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May
27
comment Does every integer $n > 2$ have an arithmetic expression involving at least two consecutive integers but excluding $n$ itself?
@DonAntonio: Similarly, $19=4!-5$ can be interpreted as using the integers $24$ and $5$, instead of two consecutive integers $4$ and $5$. Such interpretations seem perverse to me, but now I realize that's only because of my bias of trying to have the question make sense. (One can even literally interpret the wording in the title as saying that we just have to use at least two consecutive integers and then can use anything else at all—say $n=(n-3)+1+2$ where $1,2$ are consecutive—but I tend not to gravitate towards such "finding loopholes".) But I agree one's personal choice may be different.
May
27
comment Does every integer $n > 2$ have an arithmetic expression involving at least two consecutive integers but excluding $n$ itself?
@DonAntonio: (…) I'd not prefer this interpretation as it invalidates the question; I'd prefer to interpret that it's the integers $1,2,3,4$ that are used. Similarly, there is also some leeway in interpreting the question based on the examples (the problem of induction). I chose to interpret the tightest question that fits the examples (and also match the remark at the end on triangular numbers), but yes you could interpret it differently: e.g. $16=(2/3)(4/5)(6+7+8+9)$, which uses consecutive integers $2,3,4,5,6,7,8,9$ can be interpreted as using fractions, and thus not fit the question. :-)
May
27
comment Does every integer $n > 2$ have an arithmetic expression involving at least two consecutive integers but excluding $n$ itself?
@DonAntonio: Actually this discussion has been useful to me, thanks. Let me declare a bias I just realized: I prefer to interpret the examples in such a way that they fit the question. (It is possible to interpret them differently; that is your prerogative.) As the question is not specified unambiguously, nor is it specified how the examples map to the question (e.g. it doesn't say "in the example $10=1+2+3+4$, the consecutive integers used are $1,2,3,4$), there is some leeway in interpretation. E.g. you could say in $11=1-2+3\times4$ that the integers used are $1,-2,3,4$, not consecutive. (…)
May
27
comment Does every integer $n > 2$ have an arithmetic expression involving at least two consecutive integers but excluding $n$ itself?
@DonAntonio When you divide two integers, you can get a fractional number. But you cannot use that fact to conclude that the example doesn't use integers. For example, in $10 = 1+2+3+4$, we have $3+4=7$, but it would be absurd to say that because the example "uses" $7$ which is not consecutive with $1$ and $2$, the example does not satisfy the constraints of the question. :-)
May
27
comment Does every integer $n > 2$ have an arithmetic expression involving at least two consecutive integers but excluding $n$ itself?
@DonAntonio: I didn't mean to annoy you — yes I did read what you wrote, but I could ask you the same question. :-) To clarify: this discussion started when you said that "Some of the examples of the OP don't even use integers", and I pointed out that all the OP's examples could indeed be seen as using integers, true to the question title. Then again you said that the OP's examples for 16 and 17 use fractional numbers (rather than integers?), and I pointed out that that cannot be the way to interpret it, because the OP specifically mentions integers. Is something not clear?
May
27
comment Does every integer $n > 2$ have an arithmetic expression involving at least two consecutive integers but excluding $n$ itself?
Rather, $(n+1)$, $(n+2)$ and $(n+3)$ are three consecutive integers. +1.
May
26
comment Does every integer $n > 2$ have an arithmetic expression involving at least two consecutive integers but excluding $n$ itself?
@DonAntonio There is no such thing as a fractional integer, so the OP's examples for 16 and 17 must be interpreted as using the consecutive integers $2,3,4,\dots $ (with division), rather than as using arbitrary fractions like $2/3$. As for whether the consecutive integers must all be less than $n$ (as in all the examples) or not, we can await the OP's clarification, though there doesn't seem to be much doubt about which one is the less trivial question. :-)
May
26
comment Does every integer $n > 2$ have an arithmetic expression involving at least two consecutive integers but excluding $n$ itself?
@DonAntonio: The title of the question mentions "integers". I interpret the examples as always using some set of consecutive positive integers in order (and in particular each integer exactly once), but the set of allowed operations includes negation/subtraction and division.
May
26
comment Does every integer $n > 2$ have an arithmetic expression involving at least two consecutive integers but excluding $n$ itself?
I presume the OP means consecutive positive integers strictly smaller than $n$. A fact that you can use: Every number that is not a power of $2$ is a sum of consecutive positive integers (all smaller than itself). For instance, $11 = 5 + 6$, $12 = 3 + 4 + 5$, $13 = 6 + 7$, $17 = 8 + 9$, $18 = 5 + 6 + 7$, $19 = 9 + 10$, etc. (For proof, see here or here.) So it's enough to ask the (suitably modified) question for powers of $2$.
May
6
comment Proving a recurrence relationship for a sequence
Really nice solution, congratulations. :-) The main trick seems to be making sure that the $(n-1)b_{n-2}$ term gets multiplied by $x^{n-1}/(n-1)!$ rather than say $x^n/n!$ or $x^{n-2}/(n-2)!$, so that the $(n-1)$ factor cancels cleanly and we don't have to solve too-complicated differential equations.
May
5
comment Expected number of loops
Just for completeness: yes, $H_n = 1 + \frac12 + \dots + \frac1n$ (harmonic number), and the asymptotics $(\ln n)/2 + 0.98$ mentioned above does indeed give about $3.28$ for $n=100$.
May
5
comment Proving a recurrence relationship for a sequence
Yes definitely the right question is to find an EGF; note that $\exp(x + x^2/2) = 1+x+\frac{2x^2}{2!}+ \frac{4x^3}{3!} + \frac{10x^4}{4!} + \dots$ and the coefficients satisfy the recurrence.
May
5
comment Proving a recurrence relationship for a sequence
Aside: I think the question, in place of "Let $B(x)$ be the recurrence relation for the sequence $b_n$", should have "Let $B(x)$ be the [exponential] generating function for the sequence $b_n$".
Apr
29
comment Placing indistinguishable objects on a indistinguishable shelve
The answer is not $10!$ for either, because if you think about it, the fact that there are $3$ shelves surely matters and must figure in the answer.
Apr
28
comment When tossing a coin, what's the likelihood of coming up heads when it came up 999 tails consecutively?
By the way, if you see TTT, the expected number of heads and tails among the remaining $7$ is $7/2$ each, so the expected total number of heads and tails (among the ten tosses) would now be $7/2 = 3.5$ and $3 + 7/2 = 6.5$ respectively, no longer $5$ and $5$ (that's only the expected value before the results of any tosses).
Apr
28
comment When tossing a coin, what's the likelihood of coming up heads when it came up 999 tails consecutively?
This question is one duplicate, though I expect there are others.
Apr
28
comment When tossing a coin, what's the likelihood of coming up heads when it came up 999 tails consecutively?
The coin has no memory. See gambler's fallacy. This question has been asked and answered many times on this site before.
Apr
28
awarded  Good Answer
Apr
27
comment Find the probability of n different people having the same birthday month
Are there a fixed set of $n$ people? You shouldn't have the leading $n$ factor (e.g. imagine choosing the first of the $n$ people...)
Apr
25
comment Visually deceptive “proofs” which are mathematically wrong
What is the visually deceptive proof here?