 mersenneforum.org Sums of consecutive integers
 Register FAQ Search Today's Posts Mark Forums Read 2014-12-13, 18:52 #1 TheMawn   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 re-write the sums. I'm sure I can wrestle out some relation between M and K that explains everything.   2014-12-13, 19:34 #2 Batalov   "Serge" Mar 2008 Phi(4,2^7658614+1)/2 100100110010002 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  As A and B increase, B/A --> sqrt(2) You can plug  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)   2014-12-14, 18:13 #3 TheMawn   May 2013 East. Always East. 11×157 Posts Ah. That's the step that never occurred to me. Thanks. Batalov.   2014-12-19, 11:47   #4
davar55

May 2004
New York City

423310 Posts Quote:
 Originally Posted by Batalov 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  As A and B increase, B/A --> sqrt(2) You can plug  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?   2014-12-19, 12:42 #5 Nick   Dec 2012 The Netherlands 23·11·19 Posts 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  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post carpetpool Miscellaneous Math 14 2017-02-26 15:55 gophne Miscellaneous Math 54 2017-02-22 22:28 XYYXF Computer Science & Computational Number Theory 0 2014-12-05 17:32 mfgoode Puzzles 12 2007-07-24 09:43 Kees Puzzles 22 2006-07-30 15:33

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