Timeline for Proving that every term of the sequence is an integer
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| May 29, 2013 at 21:24 | comment | added | Abhinav Kumar | The line breaks are all messed up, but I'm sure you can unentangle them. | |
| May 29, 2013 at 21:23 | comment | added | Abhinav Kumar | @Barry, you can get the recurrence by doing some linear algebra. Here's some gp code to do the m = 3 line (my indices are shifted by 1, since vectors and matrices in gp start with 1 rather than 0). Then you prove it by induction, of course :-) d = 80; A = matrix(d,d); for(i=1,d,A[i,1] = 1; A[i,2] = 1; A[1,i] = 1); for(i=2,d, for(j=3,d, A[i,j] = (A[i,j-2]*A[i-1,j] + A[i,j-1]*A[i-1,j-1])/A[i-1,j-2])); v = A[4,] m = matrix(40,9,i,j,v[i+j-1]); k = matkerint(m) p = vector(9,i,x^(i-1))*k factor(p[1]) | |
| May 29, 2013 at 19:49 | comment | added | Barry Cipra | @Abhinav, can you elaborate on your comment? For $m=2$ (i.e., the third row in MHMertens's matrix), I get (and can prove, using the recursion I've given in a separate answer) a 4-term linear recurrence with characteristic polynomial $x^4-x^3-4x^2-x+1 = (x+1)^2(x^2-3x+1)$, but I don't see how to get, much less prove, corresponding recurrences for $m\gt2$. (I only looked for the 4-term recurrence because your comment said there was one, and it was easy enough to find.) | |
| May 29, 2013 at 14:49 | comment | added | Abhinav Kumar | The degrees of the linear recurrence satisfied by the sequences $(a_{m,n})_n$ are $1,2,4,8,15,26,...$ for $m = 0,1,2,3,4,5,...$, which OEIS recognizes as the Cake numbers (A000125). The characteristic polynomials seem to factor into low-degree pieces. | |
| May 29, 2013 at 14:27 | comment | added | Dietrich Burde | Even if you fix either $n$ or $m$, the resulting (integer) sequence becomes quickly unknown to "Integer Sequences". Here quickly means for $m\ge 2$ or $n\ge 4$. | |
| May 29, 2013 at 14:08 | comment | added | Barry Cipra | @Tom, once again I demonstrated my inability to do unassisted arithmetic. The $91$ was wrong, but it should have been a $59$, rather than $57$. I just corrected my correction. | |
| May 29, 2013 at 14:07 | history | edited | Barry Cipra | CC BY-SA 3.0 |
edited body
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| May 29, 2013 at 13:46 | comment | added | Tom De Medts | @Barry: I've checked the first $400 \times 400$ entries by computer, and they are all integers. | |
| May 29, 2013 at 13:43 | comment | added | Barry Cipra | @MHMertens, I took the liberty of correcting an entry in the matrix: there was a 91 that should have been a 57. Having done so, however, the next term in that row appears to be $(57\cdot8+21\cdot13)/5=145.8$. (Even if you leave the $57$ as $91$, you don't get an integer. Am I misreading things?) | |
| May 29, 2013 at 13:33 | history | edited | Barry Cipra | CC BY-SA 3.0 |
corrected an entry in the matrix
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| May 29, 2013 at 12:39 | history | edited | MHMertens | CC BY-SA 3.0 |
corrected calculation errors in the previous version.
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| May 29, 2013 at 10:24 | comment | added | MHMertens | Arrggh, sorry about that! I didn't see the Fibonacci sequence in there at first and so the two 4s in the matrix didn't disturb me. It was still early this morning... | |
| May 29, 2013 at 9:00 | comment | added | Tom De Medts | (By the way, it's also a little bit confusing that you interchanged rows and columns.) | |
| May 29, 2013 at 8:59 | comment | added | Tom De Medts | @MHMertens: Unfortunately, you did make a calculation mistake. The row $1, 4, 39/2, \dots$ should be $1, 5, 21, 91, \dots$. | |
| May 29, 2013 at 8:43 | comment | added | Dietrich Burde | But $(a_{1,n})=(1,1,2,3,5,8,...)$ should be the Fibonacci sequence, or not ? | |
| May 29, 2013 at 8:39 | history | answered | MHMertens | CC BY-SA 3.0 |