Consecutive consecutive sums of equal value

Question

Given the list of counting numbers, what is the largest amount of consecutive consecutive sums of equal value that can be found? Is there a limit?

For example,

[1, 2] and [3] are two consecutive consecutive numbers that both add up to 3.

(If somebody understands the question and has a better way of phrasing, please do help me rephrase this.)

Answer 1

You can't have more than two.
There is this pattern: $1+2=3,4+5+6=7+8,9+10+11+12=13+14+15,...$
where the first sequence goes from $n^2$ to $n(n+1)$, but it's only two in a row.
Multiply by eight, then $$8[(m+1)+(m+2)+...+n]=8\left[\frac{n(n+1)}2-\frac{m(m+1)}2\right]=(2n+1)^2-(2m+1)^2$$ So a sequence of three of these will have $$(2n+1)^2-(2m+1)^2=(2p+1)^2-(2n+1)^2=(2q+1)^2-(2p+1)^2$$ so you have four squares in arithmetic progression.
This link shows you can't have four squares in arithmetic progression.

Answer 2

Consider $$\quad N=x+(x+1)+(x+2)+\cdots +(x+n)\\ \implies N=\frac{(n^2 + 2 n x + n + 2 x)}{2} = \frac{(2x+n)(n+1)}{2}\\ $$ N is the sum of consecutive numbers if and only if $N$ is factorable as $\space 2N=(2x+n)(n+1).$

For example $$N=7 \rightarrow 14=\big(2(x)+n\big)\big(n+1\big)=\big(2(3)+1\big)\big(1+1\big), \quad 7=3+4$$

Now we solve for $x$ and, try values of $n$ where $\quad\sum_{1}^{n} k<N\quad $ and the highest integer solution where $\quad (1+2+3+\cdots+n)<N\quad$ indicates $n$ as the maximum possible number of consecutive integers sums equal to $N$.

$$N=\frac{(n^2 + 2 n x + n + 2 x)}{2} =\frac{2(n+1)x+n(n+1)}{2} \implies x=\frac{2N-n(n+1)}{2(n+1)}$$

The number $n$ will sometimes be greater than the number of sums equal to $N$ but it will never be less. Here is a table showing the results for $\quad 1\le N\le 25$

\begin{align*} (N,n)\qquad &\text{ consecutive integer sums}\\ (2,0)\qquad &2\\ (3,1)\qquad &3 = 1+2\\ (4,0)\qquad &4\\ (5,1)\qquad &5 = 2+3\\ (6,2)\qquad &6 = 1+2+3\\ (7,1)\qquad &7 = 3+4\\ (8,0)\qquad &8\\ (9,2)\qquad &9 = 4+5\quad = 2+3+4\\ (10,3)\qquad &10 = 1+2+3+4\\ (11,1)\qquad &11 = 5+6\\ (12,2)\qquad &12 = 3+4+5\\ (13,1)\qquad &13 = 6+7\\ (14,3)\qquad &14 = 2+3+4+5\\ (15,4)\qquad &15 = 7+8 \quad = 4+5+6 \quad = 1+2+3+4+5\\ (16,0)\qquad &16\\ (17,1)\qquad &17=8+9\\ (18,4)\qquad &18=5+6+7\quad =3+4+5+6\\ (19,1)\qquad &19=9+10\\ (20,4)\qquad &20 =2+3+4+5+6\\ (21,6)\qquad &21=1+2+3+4+5+6\quad =6+7+8=10+11\\ (22,3)\qquad &22=4+5+6+7+8\\ (23,1)\qquad &23=11+12\\ (24,1)\qquad &24=7+8+9\\ (25,4)\qquad &25=3+4+5+6+7\quad =12+13\\ \end{align*}