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Mar 24, 2010 at 4:23 comment added tdnoe Thank you all for answering this obvious question. Next time I will have a cup of coffee first!
Mar 24, 2010 at 0:56 comment added Gerhard Paseman According to Michael Lugo, it looks like Erdos followed my advice in 1935, although he seemed more interested in asymptotics than in arbitrarily long arithmetic progressions for this problem.
Mar 23, 2010 at 21:56 answer added Michael Lugo timeline score: 21
Mar 23, 2010 at 21:38 comment added Gerhard Paseman To finish this train of thought: 1) Show that the multiple of any abundant (here I use abundant to include perfect numbers) number is also abundant. 2) Show that for any prime p, there is an abundant number whose smallest prime factor is p. 3) Use the Chinese Remainder Theorem to show that there are arbitrarily long arithmetic progressions of abundant numbers, where you can choose the common difference d in advance. That was fun. Now back to matrices. Gerhard "Ask Me About System Design" Paseman, 2010.03.23
Mar 23, 2010 at 21:33 history edited tdnoe CC BY-SA 2.5
Corrected spelling
Mar 23, 2010 at 21:30 comment added Gerhard Paseman I think you have enough now to make your own proof. Good conjecture on your part, now see what you can do to improve it. Gerhard "Ask Me About System Design" Paseman, 2010.03.23
Mar 23, 2010 at 21:29 comment added Gerhard Paseman Similar questions might be better for this forum, e.g. what is the maximum number possible of abundant integers in any closed interval of length 5? Gerhard "Ask Me About System Design" Paseman, 2010.03.23
Mar 23, 2010 at 21:26 comment added tdnoe Not homework. Just an observation.
Mar 23, 2010 at 21:24 comment added Gerhard Paseman 12n <= 1 + 6n + 2 + 3n + 3 + 2n + 6 + n , so this should help answer the question. Is it a homework question? Gerhard "Ask Me About System Design" Paseman, 2010.03.23
Mar 23, 2010 at 21:15 history asked tdnoe CC BY-SA 2.5