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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.)

user333696
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  • I roughly understand what you're saying but some details aren't clear. Does the first sum have to start at $1$ or could it be something like $[4,5,6]$ and $[7,8]$? Also, when you say largest "amount" do you mean the sum is large or the number of blocks is large (in your example, two)? – Erick Wong May 06 '16 at 06:49
  • It can start at any given number and amount means the number of blocks. – user333696 May 06 '16 at 07:03
  • Not quite what you asked, but for two blocks you can get them arbitrarily big. For example, take $(1,a,b)$ and the blocks $1,\dots,a$, and $a+1,\dots,a+b$. Then $a,b$ are given by simple linear recurrence relations. The first few terms are $(1,14,6),(1,84,35),(1,492,204),(1,2870,1189),(1,16730,6930)$ – almagest May 06 '16 at 08:17

2 Answers2

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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.

Community
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Empy2
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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*}

poetasis
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