[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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