Red-Dog's Last Theorem

[EvilPie]

Lol

I thought I was the only saddo that wanted to work this out.

Are you sure about you 7 and 8? They look a bit low to me.

Looking at this sequence there will be no mathematical formulae available. UL

Matt, don't be silly! Of COURSE there is a formula.

Well if there is it's going to have lots of log's, ln's, <'s, >'s and upside down i's (haven't got a factorial button on here).

I honestly don't see how this sequence can have a mathematical formula to it. I agree it should have but the sequence looks too awkward.

What formula can possibly make both 7 and 8 = 4?

I'll be thinking on this for a while.

It's interesting that 3, 6 and 9 lead to 3, 6 and 9. Could be a starting point me thinks.

Really trivial example, but:

4 + (floor (n/9))

(n x 0) + 4

Even trivialler.

[tikay]

This is more complex than you could ever possibly imagine and, to my knowledge, is unsolved.

Quite probably.

In other words, the Formula for it has yet to be discovered. But it must exist.

[kinboshi]

We promised Mark @ DTD that the clever bods on here would have an answer for him tonight. 

[  ] He'll be devastated.

[tikay]

We promised Mark @ DTD that the clever bods on here would have an answer for him tonight. 

[  ] He'll be devastated.

We await Tighty. Andrew & JonMW are let downs.

[EvilPie]

This is more complex than you could ever possibly imagine and, to my knowledge, is unsolved.

Quite probably.

In other words, the Formula for it has yet to be discovered. But it must exist.

You would think so wouldn't you.

It seems fairly straight forward. You have 2 equal stacks, you intersperse them equal, half them again.......

Seems like quite a simple sequence.

With even numbered stacks I'm sure the formula would be simple but when you throw in uneven numbered stacks it gets very complex.

Add in a few prime numbered stacks and it becomes imo almost impossible.

[12barblues]

I have a few minutes free this afternoon, so I'll tackle Red Dog's Last Theorem just as soon as I've finished my proof of the Reimann Hypothesis.....

[tikay]

I have a few minutes free this afternoon, so I'll tackle Red Dog's Last Theorem just as soon as I've finished my proof of the Reimann Hypothesis.....

http://en.wikipedia.org/wiki/Riemann_hypothesis

There you go, sorted. Now sort that piffling chip-riffliing thing please.

[EvilPie]

I have a few minutes free this afternoon, so I'll tackle Red Dog's Last Theorem just as soon as I've finished my proof of the Reimann Hypothesis.....

http://en.wikipedia.org/wiki/Riemann_hypothesis

There you go, sorted. Now sort that piffling chip-riffliing thing please.

Not until he's sorted out the second 10 trillion zeros.

You can never be too sure.

[tikay]

I have a few minutes free this afternoon, so I'll tackle Red Dog's Last Theorem just as soon as I've finished my proof of the Reimann Hypothesis.....

http://en.wikipedia.org/wiki/Riemann_hypothesis

There you go, sorted. Now sort that piffling chip-riffliing thing please.

Not until he's sorted out the second 10 trillion zeros.

You can never be too sure.

Made my eyes glaze over, that!

Next, the Yang-Mills existence & mass gap.

[EvilPie]

Lol

I thought I was the only saddo that wanted to work this out.

Are you sure about you 7 and 8? They look a bit low to me.

Looking at this sequence there will be no mathematical formulae available. UL

Matt, don't be silly! Of COURSE there is a formula.

Well if there is it's going to have lots of log's, ln's, <'s, >'s and upside down i's (haven't got a factorial button on here).

I honestly don't see how this sequence can have a mathematical formula to it. I agree it should have but the sequence looks too awkward.

What formula can possibly make both 7 and 8 = 4?

I'll be thinking on this for a while.

It's interesting that 3, 6 and 9 lead to 3, 6 and 9. Could be a starting point me thinks.

There will be a formula for it, 100% guaranteed. If it can happen, there's a formula.

If chance permits, Matt, try & get a copy of, & read, "Fermat's Lost Theorem", upon which this Thread title is based. You'd absolutely love it. And you'd learn the mathematical formula for working out the true (actual) length of the bendiest of rivers on earth, simply by using pye.

I can't find it.

[tikay]

Lol

I thought I was the only saddo that wanted to work this out.

Are you sure about you 7 and 8? They look a bit low to me.

Looking at this sequence there will be no mathematical formulae available. UL

Matt, don't be silly! Of COURSE there is a formula.

Well if there is it's going to have lots of log's, ln's, <'s, >'s and upside down i's (haven't got a factorial button on here).

I honestly don't see how this sequence can have a mathematical formula to it. I agree it should have but the sequence looks too awkward.

What formula can possibly make both 7 and 8 = 4?

I'll be thinking on this for a while.

It's interesting that 3, 6 and 9 lead to 3, 6 and 9. Could be a starting point me thinks.

There will be a formula for it, 100% guaranteed. If it can happen, there's a formula.

If chance permits, Matt, try & get a copy of, & read, "Fermat's Lost Theorem", upon which this Thread title is based. You'd absolutely love it. And you'd learn the mathematical formula for working out the true (actual) length of the bendiest of rivers on earth, simply by using pye.

I can't find it.

I SO nearly bit.......even got as far as a Link to Amazon. I'm so easy to whoosh!

[david3103]

Lol

I thought I was the only saddo that wanted to work this out.

Are you sure about you 7 and 8? They look a bit low to me.

Looking at this sequence there will be no mathematical formulae available. UL

Matt, don't be silly! Of COURSE there is a formula.

Well if there is it's going to have lots of log's, ln's, <'s, >'s and upside down i's (haven't got a factorial button on here).

I honestly don't see how this sequence can have a mathematical formula to it. I agree it should have but the sequence looks too awkward.

What formula can possibly make both 7 and 8 = 4?

I'll be thinking on this for a while.

It's interesting that 3, 6 and 9 lead to 3, 6 and 9. Could be a starting point me thinks.

There will be a formula for it, 100% guaranteed. If it can happen, there's a formula.

If chance permits, Matt, try & get a copy of, & read, "Fermat's Lost Theorem", upon which this Thread title is based. You'd absolutely love it. And you'd learn the mathematical formula for working out the true (actual) length of the bendiest of rivers on earth, simply by using pye.

I can't find it.

Fermat's Last Theorem and we know where it is, it's the proof that was lost.

As to Red-Dog's problem - the issue of which way you shift the stacks for the second and subsequent riffles will be crucial for odd starting stacks, (which colour chip is on the base of the left-hand stack?) you need consistency in both this, and in the form of the riffle. Did this impact on the observed results given thus far?

[12barblues]

I have a few minutes free this afternoon, so I'll tackle Red Dog's Last Theorem just as soon as I've finished my proof of the Reimann Hypothesis.....

http://en.wikipedia.org/wiki/Riemann_hypothesis

There you go, sorted. Now sort that piffling chip-riffliing thing please.

Not until he's sorted out the second 10 trillion zeros.

You can never be too sure.

Made my eyes glaze over, that!

Next, the Yang-Mills existence & mass gap.

Blimey!  Tough audience.  Could be a busy afternoon as I have to cut the grass as well.  I suppose I could fit it in after I finish my thesis entitled 'Einstein 1 Bohr 0 :  Why The Copenhagen Interpretation Of Quantum Mechanics Is Ridic Com'.

[cod meharly]

This is gna drive me crazy...im quite a nerd with numbers and to me this looks way more complex than i could possibly imagine!
 

[TightEnd]

Not too difficult really

Just input the starting numbers into the following equations

    α0 = wz + h + j − q = 0

    α1 = (gk + 2g + k + 1)(h + j) + h − z = 0

    α2 = 16(k + 1)3(k + 2)(n + 1)2 + 1 − f2 = 0

    α3 = 2n + p + q + z − e = 0

    α4 = e3(e + 2)(a + 1)2 + 1 − o2 = 0

    α5 = (a2 − 1)y2 + 1 − x2 = 0

    α6 = 16r2y4(a2 − 1) + 1 − u2 = 0

    α7 = n + l + v − y = 0

    α8 = (a2 − 1)l2 + 1 − m2 = 0

    α9 = ai + k + 1 − l − i = 0

    α10 = ((a + u2(u2 − a))2 − 1)(n + 4dy)2 + 1 − (x + cu)2 = 0

    α11 = p + l(a − n − 1) + b(2an + 2a − n2 − 2n − 2) − m = 0

    α12 = q + y(a − p − 1) + s(2ap + 2a − p2 − 2p − 2) − x = 0

    α13 = z + pl(a − p) + t(2ap − p2 − 1) − pm = 0


The 14 equations α0, …, α13 can be used to produce a prime-generating polynomial inequality in 26 variables:

    (k+2)(1-\alpha_0^2-\alpha_1^2-\cdots-\alpha_{13}^2) > 0

ie:

    (k + 2)(1 −
    [wz + h + j − q]2 −
    [(gk + 2g + k + 1)(h + j) + h − z]2 −
    [16(k + 1)3(k + 2)(n + 1)2 + 1 − f2]2 −
    [2n + p + q + z − e]2 −
    [e3(e + 2)(a + 1)2 + 1 − o2]2 −
    [(a2 − 1)y2 + 1 − x2]2 −
    [16r2y4(a2 − 1) + 1 − u2]2 −
    [n + l + v − y]2 −
    [(a2 − 1)l2 + 1 − m2]2 −
    [ai + k + 1 − l − i]2 −
    [((a + u2(u2 − a))2 − 1)(n + 4dy)2 + 1 − (x + cu)2]2 −
    [p + l(a − n − 1) + b(2an + 2a − n2 − 2n − 2) − m]2 −
    [q + y(a − p − 1) + s(2ap + 2a − p2 − 2p − 2) − x]2 −
    [z + pl(a − p) + t(2ap − p2 − 1) − pm]2)
    > 0