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I'm pretty sure it's 40 but I'm not too sure if it's enough to show that:

$n(n+1) + 41$

$41(\frac{n(n+1)}{41} + 1)$

and the smallest composite no# will be achieved when n+1 = 41, n =40?

Am I missing anything here? --> Yes I still need to show that for n < 40, it's prime..ideas?

Thanks

smh
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    What is your argument for why the smallest composite will be achieved when $n+1=41$? – anon Oct 31 '13 at 01:45
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    I guess you want smallest positive $n$. It is indeed $40$, but one has to verify that everything below is prime. Not pleasant! – André Nicolas Oct 31 '13 at 01:46
  • It's a finite number of drudgery-filled cases. Write a computer program to check it. – ncmathsadist Oct 31 '13 at 01:51
  • @anon since we want $\frac{n(n+1)}{41}$ to be the smallest possible integer – smh Oct 31 '13 at 01:52
  • Checking modularity for each prime $p<41$? – Carlos Eugenio Thompson Pinzón Oct 31 '13 at 01:58
  • By that reasoning, $7\left(\frac{n(n+1)}{7}+1\right)$ isn't composite until $n+1=7$. But it does happen to be composite for $n=1$. – anon Oct 31 '13 at 01:59
  • @CarlosEugenioThompsonPinzón yeah I was thinking that too hmm – smh Oct 31 '13 at 02:05
  • @anon interesting point! i'll have to re-think my method – smh Oct 31 '13 at 02:06
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    Rabinowitz showed, in 1913, that $x^2+x+p$ represents the maximal number of consecutive primes if and only if $x^2+xy+py^2$ is the only (equivalence class of) positive binary quadratic form of its discriminant. This condition is called "class number one." For negative discriminants the set of such discriminants is finite, called Heegner numbers. For you, $p=41.$ see http://math.stackexchange.com/questions/289338/is-the-notorious-n2-n-41-prime-generator-the-last-of-its-type/289357#289357 – Will Jagy Oct 31 '13 at 02:29

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