about A001333, A034183, A000129 (Pell numbers)

[vdmcc]

about
A001333 1,1,3,7,17,41,99,239,577,1393
A034183   1,3,7,17,41,99,239,577,1393
A000129 0,1,2,5,12,29,70,169,408,985
A024542   1,2,5,12,29, ... now dead, deceased, gone, ... !

they are linked by the definition
"(denominators of continued fraction convergents to sqrt(2))."

f(n) = 2*f(n-1)+f(n-2), f(0)=0, f(1)=1;

nothing new, except maybe :

A001333(n)+A001333(n+1) = 2 A000129(n+1)
A000129(n)+A000129(n+1) =   A001333(n+1)

and working out the recursion gives :

A001333(n) =   ( (1-Sqrt[2])^n + (1+Sqrt[2])^n) /2
A000129(n) =   (-(1-Sqrt[2])^n + (1+Sqrt[2])^n) /2/Sqrt[2]
-------------------------------------------------------------
we meet these sequences pair-wise in :
 NestList[1+1/(#+1 )&,m,3]
 =
          1            1                1
{m, 1 + -----, 1 + ---------, 1 + -------------,   ...  }
        1 + m            1                1
                   2 + -----      2 + ---------
                       1 + m                1
                                      2 + -----
                                          1 + m
=
    2 + m  4 + 3 m  10 + 7 m  24 + 17 m  58 + 41 m  140 + 99 m  338 + 239 m
{m, -----, -------, --------, ---------, ---------, ----------, -----------}
    1 + m  3 + 2 m  7 + 5 m   17 + 12 m  41 + 29 m  99 + 70 m   239 + 169 m

with m -> 0 :
{0, 2, 4/3, 10/7, 24/17, 58/41, 140/99, 338/239}
numerator = 2 A000129(n) ; denominator = A001333(n)

and with m -> 1 :
{1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408}
numerator =  A001333(n+1) ; denominator = A000129(n+1)

so, the above sequence of rationals in m is :
( 2 a000129[n]+a001333[n] m)/(a001333[n]+a000129[n] m)

we could call this thing "the n-th iterate of function f: x -> 1+1/(x+1)  on
m"
or (f^n)(m) for short.

with of course:
Limit[a001333[n]/a000129[n],n->oo] = Sqrt[2]
and
Limit[a000129[n+1]/a000129[n],n->oo] = Limit[a001333[n+1]/a001333[n],n->oo]
= 1+Sqrt[2]

as I said, nothing new,

w.meeussen.vdmcc at vandemoortele.be
tel  +32 (0) 51 33 21 11
fax +32 (0) 51 33 21 75