# (2s1) Designs withs intersection numbers by Ionin Y. J., Shrikhande M. S. By Ionin Y. J., Shrikhande M. S.

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Extra info for (2s1) Designs withs intersection numbers

Example text

For instance, the solution to question (i) for three consecutive heads demands the solution of x 3 - x 2 - x - 1 = 0, which produces the 'Tribonacci' sequence. ;, which we denote by g2 (x), as follows. 82(x) = LFlx;, 82(x)-x i=O = 82(x)-f6x 0 -F[x 1 . Using (lla) we have (since F0 = 0) LFlx; i=2 = L(3Fl-1-Fl-2+2(-l);+i)x; i=2 =x L 3Ffx; -x2 Li=OF;2x; + 2xz Li=O(- l);x2 i=l = 3xg2(x) - x 2g2(x) 2x2 +x +-1 - . Thus g2(x)(l-3x+x 2) = 2x2 x-l+x = x-x2 l+x. It follows that 82(x) x-x 2 1 - 2x - 2x 2 + x 3 • Riordan (1962) has developed generating functions for higher powers of Fibonacci numbers.

43) i= I 10. We have n n n n i= I i= I i= I i= 1 2: Gf= 2: G;(G;+ 1-G;_ 1)= 2: G;G;+ 1- 2: G;- 1 G; that is n 2: Gf = GnGn+1 -GoG1. i= 1 (44) Relationships 44 If G; [Ch. III =F;, then n L Ff= FnFn+l· (45) i= 1 This formula can also be illustrated geometrically. We do it in Fig. V(a). (See 13X21=F7 xF8 p2 7 p2 4 52 Fig. V(a). Ch. ) As n increases, the rectangle tends to a Gold Rectangle, because Jim n= F~+i tends oo n to 't (see (101) in Chapter VIII). In Fig. V(a) we have also drawn the straight lines which connect the centres C; of alternate squares.

9. Consider Thus Gi-G'f = G3Go Gj-Gi= G4G1 and, after addition, (42) Relationships Ch. III] 21 43 x 34 3 13 x5 5 2X3- A-' 1 x 1 1 x2 x 21 8 x8 x 13 Fig. IV(b). n 2: G;+2G;-1 = G~+1 - Gf. (43) i= I 10. We have n n n n i= I i= I i= I i= 1 2: Gf= 2: G;(G;+ 1-G;_ 1)= 2: G;G;+ 1- 2: G;- 1 G; that is n 2: Gf = GnGn+1 -GoG1. i= 1 (44) Relationships 44 If G; [Ch. III =F;, then n L Ff= FnFn+l· (45) i= 1 This formula can also be illustrated geometrically. We do it in Fig. V(a). (See 13X21=F7 xF8 p2 7 p2 4 52 Fig.