I think you all ought to know that Google's Gmail is advertising restroom partitions in the margins of this message. I think they noticed our wide stance. On Sun, Jul 12, 2009 at 7:32 PM, Joerg Arndt <arndt@jjj.de> wrote:
* 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.
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