Consecutive Numbers Sum (or $CNS$) is the number of different consecutive positive integers' summation resulting to a specific number. For example;
$CNS(45)=5$
$CNS (945)=?$ and $CNS (947)=?$
Hint: You do not have to find every single of them to get the result!
Here's some math; thanks to Gauss we know that the sum of the first $n$ natural numbers is $$\sum_{i=0}^ni=\frac12n(n+1).$$ So by subtracting the first $m-1$ terms we get the sum of all consecutive integers from $m$ to $n$; $$\sum_{i=m}^ni=\frac12n(n+1)-\frac12(m-1)m=\frac12(n+m)(n-m+1).$$ To count the number of ways to write an odd number $k$ as a sum of consecutive integers, we want to find natural numbers $m$ and $n$ with $m<n$ such that $$(n+m)(n-m+1)=2k.$$ In particular this gives a factorization of $2k$. Conversely, if $2k=a\times b$ is a factorization then setting $$m:=\frac{a-b+1}{2}\qquad\text{ and }\qquad n:=\frac{a+b-1}{2},$$ gives $(n+m)(n-m+1)=2k$. By some kind of magic everything falls into place:
This shows that half of all factorizations of $2k$ give natural numbers $m$ and $n$ such that $\sum_{i=m}^ni=k$. The number of factorizations is the same as the number of divisors, and the number of divisors of $2k$ is twice the number of divisors of $k$, because $k$ is odd. So the number of ways to write $k=\sum_{i=m}^ni$ equals the number of divisors of $k$, often denoted $\sigma_0(k)$. We don't want the trivial sum $k=\sum_{i=k}^ki$, so we find that $$CNS(k)=\sigma_0(k)-1,$$ where $\sigma_0$ is the divisor counting function. Using the linked formula for $\sigma_0$, the prime factorizations $$945=3^3\times5\times7\qquad\text{ and }\qquad947=947,$$ show without any aid from a machine that \begin{align} 945&=\sigma_0(3^3\times5\times7)-1=(3+1)(1+1)(1+1)-1=15\\ 947&=\sigma_0(947)-1=(1+1)-1=1. \end{align}
I got
$CNS(945)=15$ and $CNS(947)=1$.
Short way:
From playing around with other numbers, I noticed that $CNS(n)=o(n)-1$, where $o(n)$ is the number of odd divisors of $n$. I'm not sure how to prove this, but this does give us $CNS(945)=15$ and $CNS(947)=1$ as shown below, since there are sixteen odd divisors of $945$ and only two for $947$ ($1$ and $947$, since it's prime).
Long way:
For $CNS(945)$:
For there to be an odd number of consecutive integers, the average of them (i.e. the number in the middle) must be odd. Thus, we can find all the odd factors of $945$, and what they multiply with to get $945$:
$945 = 3\times315 = 314+315+316$
$945 = 5\times189=187+188+189+190+191$
$945 = 7\times135=132+133+134+135+136+137+138$
$945 = 9\times105=101+102+\cdots+109$
$945 = 15 \times 63=56+57+\cdots+70$
$945 = 21\times 45=35+36+\cdots+55$
$945 = 27\times 35=22+23+\cdots+48$
$945 = 35\times 27=10+11+\cdots+44$
All products after $35\times27$ (i.e. $45\times21$, $63\times15$, etc.) do not work, since there aren't enough positive numbers available on either side of the average number to reach $945$.
For an even number of integers, the average will be $\textrm{something}.5$. Thus, we need to find integer solutions for $m$ and $n$ such that $\frac{945}{m}=n+0.5$. Solving for $m$ and $n$ here, we get the following sums:
$945 = 2\times 472.5=472+473$
$945 = 6\times 157.5=155+156+157+158+159+160$
$945 = 10\times 94.5=90+91+\cdots+99$
$945 = 14\times 67.5=61+62+\cdots+74$
$945 = 18\times 52.5=44+45+\cdots+61$
$945 = 30\times 31.5=17+18+\cdots+46$
$945 = 42\times 22.5=2+3+\cdots+43$
Similar to above, all products after $42\times22.5$ (i.e. $54\times17.5$, $70\times17.5$, etc.) do not work, since there again aren't enough positive numbers available on either side of the average number to reach $945$.
We have eight sums with an odd middle and seven with an even middle, thus $CNS(945)=15$.
For $CNS(947)$:
We can use a similar method to above. First we have to find all the odd factors of $947$, but since $947$ is prime, the only factors are $1$ and $947$. Therefore there are no strings of an odd number of consecutive integers that sum to $947$.
For an even number of integers, we can again use our formula, $\frac{947}{m}=n+0.5$, to find all solutions. Solving this, we find that the only solution that works is $947=2\times473.5=473+474$. This is thus the only way to sum consecutive integers to $947$, and so $CNS(947)=1$.
The sum of consecutive integers from n to m is
$\frac{ (n + m) (m - n + 1) }{ 2 } = \frac{ m^2 - n^2 + m + n }{ 2 }$
So we want to count
$\# \{ (n,m) \in \mathbb{N}^2: m^2 - n^2 + m + n = 2k \} = \text{CNS}(k)$
Isolating n,
$n^2 - n - m^2 -m + 2k = 0$
$n = \frac{ 1 + \sqrt{4m^2 + 4m - 8k + 1} }{2}$
Since $n$ must be integer, this means $4m^2 + 4m - 8k + 1$ must be a perfect square not divisible by 4.
$4m^2 + 4m - 8k + 1 = (2m + 1)^2 - 8k$
$4m^2 + 4m - 8k + 1 = s^2$, with $s$ odd.
$m = \frac{ -1 + \sqrt{8k + s^2} }{2}$
Again, $8k + s^2$ must be a perfect square: $8k + s^2 = (s+r)^2$.
Also, $8k + s^2$ must not be divisible by 4, granted because $s$ odd.
We can consider $s$ to be positive because $-s$ and $s$ give the same solution, and $0$ is divisible by 4. We can also consider $r$ to be positive.
$8k + s^2 = s^2 + 2sr + r^2$
$8k = 2sr + r^2$
$s = \frac{ 4k }{r} - \frac{r}{2}$
We will have a solution for every $r$ such that
So, for $\text{CNS}(45)$, since $4 * 45 = 2^2 * 3^2 * 5$ and $\lfloor 2 \sqrt{2*45} \rfloor = 18$
That's why $\text{CNS}(45) = 5$.
Since $4 * 945 = 2^2 * 3^3 * 5 * 7$ and $\lfloor 2 \sqrt{2*945} \rfloor = 86$,
Therefore,
$\text{CNS}(945) = 15$.
Since $4 * 947 = 2^2 * 947$ and $\lfloor 2 \sqrt{2*945} \rfloor = 87$,
Therefore,
$\text{CNS}(947) = 1$.
Let $k = 2^{a_0} {p_1}^{a_1} {p_2}^{a_2} \dots {p_l}^{a_l}$, where $2, p_1, p_2, \dots, p_l$ are $l+1$ different prime numbers.
$\text{CNS}(k) = \# \left\{ 2^{b_0} {p_1}^{b_1} \dots {p_l}^{b_l} : \begin{array}{c} b_0 \in \{ 1, a_0+2\} \\ 0 \leq b_i \leq a_i \quad \forall i=1,\dots,l \\ 2^{b_0-1} {p_1}^{b_1} \dots {p_l}^{a_l} \leq \sqrt{2*k} \end{array} \right\}$
The last constraint is annoying. But if we remove it we will get the double of solutions, so we can just divide by 2.
Therefore, the closed formula is
$\text{CNS}(k) = \frac{ 2 (a_1+1) \dots (a_l+1)}{2} - 1 = (a_1+1) \dots (a_l+1) - 1$
The $-1$ at the end is because we ignore summations of only one term.
We have to count the number of integers $k\ge 2$ for which $n$ can be written as a sum of $k$ consecutive numbers. Certainly, this is only possible if $n\ge 1+2+\dots+k=k(k+1)/2$ is large enough. The other rules are this, which are proved by pictures below.
If $k$ is odd, it can be done iff $n$ is large enough and $n\equiv 0$ (mod $k$).
If $k$ is even, it can be done iff $n$ is large enough and $n\equiv k/2$ (mod $k$).
To compute CNS(945),
Factorize it as 3•5•7•9. From this, we can see it is a multiple of the odd numbers 3, 5, 7, 9, 15, 21, 27, 35, and 45, which are all small enough, and all other odd factors are too large. For an even $k$ to work, $n/k$ must be a half integer, which occurs if $k$ is double one of its odd factors. The ones which are small enough are 2, 6, 10, 14, 18, and 30. This is 15 numbers total.
For CNS(947),
This number is prime. It's only odd factors are itself, which is too large, and 1. The only $k$ which works is then 2•1=2, so CNS(947)=1.
If an number $n$ can be written as the consecutive sum of $k$ numbers, then $n$ is the number of dots in a trapezoidal picture with $k$ rows like this:
•••••••• ^
••••••••• |
•••••••••• k rows
••••••••••• |
•••••••••••• v
Flipping the below triangle makes a rectangle with $k$ rows, showing $n$ is a multiple of $k$. The transformation can be done in reverse provided there are enough columns, which happens when $n\ge k(k+1)/2$.
•••••••• ••••••••**
••••••••• •••••••••*
•••••••••• => ••••••••••
••••••••••* ••••••••••
••••••••••** ••••••••••
If $k$ is even, the same triangle flip results in a rectangle plus half a column.
•••••••• ••••••••**
••••••••• •••••••••*
•••••••••* => •••••••••
•••••••••** •••••••••
Trick
Just add a digit to the existing series
Example
$22+923=945$
$14+15+916=945$
$7+8+9+10+911=945$
$5+6+7+8+9+910=945$
$1+2+3+4+5+6+7+8+909=945$
For the next :
May be do the same as above adding +2 to any number :)
45 = 45does not count as a 6th possibility? $\endgroup$