or 7 piles of 4, or 1 pile of 28, or 28 piles of 1. From this information you then make a pile of 1, a pile of 2, a pile of 4, a pile of 7, and a pile of 14, and this uses up all the beans. Voila!!! 28 is a Perfect Number. No symbols...simplest representation...ergo, more primal than the idea of Mersenne primes. :eek_big:

[Turtle]

Roger. The crux of my position is that perfect numbers, that is the concept or very nature-of perfect numbers, requires no symbols whatsoever for their explanation or demonstration. If you have a pile of beans, you can make another pile of any number, and procede to rearrange them into all the possible arrangements of equal piles. Say a pile of 28 beans. You can have 2 piles of 14, or 14 piles of two, or 4 piles of seven, or 7 piles of 4, or 1 pile of 28, or 28 piles of 1. From this information you then make a pile of 1, a pile of 2, a pile of 4, a pile of 7, and a pile of 14, and this uses up all the beans. Voila!!! 28 is a Perfect Number. No symbols...simplest representation...ergo, more primal than the idea of Mersenne primes. :D

 

Now carrying on with our pile of 28 beans, & knowing now 28 is a perfect number & then knowing all (even) perfect numbers are hexagonal numbers, we can rearrange our beans into a hexagonal matrix form known as a gnomon. Here it is:

 

PS Just logged out & went surfing only to find this bit of hexagonality! Amazing! Nature loves her perfect numbers as much as Fibonacci1.

SPACE.com -- Bizarre Hexagon Spotted on Saturn

[Turtle]

More you say!? :cup: OK...from the turtle's beak. All hexagonal numbers are triangular; moreover, every other triangular number is hexagonal. I'll draw up the triangular number gnomon for 28 in a wee bit.

 

Then we have what I think is at least a misleading, and at most a simply wrong statement on the mathworld page I linked to. Scroll down to line (17):

 

Perfect Number -- from Wolfram MathWorld

... In addition, all even perfect numbers are hexagonal numbers, so it follows that even perfect numbers are always the sum of consecutive positive integers starting at 1, for example, ...

 

It is triangular numbers that are the sums of the consecutive positive integers, not hexagonal numbers. ;)

[Turtle]

Nature loves her perfect numbers as much as Fibonacci.

 

This bit keeps gnawing at the back of my mind and as I can't afford to lose any more of it I have an inclination to do something about it.

 

It goes something like grouping the Fibonacci numbers according to the classes of Figurate numbers they belong to. We'll see.

 

1 1 2 3 5 8 13 21 34...

[TheBigDog]

This bit keeps gnawing at the back of my mind and as I can't afford to lose any more of it I have an inclination to do something about it.

 

It goes something like grouping the Fibonacci numbers according to the classes of Figurate numbers they belong to. We'll see.

 

1 1 2 3 5 8 13 21 34...

Here is a Fibonacci puzzle of sorts...

 

Is there any positive integer that cannot be expressed as a sum of non-repeating digits within the Fibonacci sequence? so...

 

1 = 1

2 = 2

3 = 3

4 = 3 + 1

5 = 5

6 = 5 + 1

7 = 5 + 2

8 = 5 + 3

9 = 8 + 1

10 = 8 + 2

11 = 8 + 3

12 = 8 + 3 + 1 *first to require 3 numbers

13 = 13

14 = 13 + 1

ect...

 

What is the most Fibonacci numbers needed to sum up to a number? Is there an algebraic way of approaching this? I am thinking of a brute force calculation myself, that is what I am good at...

 

And for those who have been around for a while, or have read all the interesting math posts, are there any primitive right triangles to be found?

 

Bill

[Turtle]

Here is a Fibonacci puzzle of sorts...

 

Is there any positive integer that cannot be expressed as a sum of non-repeating digits within the Fibonacci sequence? so...

 

1 = 1

2 = 2

3 = 3

4 = 3 + 1

5 = 5

6 = 5 + 1

7 = 5 + 2

8 = 5 + 3

9 = 8 + 1

10 = 8 + 2

11 = 8 + 3

12 = 8 + 3 + 1 *first to require 3 numbers

13 = 13

14 = 13 + 1

ect...

 

What is the most Fibonacci numbers needed to sum up to a number? Is there an algebraic way of approaching this? I am thinking of a brute force calculation myself, that is what I am good at...

 

And for those who have been around for a while, or have read all the interesting math posts, are there any primitive right triangles to be found?

 

Bill

 

I had to mull this over a bit, but I think I have a good response. As to the non-repeating digits question, this is a matter of which base is used to represent the number & so it isn't a valid question, only a good question. For example in base ten we have only 9 digits (not including zero) so the biggest number you can have without repeating a digit is the sum of them all:

1+2+3+4+5+6+7+8+9=45.

 

But wait! All is not lost, because this brings us back to figurate numbers in general, and triangular numbers in specific as they are the sums of consecutive integers. 45 is the ninth triangular number. Segue to your second question, which is both good and valid. :)

 

Can any number (positive integer) be represented as the sum of Fibonacci numbers? I don't know yet, but I recall that Fermat proved that any positive integer is the sum of at most 3 triangular numbers, or 4 square numbers, or 5 pentagonal numbers, etcetera.

 

More mulling required...:) :) :) :cup:

[CraigD]

Is there any positive integer that cannot be expressed as a sum of non-repeating digits within the Fibonacci sequence?

I had to mull this over a bit, but I think I have a good response. As to the non-repeating digits question, this is a matter of which base is used to represent the number & so it isn't a valid question, only a good question.

Bill, you mean “sum of non-repeating terms of the Fibonacci sequence”, correct? So the problem is independent of a particular numeral system?

[TheBigDog]

Bill, you mean “sum of non-repeating terms of the Fibonacci sequence”, correct? So the problem is independent of a particular numeral system?

 

Yes, I meant members of the set which is Fibonacci. Numbers, not digits. :P

[Turtle]

Yes, I meant members of the set which is Fibonacci. Numbers, not digits. ;)

 

 

OK, so the question is, do any numbers exist which cannot be broken into a sum of different Fibonacci numbers.

We better first decide if the two 1's beginning the set are 'different' or count as just one one. That is, the list starts

1 1 2 3 5 8 13 21 34...

and the first 1 is the first Fibonacci number and the second Fibonacci number is 1. Is this enough to say they are 'different' as we intend, or does the fact they have the same value make them the 'same' as we mean it? :P :D :)

[TheBigDog]

I asked myself the same question. :P So I decided to spurn the first 1.

 

Bill

[Turtle]

I asked myself the same question. :P So I decided to spurn the first 1.

 

Bill

 

;) Nothing like a good spurnation to get the ball rolling. :D Of course, spurnation requires justification, and I have been off fiddling the numbers and I think we may as well leave it (it being the 'different' 1:hyper: ) justifiably. To whit, there is no arguing that as far as sets go, the two are distinct & different elements; moreover, since three is a set element, removing a 1 doesn't keep you from getting a sub-total of 3 (2+1).

 

My approach is to look for counter-examples, as we need only one to make a proof, and it is giving me insight into how numbers deconstruct into sums of Fibonacci. I started with 88, as it is the sum of the first 9 Fibonacci's in the list I gave. I can say assuredly all numbers between 88 & 1 inclusive can be constructed as sums of lesser different Fibonacci numbers, and none require the extra 1 we accuse of being 'different'.

 

Shall we be remembered as mathematical biggots! :D Nay! Nay say I; all for one and one for all. :)

[TheBigDog]

It may be that the second 1 is never needed. So with the 1 unspurned we can say that we are limited to two 1's, and one two, likewise all other non-1's.

 

Bill

[freeztar]

Here is a Fibonacci puzzle of sorts...

 

Is there any positive integer that cannot be expressed as a sum of non-repeating digits within the Fibonacci sequence? so...

 

1 = 1

2 = 2

3 = 3

4 = 3 + 1

5 = 5

6 = 5 + 1

7 = 5 + 2

8 = 5 + 3

9 = 8 + 1

10 = 8 + 2

11 = 8 + 3

12 = 8 + 3 + 1 *first to require 3 numbers

13 = 13

14 = 13 + 1

ect...

 

What is the most Fibonacci numbers needed to sum up to a number? Is there an algebraic way of approaching this? I am thinking of a brute force calculation myself, that is what I am good at...

 

I've been working on this for the last two hours and I feel I'm getting close. I've already found some interesting relationships. It started out for me as fun as I haven't done these kinds of math exercises in years, but now I'm banging my head against the wall trying to express my thoughts through algebra.

Here's what I'm working with so far (assuming a single 1;)):

 

x=amount of Fibonacci numbers needed to sum to a positive integer

 

I've found that the first five in the series occur as follows:

x(1)=1

x(2)=4

x(3)=12

x(4)=33

x(5)=88(I haven't actually written out to this number yet, it is an extrapolation based on the fact that 33-12=21)

 

So I finally figured upon this formula:

phi(0)=any Fibonacci number in the sequence

phi(0)=phi(phi-1)+phi(phi-3)+phi(phi-4) {note: the "-1" etc. is a subscript; ie "phi-1" is the next lowest Fibonacci number}

 

I'm rusty in the math department so drill my post as needed. :P

[Turtle]

I've been working on this for the last two hours and I feel I'm getting close. I've already found some interesting relationships. It started out for me as fun as I haven't done these kinds of math exercises in years, but now I'm banging my head against the wall trying to express my thoughts through algebra.

Here's what I'm working with so far (assuming a single 1;)):

 

x=amount of Fibonacci numbers needed to sum to a positive integer

 

I've found that the first five in the series occur as follows:

x(1)=1

x(2)=4

x(3)=12

x(4)=33

x(5)=88(I haven't actually written out to this number yet, it is an extrapolation based on the fact that 33-12=21)

 

So I finally figured upon this formula:

phi(0)=any Fibonacci number in the sequence

phi(0)=phi(phi-1)+phi(phi-3)+phi(phi-4) {note: the "-1" etc. is a subscript; ie "phi-1" is the next lowest Fibonacci number}

 

I'm rusty in the math department so drill my post as needed. :D

 

Aha! we have drawn you into our web. ;) On with the drilling then. :D I'll type my work up.

 

Leftmost column, the consecutive positive integers, rightmost column a sum of Fibonacci numbers that equals the integer left of it, or if the integeris itself Fibonacci than just that integer.

 

Notice that I have used the least Fibonacci numbers possible to make the sum, but that other combinations are possible. For example where 5 occurs in the sum string, you can substitute a 3+2 instead. So it's not a matter of how many different Fibonacci numbers are used, but whether any combination suffices.

 

Here's the list of the first 27; more to come on the Fibonacci vs. Figurate scenario in a short bit. :P

 

1=1
2=2
3=3
4=3+1
5=5
6=5+1
7=5+2
8=8
9=8+1
10=8+2
11=8+3
12=8+3+1
13=13
14=13+1
15=13+2
16=13+3
17=13+3+1
18=13+5
19=13+5+1
20=13+5+2
21=21
22=21+1
23=21+2
24=21+3
25=21+3+1
26=21+5
27=21+5+1
.
.
.

 

PS My gut is telling me a to conjecture that all positive integers may be constructed as a sum of different Fibonacci numbers. Who's up for a pool?:)

[Turtle]

OK...the set of Fibonacci numbers & its intersection with the set of sets of Figurate/polygonal numbers.

 

Again with a scan & I'll type it up later. Right column, the set of ordered Fibonacci numbers, left column the nature of the intersection with the set(s) of figurate numbers. To clarify a bit, all numbers n are trivially figurate as the 2nd element in the n-sided set. In the list then, 21 is the first non-trivially figurate number as it is the 6th 3-sided number and 3rd 8-sided number.

 

What is most interesting to me is that the Fibonacci number 89 is non-figurate except trivially as the 2nd 89-sided number. Here's some insight on non-figurate numbers from another thread:

Following this line in view of the goal of using the generalized expression of figurate numbers, (n/2)*((s-2)*n)-s+4 when S >= 3 and n >= 3, to find non-figurate numbers,...

 

For s & n >= 3, 89 has no integer solutions in the above equation.

 

Well, that's a bit to chew over eh!? :P ;) :) Better let it simmer. :D