20141213, 18:52  #1 
May 2013
East. Always East.
11·157 Posts 
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 rewrite the sums. I'm sure I can wrestle out some relation between M and K that explains everything. 
20141213, 19:34  #2 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
10010011001000_{2} Posts 
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 [610] [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 20141213 at 19:37 Reason: (forgot the link) 
20141214, 18:13  #3 
May 2013
East. Always East.
11×157 Posts 
Ah. That's the step that never occurred to me. Thanks. Batalov.

20141219, 11:47  #4  
May 2004
New York City
4233_{10} Posts 
Quote:


20141219, 12:42  #5 
Dec 2012
The Netherlands
2^{3}·11·19 Posts 
Possibly of interest:
https://gowers.wordpress.com/2014/11...ofkthpowers/ Last fiddled with by Nick on 20141219 at 12:43 Reason: corrected link 
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