20040330, 00:59  #1 
Mar 2004
3·127 Posts 
sequence
Let's define the following sequence:
x[0] = 1 x[n] = ( 1 + x[0]^3 + x[1]^3 + .... + x[n1]^3) / n Are all of these sequence integers? 
20040402, 01:54  #2 
"Richard B. Woods"
Aug 2002
Wisconsin USA
1E0C_{16} Posts 
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? Last fiddled with by cheesehead on 20040402 at 02:02 
20040404, 05:47  #4 
Aug 2002
Ann Arbor, MI
433 Posts 
I can get it down to a recursive of sorts....
x[n]=(1+x[0]^3+x[1]^3...+x[n1]^3)/n n*x[n]=(1+x[0]^3+x[1]^3...+x[n2]^3)+x[n1]^3 n*x[n]x[n1]^3=(1+x[0]^3+x[1]^3...+x[n2]^3) (n*x[n]x[n1]^3)/(n1)=(1+x[0]^3+x[1]^3...+x[n2]^3)/(n1) (n*x[n]x[n1]^3)/(n1)=x[n1] n*x[n]x[n1]^3=(n1)*x[n1] x[n]={(n1)*x[n1]+x[n1]^3}/n edit: lil bit more x[n]=x[n1]*[n1+x[n1]^2]/n x[n]=x[n1]*{n(x[n1]1)(x[n1]+1)}/n this tells us that if n divides x[n1]1, x[n1], or x[n1]+1, then x[n] is integral. If it doesn't, then x[n] is not an integer. Last fiddled with by Kevin on 20040404 at 05:54 
20040404, 07:26  #5 
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}×3×641 Posts 
http://mathworld.wolfram.com/GoebelsSequence.html
"A sequence even more striking for assuming integer values only for many terms is the 3Göbel sequence ..." Last fiddled with by cheesehead on 20040404 at 07:33 
20040404, 23:58  #6 
Mar 2004
3×127 Posts 
Maybe I should have asked "Which is the first noninteger element of the sequence?"...

20040405, 13:52  #7 
Oct 2002
7 Posts 
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? 
20040405, 22:49  #8  
Dec 2003
23 Posts 
Quote:
Of course you could just follow the reference given in the OEIS: http://links.jstor.org/sici?sici=000...3E2.0.CO%3B2Z (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 fourth powers yields a noninteger for the 97 number, for fifth powers at the 214th position ... Last fiddled with by FeLiNe on 20040405 at 22:53 Reason: slightly more info 

20040406, 12:48  #9 
Oct 2002
7 Posts 
Is the article available somewhere else? The site seems to be limited to people or places that have existing accounts.

20040406, 18:26  #10 
Dec 2003
23 Posts 
Oh, sorry  I didn't realise that. My cableaccess is set up such that the machine appears in the IPrange 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 cubecase. 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... 
20040406, 21:37  #11 
Mar 2004
3·127 Posts 
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 ( http://mathworld.wolfram.com/GoebelsSequence.html ) 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... 
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