* rwg@sdf.lonestar.org <rwg@sdf.lonestar.org> [Jul 13. 2009 09:11]:

joerg> Let P(x) = prod(n=1, oo, (1+x^n)).

...

This is the generating function for partitions into distinct parts.

We have P(x^2)^8 + 16*x*P(x^2)^16*P(x)^8 = P(x)^16

So the number of partitions of n into distinct parts of 16 kinds (rhs.) equals the number of partitions of n into even parts of 8 kinds (1st term lhs.) plus *WHAT*?

P.S.: the equality really is k^2+k'^2=1 massaged to render a combinatorial relation.

How about: The number of partitions of n into distinct integers using 16 different fonts equals the number of partitions of n into distinct even parts using 8 fonts, plus 16 times the number of partitions of n-1 into distinct even parts using 24 fonts and distinct odd parts using 8 fonts? --rwg

With a little correction (ND='number of partitions into distinct parts'): ND of n using 16 different fonts equals ND of n using 8 fonts, plus 16 times ND of n-1 even parts using 24 fonts and ND using 8 fonts. (dropped the 'odd' with the last term) Thanks. Should have said the identity is usually given as E1^16*E4^8+16*x*E1^8*E4^16 == E2^24 where Ek = eta(x^k) [partition eta, no x^(1/24) in front] ... and this can also (trivially be rewritten into an identity for counting unrestricted partitions: 16*x/(E2^24*E1^8) + 1/(E2^24*E4^8) == 1/(E1^16*E4^16) I give this one a try, write NP for 'number of (unrestricted) partitions': NP of n into 16 fonts and multiples of 4 into 16 fonts equals 16 times NP of n-1 into even parts of 24 fonts and parts of 8 fonts plus NP of n into even parts of 24 fonts and multiples of 4 of 8 fonts. Now if some superbrain could argue _combinatorially_ that one of the identities is correct, that would be truly nice.

me braindead as usual... The relation was

P(x^2)^8 + 16*x*P(x^2)^16*P(x)^8 = P(x)^16

[...something wrong...]

Here we go (ND='number of partitions into distinct parts'): ND of n using 16 different fonts [P(x)^16] equals ND of n into even parts using 8 fonts [P(x^2)^8], plus 16 times ND of n-1 [16*x* ] into even parts using 16 fonts [ *P(x^2)^16* ] and into arbitrary parts using 8 fonts [ *P(x)^8]. @ Dirk: 'fonts' := 'kinds of parts' (really: kinds of letters) @ RWG: I now see your odd 'odd' comes from this: P(x)/P(x^2) = prod(k=0, 00, 1 + x^(2k+1) ) =: O(x), so P(x^2)^16*P(x)^8 == P(x^2)^24 * O(x^8) So you secretely write the relation as

P(x^2)^8 + 16*x*P(x^2)^24*O(x^8) = P(x)^16

and you interpretation is correct as given: The number of partitions of n into distinct integers using 16 different fonts equals the number of partitions of n into distinct even parts using 8 fonts, plus 16 times the number of partitions of n-1 into distinct even parts using 24 fonts and distinct odd parts using 8 fonts.

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