Vtyjkyuk

All the positive numbers can be expressed as a sum of one, two or more consecutive positive integers. For example 9 can be expressed in three such ways, 2+3+4, 4+5 or 9. In how many ways can a number be expressed as a sum of consecutive numbers?

In How many ways can this work for 65

Here for 9 answer is 3, for 10 answer is 3, for 11 answer is 2.

Here's one more way to calculate this, from my answer to this question on another site:

An integer $n$ is expressible as the sum of $m$ consecutive positive integers if and only if either:

• $m$ is odd and $frac nm$ is an integer, or


• $m$ is even and $frac nm + frac12$ is an integer,

and $frac nm ge frac m2$ (or else some of the integers in the sum would be zero or negative).

These conditions follow from the fact that the sum of an arithmetically increasing sequence of $m$ numbers equals $m$ times the mean of the numbers.

The last condition can be rewritten as $m le sqrt2n$. Thus, it's sufficient to iterate over all integers $m$ from $1$ to $lfloor sqrt2n rfloor$ and check whether $frac nm + frac m2 + frac12$ is an integer.

A sum of consecutive numbers is a difference of triangular numbers. The paper below gives a solution for the case of nonconsecutive triangular numbers.

Nyblom, M. A.
On the representation of the integers as a difference of nonconsecutive triangular numbers.
Fibonacci Quart. 39 (2001), no. 3, 256–263.

The main result is that the number of distinct representations of a nonzero integer $m$ as a difference of nonconsecutive triangular numbers is given by $d−1$, where $d$ is the number of odd divisors of $m$.

Fix $k$. Is there a way that a number $N$ can be written in more than one way as a sum of $k$ consecutive number? Certainly not because
$$
a+(a+1)+cdots+(a+k-1)neq
b+(b+1)+cdots+(b+k-1)
$$
if $aneq b$. On the other hand $N$ is the sum of $k$ consecutive number if and only if $N$ is the form
$$
N=frac12left[(n+k)(n+k+1)-n(n+1)right]
$$
for some $n$. Does that help?

Factorise the number and find the number of odd factors .
Total number of odd factors (except 1) is the answer.

Express N in terms of prime factors

$N = a^p . b^q . c^r$

If a = 2 .
Number of odd factors = (q+1)(r+1) - 1 .
Note :
1 is subtracted because 1 cannot be answer as consecutive terms means greater than 1 term.

For eg.

$100 = 2^2 . 5^2 $

So Number of odd factors = (2+1) - 1 = 2 = Number of ways of writing 100 as sum of 2 or more consecutive integers .
They are

18, 19, 20, 21, 22

9,10,11,12,13,14,15,16

ANSWER:


Number of ways of writing N as sum of consecutive positive integers is Number of odd factors in that number (except 1).


Also see : http://mathblag.wordpress.com/2011/11/13/sums-of-consecutive-integers/

I thought of this same question several months ago during one of my classes and I worked out the solution during my lunch break, the same sort of argument could be used to find the number of representations of $n$ in any arithmetic sequence modulo a positive integer. I got that if $S(n)$ denotes the number of representations of $n$ as a sum of successive natural numbers with $nge 1$ then that:

$$S(n)=d(fracn2^v_2(n))$$

Where $v_2(n)$ is the $2$-adic order of $n$, what I did was used the fact that:

$$sum_substacka^2+ab=n\(a,b)in mathbbN^2f(a,b)=sum_substackb=fracna-a\(a,b)in mathbbN^2f(a,b)=sum_substackdmid n\d<sqrtnf(d,fracnd-d)$$

To rewrite: $$S(n)=sum_substacka+(a+1)+(a+2)+dots +(b-1)+b=n\bge a\(a,b)in mathbbN^21=sum_substack(a+b)(a-b+1)=2n\bge a\(a,b)in mathbbN^21=sum_substacka^2-b^2+a+b=2n\bge a\(a,b)in mathbbN^21$$

And then simplified the resulting sum by swapping the summation indices several times and by setting $b=a-1+k$ with $kin mathbbN$ since we have that $bge a$.

This was the proof I scribbled down, where I used $chi_2$ to denote the Dirichlet character modulo $2$.

Sorry if it's kind of messy:

It is a well known fact that $$1+2+cdots+a=tfrac12 a(a+1)$$
Thus, $$b+(b+1)+cdots+a=(1+2+cdots +a)-(1+2+cdots (b-1))=tfrac12(a(a+1)-(b-1)b)$$
Where $a>b>0$. How many solutions does $$2n=a(a+1)-b(b-1)=(a+b)(a-b+1)$$ have? Well, taking two divisors $i,j$ of $2n$ such that $ij=2n$, we want to solve $$a+b=i$$ $$a-b+1=j$$The solutions of this are easy to obtain: $$a=frac12(i+j-1)$$ $$b=frac12(i-j+1)$$ For this to be integer solutions, we need $i+j$ to be odd (which also makes $i-j$ odd) Thus, we want to choose $i$ and $j$ in such a way that one is odd and the other is even (thus, the even one must contain all factors $2$ in $2n$). Let now $2^km=2n$ with $m$ odd, then there are exactly as many solutions as there are divisors for $m$ - that is, if we count the number itself as one consecutive integer. Not counting that one, we arrive at the final answer $$sigma_0(m)$$ where $sigma_0$ denotes the Divisor Function. Note that $sigma_0(p_1^e_1p_2^e_2cdots p_k^e_k)=(e_1+1)(e_2+1)cdots (e_k+1)$ where all $p_i$ are distict prime factors.

The number of ways of representing a number by consecutive integers is equal to the number of divisors of the largest odd divisor of that integer minus 1. (If you count the number itself as a representation, you don't need to subtract 1). The number of divisors of a number $n=p_1^a_1p_2^a_2...p_k^a_k$, is $d(n)=(a_1+1)(a_2+1)...(a_k+1)$. In other words, just add 1 to each power in the prime factorization and multiply them all together.

So if you want to find the number of ways to write a number $n$ as a sum of consecutive integers, factor $n$ into powers of primes, $n = p_1^a_1p_2^a_2...p_k^a_k$, then, using only the powers of odd primes, compute $(a_1+1)(a_2+1)...(a_k+1)$. (If one of the $p_i^a_i$ are a power of 2, delete that term).