Timeline for many expected streaks imply high probability for a streak
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| Feb 20, 2013 at 7:29 | comment | added | Brendan McKay | Ooops, I interpretted the large term in your denominator as having alternating signs, but now I see the signs are negative except for the leading term. The method still applies except that partial fractions approach will be harder. The smallest zero of the large term is still greater than 1, which is what you need. | |
| Feb 20, 2013 at 1:32 | comment | added | Brendan McKay | To finish the other 1% of the proof first show that the large factor in the denominator has no zeros of modulus $\le 1$. I think they all have modulus 2. Then note that the coefficients are asymptotically equal to those of $x^k/(1-x)$, namely 1, since the difference between your generating function and $x^k/(1-x)$ has radius of convergence greater than 1. This shows the coefficients are exponentially close to 1. (This is Darboux's method.) I think that in fact the denominator factors into linear factors involving roots of unity, so you can use partial fractions and get a full expansion. | |
| Feb 19, 2013 at 18:16 | history | edited | Steven Landsburg | CC BY-SA 3.0 |
Corrected typo: A k becomes an n
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| Feb 19, 2013 at 17:14 | history | answered | Steven Landsburg | CC BY-SA 3.0 |