Question

Given n, how many ways can we write n as a sum of one or more positive integers a_1 ≤ a_2 ≤ ... ≤ a_k with a_k - a_1 = 0 or 1

Answer 1

\[a_1 ≤ a_2 ≤ ... ≤ a_k \]
with
\[a_k - a_1 = 0 or 1\]

Answer 2

@Hero

Answer 3

@satellite73

Answer 4

@waterineyes

Answer 5

@Yahoo!
@hartnn

Answer 6

you didnt call me panlac thats the problem

Answer 7

@nickhouraney

Answer 8

oh i dont know

Answer 9

@ash2326
@.Sam.

Answer 10

@hartnn

Answer 11

even i did not understand the question....

Answer 12

Don't know sorry..

@mukushla may help you...

Answer 13

@mathslover

Answer 14

Allow me to do some thinking here. First off, we have some small examples:

\(1\longrightarrow 1\) way.
\(2\longrightarrow 1\) way.
\(3\longrightarrow 2\) ways.
\(4\longrightarrow 3\) ways.
\(5\longrightarrow 3\) ways.
\(6\longrightarrow 5\) ways.
\(7\longrightarrow 5\) ways.
\(8\longrightarrow 6\) ways.

Answer 15

I believe that your solution will be either this, or something very similar. \[\Large \sum_{k=1}^{\left\lfloor \frac{n}{2} \right\rfloor}\left\lceil\frac{n}{k}\right\rceil\]

Answer 16

It's not quite that, but it's similar. It might be\[\Large \sum_{k=1}^{\left\lfloor \frac{n}{2} \right\rfloor}\left\lfloor\frac{n}{k+1}\right\rfloor\]If it is this, then for \(n=9\), it predicts that the solution will be \[4+3+2+1=10\]

Answer 17

If we work it out, we get \[1+1+1+1+1+1+1+1+1\]\[1+1+1+1+1+1+1+2\]\[1+1+1+1+1+2+2\]\[1+1+1+2+2+2\]\[1+2+2+2+2\]\[2+2+2+3\]\[3+3+3\]\[4+5\]Which is 8 ways. (Also, \(n=8\) actually has 7 ways and not 6.)

Answer 18

This is an interesting problem. I think I'm close, but I'm not there yet.

Answer 19

I'll have to come back to this later.

Answer 20

okay, thanks

Answer 21

@mukushla

Answer 22

ahh...misread it the other day...let me think again

Answer 23

i believe that answer is \(n\) way ... i'll post my solution later.

Answer 24

you're right

Answer 25

this was just a tricky one

Answer 26

I agree as well. I'll wait for @mukushla's solution though since I don't have an actual proof yet, and his is probably better.

Answer 27

there is only one way for each \( 1\le k\le n\).

we can rewrite the equation as\[a_1+a_2+...+a_k=n\]\[ma_1+(k-m)(a_1+1)=n\ \ \ \ \ \ \ \ \ 1\le m \le k \]\[a_1=\frac{n+m}{k}-1\]
if \(k \mid n \) then it is immediate that \(m=k\) and \(a_1=\frac{n}{k}\)...exactly one way for this case.

if \(k \nmid n\) then we have\[\frac{n}{k}-1< \frac{n}{k}+\frac{1}{k}-1=\frac{n+1}{k}-1\le a_1 \le \frac{n+k-1}{k}-1=\frac{n}{k}-\frac{1}{k} <\frac{n}{k}\]\[\frac{n}{k}-1 <a_1<\frac{n}{k}\]it gives \(a_1=\left \lfloor \frac {n}{k}\right \rfloor\) and \(m=k(\left \lfloor \frac {n}{k}\right \rfloor+1)-n \) ...one way for this case.