[FOM] arithmetic without product

[Joe Shipman]

I don’t know where Sylvester’s argument was published but it is practically trivial to see the following:

If j is even, n is expressible as the sum of j consecutive integers iff n mod j = j/2, and for sufficiently large n all the integers in the sum are positive 

If j is odd, n is expressible as the sum of j consecutive integers iff n mod j = 0, and for sufficiently large n all the integers in the sum are positive.

Therefore, if we want an integer expressible in at least k different ways as a sum of positive integers, (2k-1)(2k-3)(2k-5)...(5)(3)(1) will suffice (it has the right divisibility properties and is easily seen to be large enough).

For k>=4, we can omit the first factor (2k-1) because the product will  also, being odd, be expressible as the sum of two consecutive positive integers: 5*3*1 = 1+2+3+4+5 = 4+5+6 =7+8 = 15 etc.

— JS

Sent from my iPhone

> On Jul 22, 2018, at 5:26 PM, Cédric Doucet <doucetced at gmail.com> wrote:
> 
> Could you provide where Sylvester's argument was published please?
> 
> 2018-07-21 10:20 GMT+02:00 José Manuel Rodriguez Caballero <josephcmac at gmail.com>:
>> Consider the following theorem:
>> 
>>> For any positive integer k, there is a positive integer n having at least k representations as sum of consecutive positive integers.
>> 
>> For example, for k = 3 we have n = 9 = 4+5 = 2+3+4.
>> 
>> For the statement of this theorem we need the addition. If we also have the multiplication, then our theorem can be proved using a well-known argument due to J. J. Sylvester. On the other hand, if we have not multiplication and we restrict ourselves to first order logic, the existence of a proof of this result is not so clear. Is there an "arithmetic with just addition" where this theorem can be stated but it cannot be proved?
>> 
>> Kind Regards,
>> Jose M.
>> 
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