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 biwema 2004-03-30 00:59

sequence

Let's define the following sequence:

x[0] = 1

x[n] = ( 1 + x[0]^3 + x[1]^3 + .... + x[n-1]^3) / n

Are all of these sequence integers?

Hmmm ... Let's see:

x[0] = 1

x[1] = (1 + 1 ) / 1 = 2

x[2] = (1 + 1 + 8) / 2 = 5

x[3] = (1 + 1 + 8 + 125) / 3 = 45

x[4] = (1 + 1 + 8 + 125 + 91125) / 4 = 22815

Lazy suggestion: Someone want to check OEIS?

 FenwayFrank 2004-04-04 05:24

[QUOTE=cheesehead]Lazy suggestion: Someone want to check OEIS?[/QUOTE]

[URL=http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A005166]This[/URL] says that it's [B]not[/B] always an integer. It still is after 12 terms; calculating the thirteenth is bringing my machine to its knees.

 Kevin 2004-04-04 05:47

I can get it down to a recursive of sorts....

x[n]=(1+x[0]^3+x[1]^3...+x[n-1]^3)/n
n*x[n]=(1+x[0]^3+x[1]^3...+x[n-2]^3)+x[n-1]^3
n*x[n]-x[n-1]^3=(1+x[0]^3+x[1]^3...+x[n-2]^3)
(n*x[n]-x[n-1]^3)/(n-1)=(1+x[0]^3+x[1]^3...+x[n-2]^3)/(n-1)
(n*x[n]-x[n-1]^3)/(n-1)=x[n-1]
n*x[n]-x[n-1]^3=(n-1)*x[n-1]
x[n]={(n-1)*x[n-1]+x[n-1]^3}/n

edit: lil bit more
x[n]=x[n-1]*[n-1+x[n-1]^2]/n
x[n]=x[n-1]*{n-(x[n-1]-1)(x[n-1]+1)}/n
this tells us that if n divides x[n-1]-1, x[n-1], or x[n-1]+1, then x[n] is integral. If it doesn't, then x[n] is not an integer.

[url]http://mathworld.wolfram.com/GoebelsSequence.html[/url]

"A sequence even more striking for assuming integer values only for many terms is the 3-Göbel sequence ..."

 biwema 2004-04-04 23:58

Maybe I should have asked "Which is the first noninteger element of the sequence?"...

 FenwayFrank 2004-04-05 13:52

I can tell you that it's >15 (2.148- e 707870).
I have to shut the machine down to go into work, otherwise I'd keep it running.

JooC: do you know the answer?

 FeLiNe 2004-04-05 22:49

[QUOTE=biwema]Maybe I should have asked "Which is the first noninteger element of the sequence?"...[/QUOTE]

Of course you could just follow the reference given in the OEIS:

(Example 25).

It's quite a worthwhile and enjoyable read.

The answer is that the 89th(!) number is not an integer.

EDIT: You'll also find that the same problem for [i]fourth[/i] powers yields a non-integer for the 97 number, for [i]fifth[/i] powers at the 214th position ...

 FenwayFrank 2004-04-06 12:48

Is the article available somewhere else? The site seems to be limited to people or places that have existing accounts.

 FeLiNe 2004-04-06 18:26

Oh, sorry - I didn't realise that. My cable-access is set up such that the machine appears in the IP-range of CalTech and is, in effect, a member f the CalTech network for all intents and purposes -- with the attendant benefit that I can get to all kinds of online journals and such.

To outline the procedure: you can do the whole operation modulo 89. There's the usual rules of modular arithmetic to observe, but this keeps the size of all numbers extremely manageable.

The problem is presented for the case where the individual terms are squares and cubes (like the one in this thread) and the solution is shown for the case of squares (where the 43rd term is not an integer) and then a comment is made that the same procedure working mod89 shows that the 89th term isn't integer in the cube-case.

If it helps, the author (Richard K. Guy) gives as a reference his own book "Unsolved problems in number theory" (Springer 1981), E15. If you're lucky, that might be available at your local library...

 biwema 2004-04-06 21:37

Well done, Feline.

the 89th element, that is also the answer I have. That numebr is really big (10^(10^40)), so it is not possible to compute the whole number.

As already sugessted by FeLiNe, the easiest wat to compute the first noninteger element, is to do the whole Equation modulo a prime number. starting witrh 3, 5, 7 etc.
If you have a multiprecision tool, you can also calculate it mod 100! (factorial).

There is also a simplification at mathworld (wolfram research) about Göbel's Sequence ( [url]http://mathworld.wolfram.com/GoebelsSequence.html[/url] )

they transform the sequence in a related one, which is the already the sum of all beginning elements.

So

s[n] = n * x[n]

s[n+1] = s[n] + ((s[n]^3) / n^3)

here the modulo operation is not that difficult.

maybe there is a power, where it takes even longer to meet the first noninteger element...

 FeLiNe 2004-04-07 00:38

[QUOTE=biwema]
maybe there is a power, where it takes even longer to meet the first noninteger element...[/QUOTE]

Quoting from example 26 in that same article (which was the generalization to powers higher than than 3):

[quote]
and found the first noninteger term, x[sub]n[/sub], in the sequence involving [i]k[/i]th powers, to be
[pre]
k 2 3 4 5 6 7 8 9 10 11
n 43 89 97 214 19 239 37 79 83 239
[/pre]
[/quote]

:cat:

 Terrence Law 2004-06-11 02:03

[QUOTE=biwema]Let's define the following sequence:

x[0] = 1

x[n] = ( 1 + x[0]^3 + x[1]^3 + .... + x[n-1]^3) / n

Are all of these sequence integers?[/QUOTE]

This is a mind-boggling tool and you have to use inequality to draw a line or BEDMAS or asymptotic function. Draw and underline a graph and use subscripts.

Beware, subscript out of range in line 0, write every digit on a square.

:huh: :coffee: :rolleyes:

 Terrence Law 2004-06-11 02:05

"A sequence even more striking for assuming integer values only for many terms is the 3-Göbel sequence ..."[/QUOTE]

Let's use hints and pinpointers and just cheat and look at the answers of the books!

:smile: You are smirking!

Just add more references and publish and print the page neatly into a binder and stick it to the wall.

This sequence is challenging, you can write forever, keep writing on foolscap brown paper.

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