A curious sum

[franktaw at netscape.net]

Actually, this is pretty straightforward.  For
A002620(k) < m < A002620(k+1), m occurs once in the group
ending with A002620(k) (which is group number k-1) and in the
next group; and A002620(k) occurs in its own group and the
second one after.

I'm inclined to submit both sequences: the signed A002260, and
the cumulative sums.  Any suggestions or other comments?

Franklin T. Adams-Watters

-----Original Message-----
From: franktaw at netscape.net

Consider first the sequence which counts up to k (A002260):

1, 1,2, 1,2,3, 1,2,3,4, 1,2,3,4,5, 1,2,3,4,5,6, ...

Now, alternate the signs in  each group so that the final sign is
positive:

1, -1,2, 1,-2,3, -1,2,-3,4, 1,-2,3,-4,5, -1,2,-3,4,-5,6 ...

Take the cumulative sum of this sequence:

1, 0,2, 3,1,4, 3,5,2,6, 7,5,8,4,9, 8,10,7,11,6,12, ...

It appears that every positive integer occurs exactly twice in this
sequence (with 0 occurring only once).

(Note that the last elements of each group are the quarter-squares
(A002620), and the next to last elements are the same sequence, shifted
left by two.  These next to last elements are the minimum in the 
group.)

Franklin T. Adams-Watters



Franklin,

Your sequence is interesting!

You might consider cross-referencing your sequence to these others, also
related to A002260:

A014370 If n = binomial(b,2)+binomial(c,1), b>c>=0 then a(n) =
binomial(b+1,3)+binomial(c+1,2).

A051678 Square-pyramid-tree numbers: a(n)=sum(b(m),m=1..n), b(m)=1^2,
1^2,2^2, 1^2,2^2,3^2,..=(A002260)^2.

Note these also have the same parity as your sequence:

1,0,2,3,1,4,3,5,2,6,7,5,8,4,9,8,10,7,11,6,12,13,11,14,10,15,9,16,15,17,14,18
,13,19,12,20,21,19,22,18,23,17,24,16,25,24,26,23,27,22,28,21,29,20,30,

1,0,0,1,1,0,1,1,0,0,1,1,0,0,1,0,0,1,1,0,0,1,1,0,0,1,1,0,1,1,0,0,1,1,0,0,1,1,
0,0,1,1,0,0,1,0,0,1,1,0,0,1,1,0,0,

The run lengths in the parity sequence are:

1,2,2,1,2,2,2,2,1,2,2,2,2,2,2,1,2,2,2,2,2,2,2,2,1, ...

(not in the OEIS).

Regards,
Jeremy Gardiner

On 26/5/08 22:06, "franktaw at netscape.net" <franktaw at netscape.net> wrote:

> Consider first the sequence which counts up to k (A002260):
> 
> 1, 1,2, 1,2,3, 1,2,3,4, 1,2,3,4,5, 1,2,3,4,5,6, ...
> 
> Now, alternate the signs in  each group so that the final sign is
> positive:
> 
> 1, -1,2, 1,-2,3, -1,2,-3,4, 1,-2,3,-4,5, -1,2,-3,4,-5,6 ...
> 
> Take the cumulative sum of this sequence:
> 
> 1, 0,2, 3,1,4, 3,5,2,6, 7,5,8,4,9, 8,10,7,11,6,12, ...
>