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Given that:

$p(n)=p(n-1)+p(n-2)-p(n-5)-p(n-7)+⋯$

Why can one not state:

$p(n)≥p(n-1)+p(n-2)-p(n-5)-p(n-7)$

Here is the logic: the subsequent 2 terms of the relation are additive and the 2 following one are negative, etc. yet the magnitude of p(k) steadily decreases hence the resultant value most be slightly positive for all of the following double pairs and for an $n$ large enough, it is not even necessary to consider the last few terms of the sum in terms of their potential to tip the inequality.

Where is the mistake?

Just_a_fool
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    It's a very informal argument, and to make it into a proof you will have to do some work. For example: Why does the $++--$ pattern persist? That is, why does the sequence of the $1$st, $-1$st, $2$nd, $-2$nd, etc. pentagonal number increase? Furthermore, why does $p(k)$ increase with $k$ ? Also, you probably want to assume $n \geq 7$ or something like that. – darij grinberg Feb 16 '15 at 06:19
  • Thanks for your reply. Though, I have, in a little thing I'm trying to write, precisely done all of what you stated, lol. The problem is, this argument allows me to a develop a lower bound for p(n), namely 1.7^n, which is incorrect. – Just_a_fool Feb 16 '15 at 06:23
  • One problem is that it's false for some small values of $n$. – Gerry Myerson Feb 16 '15 at 06:27
  • Try looking for your mistake in the derivation of the lower bound. The inequality in your original post is correct, as far as I can tell (and Sage confirms me for all $7 \leq n \leq 20000$). – darij grinberg Feb 16 '15 at 06:28
  • Oh... I just realized I stated A>B, C>D and A>C and concluded A-C>B-D – Just_a_fool Feb 16 '15 at 06:35

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