Riffs & Rotes & A061396

David W. Wilson wilson at aprisma.com
Thu Jun 21 00:08:58 CEST 2001


Jon Awbrey wrote:
> 
> ¤~~~~~~~~~¤~~~~~~~~~¤~~~~~~~~~¤~~~~~~~~~¤~~~~~~~~~¤
> 
> Hello, I am new to the SeqFan list.  I would like
> to report on some work that I did as a beginning
> and continuing graduate student in the years from
> 1976 to 1986.  Since this is not currently my area
> of active work, I am afraid that I will have to be
> somewhat dependent on my old notes, such as I find
> them.  I tried to organize this stuff in a logical
> order, but that was hopeless, so I will just have
> to take things up in a somewhat random order.
> 
> So let me introduce the sequence formally known as "A061396".
> 
> ¤~~~~~~~~~¤~~~~~~~~~¤~~~~~~~~~¤~~~~~~~~~¤~~~~~~~~~¤
> 
> Riffs & Rotes
> 
> 1, 2, 6, 20, 73, 281, 1124, 4488
> 
> ¤~~~~~~~~~¤~~~~~~~~~¤~~~~~~~~~¤~~~~~~~~~¤~~~~~~~~~¤
> 
> ID:       A061396
> 
> Seq:      1,1,2,6,20,73,281,1124,4488
> 
> Name:     Number of "rooted index-functional forests" (Riffs) on n nodes.
>           Number of "rooted odd trees with only exponent symmetries" (Rotes)
>           on 2n+1 nodes.
> 
> Ref:      J. Awbrey, personal journal, circa 1978.
>           Letter to N. J. A. Sloane, 1980-Aug-04.
> 
> Formula:  G.f.  A(x) = 1 + x + 2*x^2 + 6*x^3 + ... satisfies
>           A(x) = Product_{j = 0 to infinity} (1 + x^(j+1)*A(x))^a_j.
> 
> Example:  These structures come from recursive primes factorizations
>           of natural numbers, where the recursion proceeds on both
>           the exponents (^k) and the indices (_k) of the primes
>           invoked in the factorization:
> 
>           2 = (prime_1)^1 = (p_1)^1, briefly, p,   weight = 1 node  => a(1) = 1.
>           3 = (prime_2)^1 = (p_2)^1, briefly, p_p, weight = 2 nodes and
>           4 = (prime_1)^2 = (p_1)^2, briefly, p^p, weight = 2 nodes => a(2) = 2.
> 
> Keys:     nice,nonn,new,easy,more
> 
> Offset:   0
> 
> Author:   Jon Awbrey (jawbrey at oakland.edu), Jun 09 2001

I computed this sequence from the g.f.  I get

1 1 2 6 20 73 281 1124 4618 19387 82765 358245 1568458 6933765 30907194
138760603 626898401 2847946941 13001772692 59618918447 274463781383
1268064807513 5877758071073 27325789133590 127384553287387 595318140045752

My a(8) = 4618 disagrees with the published a(8) = 4488.  I computed the
sequence using two programs, with the same result, and I am pretty sure
it is correct (some corroboration would be welcomed).  In defense of my
value, notice that the relative differences of my terms increase nicely: 

               1/1                = 1.000000
               2/1                = 2.000000
               6/2                = 3.000000
              20/6                = 3.333333
              73/20               = 3.650000
             281/73               = 3.849315
            1124/281              = 4.000000
            4618/1124             = 4.108541
           19387/4618             = 4.198138
           82765/19387            = 4.269098
          358245/82765            = 4.328460
         1568458/358245           = 4.378171
         6933765/1568458          = 4.420753
        30907194/6933765          = 4.457491

whereas a(8) = 4488 produces a sudden decrease of the relative difference,
which would be unexpected in this context.

               1/1                = 1.000000
               2/1                = 2.000000
               6/2                = 3.000000
              20/6                = 3.333333
              73/20               = 3.650000
             281/73               = 3.849315
            1124/281              = 4.000000
            4488/1124             = 3.992883





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