[seqfan] Re: New sequence.

[israel at math.ubc.ca]

On Sep 19 2012, Ed Jeffery wrote:

>I have a new sequence and would like someone to work out the details (and
>proofs) followed by submitting it to OEIS, because I am not able to work on
>it. The sequence is described as follows, along with some thoughts on
>modifying the definitions of a couple of other existing OEIS sequences.
>
>First, and this is probably a known result, let M_{n,k} be the 2 X 2 matrix
>
>M_{n,k} = [0, 1; k, n].
>
>Prove that the characteristic equation of this matrix is of the form
>
>x^2 - n*x - k = 0,
>
>and with roots a_j (say), j = 1,2, such that
>
>a_1 = (n + sqrt(n^2 + 4*k))/2  and  a_2 = (n - sqrt(n^2 + 4*k))/2.

So far, scompletely routine.

>Second, taking the expressions in the above radical, form the triangle
>
>T(n,k) = n^2 + 4*k, 0 <= k <= n,
>
>which begins as
>
> 0,
> 1,  5,
> 4,  8, 12,
> 9, 13, 17, 21,
>16, 20, 24, 28, 32,
>25, 29, 33, 37, 41, 45,
>36, 40, 44, 48, 52, 56, 60,
>...
>
>(This is the sequence to be submitted, since it is not in OEIS.)
>
>Prove or disprove that the row sums of T are given by A007531 [1] (up to an
>offset).

sum_{k=0}^{n} (n^2 + 4 k) = n^2 (n+1) + 4 n (n+1)/2 = n (n+1) (n+2)

>Third, because of the definition of T, the definition and offset for
>A028884 [2] could be changed to n^2 + 4*(n - 1) with offset 1,2.
>
>Fourth, the outer diagonal {0, 5, 12, ...} of T is A028437 [3] in disguise
>so, because of the definition of T, the definition of A028437 could be
>changed from n^2 - 4 (offset 2,2) to n^2 + 4*n, or n*(n + 4), with offset
>0,2.
>
>Finally (this seems obvious), prove or disprove that the entries of T are
>distinct and that A003656 [4] (taken as a set) is a subset of T (taken as a
>set).

n^2 + 4 k = m^2 + 4 j iff 4 (k-j) = m^2 - n^2 = (m+n)(m-n)
Suppose m>n.  Now we must have m==n mod 2, so m-n >= 2, and
k >= k - j = (m+n)(m-n)/4 >= (m+n)/2 > n.  So yes, they are all distinct.

The members of T (as a set) are all the integers congruent to 0 or 1 mod 4.
Indeed, for any positive integer j, if n = floor(sqrt(j)) so 
n^2 <= j < (n+1)^2, we have 4 j = (2 n)^2 + 4 (j - n^2) with 
0 <= j - n^2 <= (n+1)^2 - 1 - n^2 = 2n  
while if m^2 + m <= j < (m+1)^2 + (m+1) we have
4 j + 1 = (2m+1)^2 + 4 (j - (m+1) m) with
0 <= j - (m+1) m <= (m+1)^2 + m - (m+1) m = 2m+1


Robert Israel
University of British Columbia


>If someone will submit this sequence, then please let me know the A-number
>so I can find it.
>
>Thanks,
>
>LEJ
>
>REFERENCES
>
>[1] OEIS sequence A007531, https://oeis.org/A007531.
>[2] OEIS sequence A028884, https://oeis.org/A028884.
>[3] OEIS sequence A028437, https://oeis.org/A028347.
>[4] OEIS sequence A003656, https://oeis.org/A003656.
>
>_______________________________________________
>
>Seqfan Mailing list - http://list.seqfan.eu/
>