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In how many ways can $63$ be written as the sum of an increasing sequence of at least two consecutive integers, where all numbers in the sequence are less than $63$?

So I have to express $63$ as $$63=x_1+x_2+x_3+\dots + x_n$$ where $x_i < x_{i+1}<63$.

I tried to compute some values and got for example $$\begin{align} 63&=1+2+29+31 \\63&=1+2+28+32 \\ 63&=1+2+27+33 \\ \vdots \\ 63&=1+2+3+57 \end{align}$$ and it seems that there is $26$ of these sums already which would seem to satisfy the conditions. I don't see how I can get further from here. Is there some algebraic identity I should use here related to sequences or what's the trick?

Jen
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  • I hope my english is right, but $1,2$ are consecutive? And since it says at least two this would be satisfactory? – Jen Nov 20 '21 at 11:12
  • Yes, $1, 2$ are consecutive. However, $1 + 2 + 29 + 31$ does not satisfy the condition that the integers are consecutive since $29 \neq 2 + 1$ and $31 \neq 29 + 1$. On the other hand, $31 + 32$ does, as does $20 + 21 + 22$. – N. F. Taussig Nov 20 '21 at 11:28
  • In your example, you used only positive integers. Do you mean integers or positive integers? – N. F. Taussig Nov 20 '21 at 11:33
  • It can be also negative integers. Also doesn't the problem state that at least two of them must be consecutive not all of them? – Jen Nov 20 '21 at 11:39
  • What they mean by at least two consecutive integers is a sum such as $(-31) + (-30) + (-29) + \cdots + (-1) + 0 + 1 + 2 + \cdots + 31 + 32 + 33$ in which all the integers are consecutive. – N. F. Taussig Nov 20 '21 at 11:44
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    Think about this: if you add consecutive numbers, you always get a multiple of the middle number (or the sum of the two middle numbers, if you have an even number of summands). So it really depends on how many ways there are to express $63$ as a product. – Gerry Myerson Nov 20 '21 at 12:14
  • Anyway, I bet this question is a duplicate. Should be worth doing a search for it. e.g., https://math.stackexchange.com/questions/296848/sum-of-consecutive-integers and https://math.stackexchange.com/questions/139842/in-how-many-ways-can-a-number-be-expressed-as-a-sum-of-consecutive-numbers and surely many others. – Gerry Myerson Nov 20 '21 at 12:15
  • @GerryMyerson, the links you gave are to versions that specify (in comments, at least) the OPs there are restricting to positive integers, so this is maybe not an exact duplicate. – Barry Cipra Nov 20 '21 at 12:30
  • The "sequence of ... consecutive integers" means "consecutive" applies to the entire sequence. Requiring just two to be consecutive would be something like "an increasing sequence of integers, at least two of them consecutive". – aschepler Nov 20 '21 at 12:42
  • @Barry, true, but there's a simple correspondence between solutions in positive integers and solutions that use negative integers. – Gerry Myerson Nov 20 '21 at 21:57

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If the number of integers in the sum is odd, we can write the sum in the form $$(x - k) + \cdots + (x - 2) + (x - 1) + x + (x + 1) + (x + 2) + \cdots + (x + k) = 63$$ where $x$ is the average value of the summands. The above expression simplifies to $$(2k + 1)x = 63$$ Thus, $$x = \frac{63}{2k + 1}$$ where $2k + 1$ is a positive integer factor of $63$. If $2k + 1 = 1$, $x = 63$. Thus, $2k + 1 = 3, 7, 9, 21, 63 \implies k = 1, 3, 4, 10, 31$, giving five possible values for $x$, namely $x = 21, 9, 7, 3, 1$. The corresponding sequences are

$x = 21, k = 1$: $(20, 21, 22)$

$x = 9, k = 3$: $(6, 7, 8, 9, 10, 11, 12)$

$x = 7, k = 4$: $(3, 4, 5, 6, 7, 8, 9, 10, 11)$

$x = 3, k = 10$: $(-7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13)$

$x = 1, k = 31$: $(-30, -29, -28, \ldots, -2, -1, 0, 1, 2, 3, \ldots, 30, 31, 32)$

If the number of integers in the sum is even, we can write the sum in the form $$(x - k) + \cdots + (x - 2) + (x - 1) + x + (x + 1) + (x + 2) + \cdots + (x + k) + (x + k + 1) = 63$$ The above expression simplifies to $$(2k + 2)x + k + 1 = 63$$ so \begin{align*} (2k + 2)x + (k + 1) & = 63\\ (k + 1)(2x + 1) & = 63\\ 2x + 1 & = \frac{63}{k + 1} \end{align*} where $k + 1$ is a positive integer factor of $63$. Thus, $k + 1 = 1, 3, 7, 9, 21, 63 \implies k = 0, 2, 6, 8, 20, 62$, giving six possible values of $x$, namely $31, 10, 4, 3, 1, 0$. However, if $k = 62$ and $x = 0$, $x + k + 1 = 63$. Thus, $x = 31, 10, 4, 3, 1$. The corresponding sequences are

$x = 31, k = 0$: $(31, 32)$

$x = 10, k = 2$: $(8, 9, 10, 11, 12, 13)$

$x = 4, k = 6$: $(-2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11)$

$x = 3, k = 8$: $(-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)$

$x = 1, k = 20$: $(-19, -18, -17, \ldots, -2, -1, 0, 1, 2, 3, \ldots, 19, 20, 21, 22)$

My thanks to @Unit for pointing out an error in my initial calculation for an even number of summands.

N. F. Taussig
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  • The final sequence is off - the sum is $93$. I think you made an error summing the "even number of terms" case - it should be $(2k+2)x + k + 1 = (k+1)(2x+1) = 63$. – Unit Nov 20 '21 at 13:29
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    @Unit Thanks for pointing out the error. I have edited accordingly. – N. F. Taussig Nov 20 '21 at 13:57
  • @N.F.Taussig but doesn't the question ask for an increasing sequence where at least 2 consecutive integers, meaning 2 have to be consecutive, not necessarily all. – D S Jan 11 '23 at 13:38
  • @DS The question asks for an increasing sequence of at least two consecutive integers, which means the sequence is increasing and has length at least two. – N. F. Taussig Jan 11 '23 at 15:16
  • @N.F.Taussig Although that does not match the examples given by the OP, upon more thought, I suppose your reasoning is more valid. – D S Jan 11 '23 at 16:49