[seqfan] Operational recurrences

Olivier Gerard ogerard at ext.jussieu.fr
Fri Apr 23 18:06:05 CEST 1999


At 17:33 +0200 99.04.23, I wrote:

that A00629
->
->Can be obtained by this process (quite similar to Gould)
->
->u[1]=x
->u[n+1] = F ( d/dx  ( G( u[n] ) ) )
->
->where G is the transformation  x->Sin[x]
->and F is the transformation    Cos[x]->1-x  and   Sin[x]->x
->
->
There were several mistakes:

u[1]=x
u[n+1] = F ( x (d/dx) ( G( u[n] ) ) )

where G is the transformation  x->Sin[x]
and F is the transformation    Cos[x]->1+x  and   Sin[x]->x

and getting the values of the resulting polynomial at x=1 gives
1,2,6,26,150,1082,9366, ... (A00629)

Of course this process can be transformed into the iteration
of only one operator on a particular function, and this is
one of the definitions of A00629 anyway.

A further note on the sequence taken by NJAS from Gould's article.
As the degree of the polynomial created at step n is Fibonacci[n], one
should perhaps look for sequences of the form:

1,2,6, ?, 60, ?, ?, 2880 in the database to search for it.

regards,

Olivier







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