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Old 2014-12-13, 18:52   #1
TheMawn
 
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Default Sums of consecutive integers

I came up with this a good while ago now. I think the big inspiration came out of playing cribbage and noticing that you get a fifteen out of 4 5 6 or out or 7 8 (hence 4 + 5 + 6 = 7 + 8).

Of course one easily knows 1 + 2 = 3. I had eventually come up with 2 + 3 + 4 + 5 + 6 + 7 = 8 + 9 + 10, but it was starting to be too much to just think about, so I had written up a piece of code to find more. In essence, I was looking for numbers N, M and K which satisfied

N + (N + 1) + (N + 2) + ... + (N + M) = (N + M + 1) + (N + M + 2) + ... + (N + M + K)

I had made some simplifications to the formula using sums to make them manageable and I started looking for combinations of N, M and K which satisfied the equality. Having far too many to make any sense from, I focused on cases where N = 1.

1 + 2 + ... + 14 = 15 + ... + 20
1 + 2 + ... + 84 = 85 + ... + 119
1 + 2 + ... + 492 = 493 + ... + 696
1 + 2 + ... + 2870 = 2871 + ... + 4059
1 + 2 + ... + 16730 = 16371 + ... + 23660
1 + 2 + ... + 97512 = 97513 + ... + 137904

The ratio of the final RHS term to the final LHS term approaches the square root of two:

(20 / 14)2 = 2.0408...
(119 / 84)2 = 2.00694...
...
(137904 / 97512)2 = 2.000035...


The ratio of the final term of either sequence to the final term of the previous respective sequence approaches (1 + sqrt(2))2 = 5.82842...

84 / 16 = 6
492 / 84 = 5.857...
...
97512 / 16730 = 5.82857...

137904 / 23660 = 5.82857...


This was as much time as I was willing to put into this while I was in school. This is the first I've thought of it since, and I thought I would share. I will likely stumble into a simple explanation. My first order of business will be to re-write the sums. I'm sure I can wrestle out some relation between M and K that explains everything.
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Old 2014-12-13, 19:34   #2
Batalov
 
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Yes, of course, it would approach sqrt(2).
1+...+A = (A+1)+...+B
add LHS to RHS and simplify
2*(1+...+A) = 1+...+B
A*(A+1) = B*(B+1)/2
(A+1/2)^2 - 1/4 = ((B+1/2)^2 - 1/4)/2 [1]
As A and B increase, B/A --> sqrt(2)
You can plug [1] into [6-10] [URL="http://mathworld.wolfram.com/DiophantineEquation2ndPowers.html"]here[/URL] and get the parametric solution.
It is more fun to do it yourself. There will be a single series for all A and B's as a function of a single integer.


There is a Euler problem for square sums.

Last fiddled with by Batalov on 2014-12-13 at 19:37 Reason: (forgot the link)
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Old 2014-12-14, 18:13   #3
TheMawn
 
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Ah. That's the step that never occurred to me. Thanks. Batalov.
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Old 2014-12-19, 11:47   #4
davar55
 
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Quote:
Originally Posted by Batalov View Post
Yes, of course, it would approach sqrt(2).
1+...+A = (A+1)+...+B
add LHS to RHS and simplify
2*(1+...+A) = 1+...+B
A*(A+1) = B*(B+1)/2
(A+1/2)^2 - 1/4 = ((B+1/2)^2 - 1/4)/2 [1]
As A and B increase, B/A --> sqrt(2)
You can plug [1] into [6-10] [URL="http://mathworld.wolfram.com/DiophantineEquation2ndPowers.html"]here[/URL] and get the parametric solution.
It is more fun to do it yourself. There will be a single series for all A and B's as a function of a single integer.


There is a Euler problem for square sums.
What about cubes, fourth powers, nth powers?
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Old 2014-12-19, 12:42   #5
Nick
 
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Possibly of interest:
https://gowers.wordpress.com/2014/11...of-kth-powers/

Last fiddled with by Nick on 2014-12-19 at 12:43 Reason: corrected link
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