561 
or 
25 
n* + 35n + Ge 
Dee nea OET Lol 
; 2000 (1—e) (1—9’) 
Eyres = o-" (20 + 607) 
2000 ~ (Ie) de 
_ 1 yer B4o_~ 989" — 429" — 11494) 
2000 (1 —e)(1 —e7) 
With the same notations as in the preceding case, we obtain 
Se SS EES ee De 
and generally 
ee ete Las pea) 
According to the values 
n= dp, 5p+1, 5p+ 2, 5p+ 3, 5p4+4 
we find 
Er zele == Zo" (1 tete = 
Tek Ge 
1 
Bs eae L4n42— Dant3 — Dani) = (1. 0, t =a =) 
„e "(2e + 69’) 1 
EME 5 
„2 "(B4g + 9807 — 429*— 114%) 1 
a (eo) 07); 25 
+ 32 Zine + 162 Dy,43 + 352 Dan) = 
(2 Zan43 ap é Zana — 6 Ean) == (= 6, 0, 0, 2, 4) 
(— 588 Dan — 8 Zan + 
(2069) AN 16.481, 176) 
or} Do 
and therefore 
l 
= 3 N \ ng Bl 
s 18000 (24n? + 840n + 6300) (1, 0, 1, il ) 
l 
+ ——. , 3, 0, 0, —1,.— 2 
48000 48n ( ) 
1 _48 
F—_ ___ (269, 4, —16, —81, — 176). 
48000 5 
In the same way we obtain, according to 
n==10p, 10p+1, 10p42,... 10p +9 
I 
W,,=— 120 n + 1800) (—1, 0, 1, 8, 3, 1, 0, —1, —3, — 3 
ze 45000 ° de, )( , 
4 . 60(5, 16, 21, 29, 19, —5, — 16, 21, 29, 19) 
48000 
and according to 
