Right and left factorials

[Artur]

Dear Thomas,
Left factorial are defined in A003422 
<http://www.research.att.com/%7Enjas/sequences/A003422>
You have to give another name for your sequence
BEST WISHES
ARTUR


Thomas Copeland pisze:
> To the interested seqfan:
>
> I've submitted the following comments to A094638.
>
> (Start)
> Consider c(t) = column vector(1, t, t^2, t^3, t^4, t^5,...) and T = A094638.
> Starting at 1 and sampling every integer to the right, we obtain
> (1,2,3,4,5,...). It's true that
> T * c(1) = (1, 1*2, 1*2*3, 1*2*3*4,...), giving n! for n>0. Call this
> sequence the right factorial (n+)! .
> Starting at 1 and sampling every integer to the left, we obtain
> (1,0,-1,-2,-3,-4,-5,...). And,
> T * c(-1) = (1, 1*0, 1*0*-1, 1*0*-1*-2,...) = (1, 0, 0, 0,...) . Call
> this the left factorial (n-)! .
> Sampling every other integer to the right, we obtain (1,3,5,7,9,...).
> T * c(2) = (1, 1*3, 1*3*5, ...) = (1,3,15,105,945,...) , giving
> A001147 for n>0, the right double factorial, (n+)!! .
> Sampling every other integer to the left, we obtain(1,-1,-3,-5,-7...).
> T * c(-2) = (1, 1*-1, 1*-1*-3, 1*-1*-3*-5,...) =
> (1,-1,3,-15,105,-945,...) = signed A001147, the left double factorial,
> (n-)!! .
> Sampling every 3 steps to the right, we obtain (1,4,7,10,...).
> T * c(3) = (1, 1*4, 1*4*7,...) = (1,4,28,280,...) , giving A007559 for
> n>0, the right triple factorial, (n+)!!! .
> Sampling every 3 steps to the left, we obtain (1,-2,-5,-8,-11,...), giving
> T * c(-3) = (1, 1*-2, 1*-2*-5, 1*-2*-5*-8,...) = (1,-2,10,-80,880,...)
> = signed A008544, the left triple factorial, (n-)!!! .
> The list partition transform A133314 of [1,T * c(t)] gives [1,T *
> c(-t)] with all odd terms negated; e.g., LPT[1,T*c(2)] =
> (1,-1,-1,-3,-15,-105,-945,...) =(1,-A001147) .
> And, e.g.f.[1,T * c(t)] = (1-xt)^(-1/t) .
> The above results hold for t any real or complex number.
> With P(n,t) = sum(k=0,...,n-1) T(n,k+1) * t^k =
> 1*(1+t)*(1+2t)...(1+(n-1)*t)  and P(0,t)=1 , exp[P(.,t)*x] =
> (1-tx)^(-1/t) .
> T(n,k+1) =  (1/k!) (D_t)^k (D_x)^n [ (1-tx)^(-1/t) - 1 ] eval. at t=x=0 .
> (1-tx)^(-1/t) - 1 is the e.g.f. for a plane m-ary tree when t= (m-1) .
> See Bergeron et al. "Varieties of Increasing Trees" (A132056) .
> (End)
>
>
> The above relations reveal the intimate connections, through T or LPT
> or sampling, between the factorials (n+)(!^m) and (n-)(!^m). The pairs
> also have conjugate interpretations as trees, ignoring signs, which
> Callan and Dieter have noted in several of the OEIS entries above.
>
> I'm not a combinatorialist and don't quite understand the
> interpretations, so can someone recommend a book/article/link similar
> to something like "Trees for Dummies" so that I can understand these
> connections more deeply and maybe write semi-intelligently about them?
>
> Tom Copeland
>
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