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I'm having difficult with the following question :

show that the number of partitions of n into parts of size 3,5,7,9,... equals to the number of partitions of n into different parts which are not 1,2,4,8,...

jony89
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    Part A is unclear. The first half of the sentence says "partitions of $n$ to sizes of", the second half says "partitions of $n$ to partitions of size". It's not clear (to me) what the difference between these two formulations is intended to express -- but it seems some difference must be intended, since otherwise the statement would clearly be false. It would help if you could reexpress the problem in terms of the standard formulation "partition into $k$ parts of size ...". – joriki Aug 26 '15 at 06:43
  • ive edited the question, also notice that both series doesnt enclude the number 6 for e.g – jony89 Aug 26 '15 at 07:35
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    It's not clear to me why you've now included two different formulations of the question; that doesn't make it in any clearer. In both formulations there are still subtle differences between the first and second halves of the sentences, but it's still not clear what, if any, distinction these differences are intended to express. The expression "partition to" is non-standard; a number is usually partitioned into parts. If the question was originally posed in a different language, it might help if you quote it in that language so that someone can reexpress it in clearer English. – joriki Aug 26 '15 at 07:41
  • i changed it to 'into'. it basically says that given number of n, show that the number of possible partitions into the numbers 3,5,7,9,... equals to the number of partitions to the same n into the numbers that are NOT EQUAL to 1,2,4,8,... i think this is clear enough .. – jony89 Aug 26 '15 at 08:23
  • No, it's not clear enough, but never mind. I get the impression that all these differences are just due to sloppiness on your part and are not intended to express any distinctions. If so, the statement is obviously false, since the numbers $3,5,7,9,\ldots$ (assuming that this refers to all odd numbers except $1$) are a subset of the numbers not equal to $1, 2, 4, 8, \ldots$ (assuming that this refers to all non-negative powers of $2$), and there are obviously partitions that include parts included in one set and not in the other, e.g. $6$. – joriki Aug 26 '15 at 08:28
  • ok maybe i mistranslated it. last try tell me if this is ok ( ive edited the question ) thanks for the patience . – jony89 Aug 26 '15 at 08:42
  • Now it makes sense. I'd be interested to hear which language has such subtle markers for the concept expressed in English by "different". – joriki Aug 26 '15 at 08:44
  • Hebrew, but indeed i translated it wrong. – jony89 Aug 26 '15 at 10:03
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    חבל, I know Hebrew, I could have translated it for you if you'd written it in Hebrew like I suggested :-) – joriki Aug 26 '15 at 10:04

1 Answers1

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For part A, for each odd integer $2k+1$ with $k$ a positive integer, consider the integers $(2k+1)2^m$ with $m$ a non-negative integer. The admissible parts in the first case include only $2k+1$ (i.e. $m=0$), whereas the admissible parts in the second case include all these integers for all $m$. In a given partition, we can group all parts with a given $k$ together and establish a one-to-one correspondence between the possible groups in the first and second cases. In the first case, the group can contain any number of parts with $m=0$, but none with $m\gt0$, whereas in the second case, the group can contain at most one part for each $m$. For any non-negative integer $\ell$, a sum of $\ell(2k+1)$ of all parts of the group can be achieved in exactly one way – in the first case by having exactly $\ell$ parts with $m=0$, and in the second case by choosing the parts according to the binary representation of $\ell$, i.e. having a part with a given $m$ iff the $m$-th bit of the binary representation of $\ell$ is $1$.

Note that this is a variant of the more well-known result that The number of partitions of $n$ into distinct parts equals the number of partitions of $n$ into odd parts, which you get if you include $1$ in the list of admissible odd parts in the first case and drop the exclusion of non-negative powers of two in the second case.

Questions shouldn't contain several unrelated questions; please post a separate question for part B.

Edit in response to the comment: The generating function for the first case is

$$ \prod_{k=1}^\infty\frac1{1-x^{2k+1}}\;. $$

The generating function for the second case is

$$ \prod_{j=1\atop j\ne2^m}^\infty\left(1+x^j\right)=\prod_{k=1}^\infty\prod_{m=0}^\infty\left(1+x^{(2k+1)2^m}\right)=\prod_{k=1}^\infty\prod_{m=0}^\infty\frac{1-x^{(2k+1)2^{m+1}}}{1-x^{(2k+1)2^m}}=\prod_{k=1}^\infty\frac1{1-x^{2k+1}}\;, $$

where the last equality holds because the product telescopes and only the denominator of the $m=0$ factor survives.

This is analogous to the generating function argument given at the end of the answer to the question I linked to above.

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