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I am looking for an efficient way to find the following numbers numbers $a, b$

Let $x$ be 63 (as example) $$\sum_{k = a}^b k = \sum_{l = 1}^b l - \sum_{j=1}^{a-1} j = \frac{b*(b+1)}{2}-\frac{a*(a-1)}{2} = x = 63$$ in this case for example $$63 = 8+9+10+11+12+13 = \sum_{k = 8}^{13} k $$

servabat
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Entimon
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2 Answers2

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Given positive integer $x$, you want positive integers $a,b$ such that $$ 2x = b(b+1) - a(a-1) $$ If $c = 2a-1$ and $d=2b+1$, this says $$ (d-c)(d+c) = d^2 - c^2 = 8x $$ (with $c$ and $d$ both odd). So look at factorizations of $8x$. If $8x = uv$ where $u$ and $v$ are positive integers, $u < v$, we want $d-c = u$ and $d+c = v$, and then $d = (u+v)/2$, $c = (v-u)/2$. Note that we can take $u \equiv 2 \mod 4$ and $v \equiv 0 \mod 4$ (or vice versa), and get $c$ and $d$ odd.

In your example, $x=63$, $8x = 2^3 \times 3^2 \times 7$, and the solutions are:

  1. $u = 2$, $v = 252$, $c = 125$, $d = 127$, $a = 63$, $b = 63$
  2. $u = 6$, $v = 84$, $c=39$, $d=45$, $a = 29$, $b=22$
  3. $u = 14$, $v = 36$, $c=11$, $d=25$, $a = 6$, $b=12$
  4. $u = 18$, $v = 28$, $c=5$, $d=23$, $a = 3$, $b=11$
  5. $u = 12$, $v = 42$, $c=15$, $d=27$, $a = 8$, $b=13$
  6. $u = 4$, $v = 126$, $c=61$, $d=65$, $a = 31$, $b=32$
Robert Israel
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Expanding your sums, you have :

$$\frac{b^2+b}{2}-\frac{a^2-a}{2} = x$$ $$\iff (b+a)(b-a+1)=2x$$

Here, you have one of the two terms which is odd and then other even... Wait I guess that should be useful but..

The most obvious solution I see is : $a = \frac{x-1}{2}, b=\frac{x+1}{2}$ but it will only work for an odd $x$. But it's not really satisfying I guesS.

servabat
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