# How many sequences of consecutive integers sum to a particular positive integer greater than 1?

By the fundamental theorem of arithmetic, every positive integer greater than 1 can be written as a product of primes. Thus, we can write every positive integer as $$2^{ k }p_{ 1 }^{ q_{ 1 } }p_{ 2 }^{ q_{ 2 } }...p_{ m }^{ q_{ m } }$$ for some non-2 primes $$\{ p_{ i }\} _{ i=1 }^{ m }$$, some non-negative integer $$k$$ and some positive integers $$\{ q_{ i }\} _{ i=1 }^{ m }$$. Let's say that the sequence of consecutive integers starts at some integer $$x$$ and ends at some integer $$x+n$$ where $$n$$ is some positive integer (thus there are $$n+1$$ terms in the sequence). Note that this excludes sets with just one member. The sum of the consecutive integers is then given by $$\sum _{ k=0 }^{ n }{ (x+k) }$$, and thus we need $$\sum _{ k=0 }^{ n }{ (x+k) } =2^{ k }p_{ 1 }^{ q_{ 1 } }p_{ 2 }^{ q_{ 2 } }...p_{ m }^{ q_{ m } }$$ for the particular $$x$$ and $$n$$ to constitute a valid sequence of $$n+1$$ members. This gives that $$(n+1)x+\frac { n(n+1) }{ 2 } =2^{ k }p_{ 1 }^{ q_{ 1 } }p_{ 2 }^{ q_{ 2 } }...p_{ m }^{ q_{ m } }\quad \Leftrightarrow \quad x+\frac { n }{ 2 } =\frac { 2^{ k }p_{ 1 }^{ q_{ 1 } }p_{ 2 }^{ q_{ 2 } }...p_{ m }^{ q_{ m } }\quad }{ n+1 }$$, and thus $$x=\frac { 2^{ k }p_{ 1 }^{ q_{ 1 } }p_{ 2 }^{ q_{ 2 } }...p_{ m }^{ q_{ m } }\quad }{ n+1 } -\frac { n }{ 2 }$$.

So, the question becomes, how many positive integers $$n$$ can we choose such that the right-hand side of that equation is an integer and thus, so too, is $$x$$.

Well, there are two potential cases, that $$n$$ is even and that $$n$$ is odd. If $$n$$ is even, we have that $$\frac { n }{ 2 }$$ is an integer and thus, so too, must $$\frac { 2^{ k }p_{ 1 }^{ q_{ 1 } }p_{ 2 }^{ q_{ 2 } }...p_{ m }^{ q_{ m } }\quad }{ n+1 }$$ be. Therefore, we have that $$(n+1)|2^{ k }p_{ 1 }^{ q_{ 1 } }p_{ 2 }^{ q_{ 2 } }...p_{ m }^{ q_{ m } }$$. If that is the case, along with n being even, $$x$$ must then be an integer and $$x$$ and $$n$$ constitute a valid pair. Thus, $$n+1$$ must be any divisor of $$2^{ k }p_{ 1 }^{ q_{ 1 } }p_{ 2 }^{ q_{ 2 } }...p_{ m }^{ q_{ m } }$$ such that $$n$$ is even, or, equivalently that $$n+1$$ is odd. Thus, we can let $$n+1$$ be any odd, positive divisor (except 1) of $$2^{ k }p_{ 1 }^{ q_{ 1 } }p_{ 2 }^{ q_{ 2 } }...p_{ m }^{ q_{ m } }$$ giving $$(q_{ 1 }+1)(q_{ 2 }+1)(q_{ 3 }+1)...(q_{ m }+1)-1$$ valid pairs $$(x,n)$$ which have even $$n$$ (we wish to exclude the case where $$n+1=1$$, as $$n$$ must be a positive integer). Next, we have the case that $$n$$ is odd. Going back to $$x+\frac { n }{ 2 } =\frac { 2^{ k }p_{ 1 }^{ q_{ 1 } }p_{ 2 }^{ q_{ 2 } }...p_{ m }^{ q_{ m } }\quad }{ n+1 }$$, we have, multiplying both sides by 2, that $$2x+n=\frac { 2^{ k+1 }p_{ 1 }^{ q_{ 1 } }p_{ 2 }^{ q_{ 2 } }...p_{ m }^{ q_{ m } } }{ n+1 }$$. Noting that because $$n$$ is odd and $$2x$$ is even, the left-hand side is an odd integer, and thus, so too, must the right-hand side be. Thus, $$n+1$$ must be a divisor of $$\frac { 2^{ k+1 }p_{ 1 }^{ q_{ 1 } }p_{ 2 }^{ q_{ 2 } }...p_{ m }^{ q_{ m } } }{ n+1 }$$ such that $$2^{ k+1 }|(n+1)$$. Therefore, we have that $$n+1=2^{ k+1 }d$$ where $$d$$ can be any positive divisor of $$p_{ 1 }^{ q_{ 1 } }p_{ 2 }^{ q_{ 2 } }...p_{ m }^{ q_{ m } }$$, as if $$d$$ is a positive integer and $$k$$ is a non-negative integer $$2^{ k+1 }d-1$$ will always be a positive integer. Thus, because each $$d$$ corresponds to $$1$$ valid odd $$n$$ we have $$(q_{ 1 }+1)(q_{ 2 }+1)(q_{ 3 }+1)...(q_{ m }+1)$$ valid odd $$n$$ and, thus, $$2(q_{ 1 }+1)(q_{ 2 }+1)(q_{ 3 }+1)...(q_{ m }+1)-1$$ total valid $$n$$, corresponding to exactly that many sequences of consecutive integers which sum to our original positive integer (excluding the sequence of just that positive integer)!

Note by Chris Callahan
1 year, 11 months ago

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This is an unusually interesting setup. I think having a problem that applies these stuffs might be worthwhile. Would you like to post one?

- 1 year, 9 months ago