partitions and the formal power series

But to review, you have given p(x) = polcoeff partition series = A(x)/A(x^2), = B(x)/ B(x^3)....; such that A(x), B(x),...etc for any k = Euler transform of the generalized ruler sequences ("p-adic valuations of kn"). Below we have an important algebraic operation and identity.It states that given A(x), B(x), etc,....for any k = 2, 3,...; A(x) = p(x) * p(x^2) * p(x^4) * p(x^8) * p(x^16) *....; andB(x) = p(x) * p(x^3) * p(x^9) * p(x^27) * p(x^81) * ...etc. Gary
Date: Wed, 17 Mar 2010 17:16:59 -0700From: qntmpkt@yahoo.comSubject: Fw: [seqfan] Re: need algebra proofTo: qntmpkta@gmail.com; qntmpkta@hotmail.com
Conjecture:> Given polcoeff S(x), by way of example given A000041, the partition sequence = (1, 1, 2, 3, 5, 7, 11, 15, 22,...); then> S(x) = (1 + x + 2x^2 + 3x^3 + ...).>> Then S(x) = some A(x) such that S(x) = A(x) / A(x^2).
If you replace the somehow obscure " S(x) = some A(x) " (?)by "there exists A(x)", then this is indeed true,given the hypothesis that S(0) = 1.(Else such an A cannnot exist, because A(0) / A(0) = 1 (or ill definedif A(0)=0).)
> Our conjecture states that A(x) = S(x) * S(x^2) * S(x^4) * S(x^8) * S(x^16) * ...).
Obviously, A(x) is not uniquely defined, so you should say "onesolution is given by ...".The choice of this solution is equivalent to the assumption (or choice) A(0)=1.You could multiply this A by any nonzero constant and you would againget the same A(x) / A(x^2).
The infinite product converges in the topology of formal power series(which means that the coefficients up to a given orderare only affected by a finite number of terms, cf.http://en.wikipedia.org/wiki/Formal_power_series )so it is perfectly well defined.To exclude any ambiguity, you can specifythat the expression ... / A(x^2)means ... * ( A(x^2) )^(-1) ,where ( 1 - T(x) )^(-1) = 1 + T(x) + T(x)^2 + ...(where T(x) is any power series with T(0)=0).
So you can check that the initial formula is true(which means : coefficients of any given power are the sameon the l.h.s. and on the r.h.s.)by writing out the relevant factors of the infinite productsand cancelling those which correspond.
> Similarly, for S(x) = B(x) / B(x^3), our conjecture states that:>> B(x) = S(x) * S(x^3) * S(x^9) * S(x^27) * S(x^81) * ...>> Generally, for any k, in S(x) = C(x) / C(x^k).>
Yes, this is the same, the only necessary condition is again that S(0)= 1, and assuming the choice B(0) = C(0) = 1.
Maximilian
PS: I think Richard Mathar already pointed out roughly the same thing