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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,090

For VI)

7 + 2 + 99 + 2 = 110

4 + 9 + 71 + 2 + 6 = 92

(9 + 7 + 53) / 3 = 23

( 59 + 95 + 2 ) / 3 = 52

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

Hi meanmaths

*Last edited by anonimnystefy (2013-08-04 07:26:34)*

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**meanmaths****Member**- Registered: 2013-08-01
- Posts: 51

Congrats bobbym for the splittings. Only one left.

I think you'll enjoy the game. Go get it :-)

Number Splitter #1

GameMaster of Mean Sumurai @ Google Play

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**meanmaths****Member**- Registered: 2013-08-01
- Posts: 51

anonimnystefy,

that was super-fast. Much faster than expected.

Truth is I can't even say right now whether your answer is correct or not.

Do you have a recursive form? It would be easier for me to check.

My gut feeling tells me that your formula is right.

I'm quite busy right now so give me 1-2 days to get back to you.

Sorry :-(

I have to ask: What is your educational background?

Number Splitter #1

GameMaster of Mean Sumurai @ Google Play

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Hi

That's okay, take your time.

This one I did with the help of the Mathematica package.

In September I'm starting 4th year of high school, which would be 12th grade, I guess.

*Last edited by anonimnystefy (2013-08-05 19:53:05)*

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,090

Use this for a recurrence.

Do you have a recursive form? It would be easier for me to check.

last solution for VI)

( 71+13+3+8+4+3 ) / 6 = 17

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

Hi meanmaths

I have been able to greatly simplify my formula from post #27.

It does look much nicer than before and it agrees with bobbym's recursion.

*Last edited by anonimnystefy (2013-08-07 01:01:08)*

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**meanmaths****Member**- Registered: 2013-08-01
- Posts: 51

Terrific.

Congrats to both of you. The formula I had looked like the one from the post #27 but used intermediate variables for (1+sqrt(5))/2 and (1-sqrt(5))/2 so confirming post #27 could not be done by a quick look.

And yeah, anonimnystefy, your new formula does look much nicer than the previous one.

I don't know bobbym's education but both of you are obviously gifted people. There's still V-b to solve but truth is I'm mighty busy these times so if you have an answer, don't expect a quick confirmation. I also have one (may be 2) related problems. I will present it (them) straight away after an answer for V-b

One last point: I presented this number splitting formulation on the math section of Reddit (don't trust Reddit caricature in the media, many of the guys in that forum are maths Ph.D.s) and it may turn out that you guys, along with a few other people close to me, have been working on a new and possibly research-worthy problem. Now, if one of us could spot an interesting link with any sort of existing concept or phenomenon, the whole thing could become very exciting. Keep your eyes open.

Best,

Number Splitter #1

GameMaster of Mean Sumurai @ Google Play

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,090

Hi;

I don't know bobbym's education but both of you are obviously gifted people.

He is gifted, I am a bumpkin .

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Bobbym's gifted as much as he denies it.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Hi

I found the generating function for V in the case that m digits are not considered for the splitting:

EDIT: Found a formula for V b):

*Last edited by anonimnystefy (2013-08-08 01:45:22)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Hi;

For convenience the recurrence is:

a[n] = 3 a[n - 1] - 5 a[n - 3] + 3 a[n - 5] + a[n - 6]

with

a[0] = 0, a[1] = 0, a[2] = 1, a[3] = 3, a[4] = 9, a[5] = 22

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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I wonder if a general formula is achievable for k unused numbers.

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**meanmaths****Member**- Registered: 2013-08-01
- Posts: 51

A quick post to present the (probably) last parts.

I'll come back to your previous answers later today or tomorrow

____Part VII____

I no longer think it is that hard, compared to the "noise" digits.

It now strikes me as relatively easy (at least the simplest versions)

Anyway, here it is

Given the sequence, there is now some order involved because we want

to tag the first split (it will be the X of previous parts). You can see it as if any

split now has a boolean/binary attribute indicating whether it is the first or not.

Only one of the split can have this boolean at true.

For instance, with a sequence of length 3 and positions 123

1 2 3 --> 1 or 2 or 3 could be the first split

1 23 --> 1 or 23 could be the first split

12 3 --> 12 or 3 could be the first split

For now, let's keep it that simple

(actually, if the first split has to be the X, there are some

obvious constraints on the lengths of the splits)

The first split can be of length

(a) 1 or 2

(b) 1 or 2 or 3 (all the other splits would still be made of 2 digits at most)

___Part VIII__

Example: 32017 --> The first split defines X=20,

the sequence becomes 3[20]17 --> the splits 3 and 17 complete the splitting.

Note that 31 cannot be a valid split: 20 is in the way

The first split defines the target sum

8995383

115161845

The first split defines the target mean

369792

195457

There are much more difficult instances in Mean Sumurai but

these will be fine for a text-version (plus, I have to hurry)

later

*Last edited by meanmaths (2013-08-08 09:57:30)*

Number Splitter #1

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Hi

For the first part 1 of VII the recurence is the same as for part 1 of V, but the initial conditions a[0]=0,a[1]=1,a[2]=3,a[3]=7.

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**bobbym****Administrator**- From: Bumpkinland
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8 + [99] + 5 + 3 + 83 = 99

[115] 1 + 61 + 8 + 45 = 115

[36]9792 -> ( 9 + 7 + 92 ) / 3 = 36

[19]5457-> (5 + 45 + 7) / 3 = 19

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Formula:

*Last edited by anonimnystefy (2013-08-08 13:11:24)*

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**bobbym****Administrator**- From: Bumpkinland
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Hi anonimnystefy;

I am not always getting integers for that when I use n = 0,1,2,3,4...

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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I have edited my post. Apparently F[n-1]+F[n]=F[n+2], according to me.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Hi;

I meant post #42.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

Good, we are talking about the same post. Have you seen the edited version of it?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Edited?

Looks the same.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Nope. The index of F changed. It was n+2 before.

*Last edited by anonimnystefy (2013-08-08 14:42:57)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Okay, it is working fine now.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

Great then. Still need to get the second part.

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