/ AICHI UNIVERSITY OF EDUCATION
A { z = x + iy 0.100 <x<0.700, 0.000 <y<0.600 }. z F (z) F n (z). 4.2
B { z = x + iy 0.700 <x<1.500, 0.600 <y<1.200 }. z F (z) F n (z). 4.2 i
ii
m n (1) (2) (3) m (4) (5) -. 2002 3 iii
1 1 1.1..................... 1 1.2 f(x) =y................... 5 1.3........... 7 1.4.... 11 2 17 2.1.................... 17 2.2......... 22 3 27 3.1.. 27 3.2 2-.................. 30 3.3.................... 34 4 39 4.1... 39 4.2 F................. 42 50 iv
1 1.1 Syracuse Hasse Ulam 3n +1 Lothar Collatz(1910-1990) 1930 1 1 4 2 1 1
6 6 3 10 5 16 8 4 2 1 7 7 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1 27 27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1 111 1 y y y 1 y 2 y n 1 y n y 1 4 2 1 1 4 2 2
1 4 2 1 27 61 61 23 35 53 5 1 ( : x f f f(x) 3x +1 2 k 2 k+1 k 3x +1 y = 2 k f y = f(x) f f(x) 3x +1=2 k y y f. x =53 3x + 1 = 160 160 32 = 2 5 f(53) = 5 3
f(x) f 2 (x) = f(f(x)) f(f(x)) f(x) f. f 3 (x) = f(f(f(x))) { n }} { f n (x) = f(f n 1 (x)) = f(f( (f(x)) )) f n (x), n =2, 3, f n (x) f(x) n f(x) n = n {}}{ f(x) f(x) f(x) f n (x) (n =2, 3, ) f(x) f 2p +1 n f n (2p +1)=1, f f 4
x f(4x +1)=f(x) 3(4x +1)+1=12x +4=2 2 (3x +1). 1.2 f(x) =y x, y f(x) =y x x, y k 3x +1=2 k y x = 2k y 1 3 3m +1,3m +2,3m +3 (m =0, 1, 2, ) y =3m + r (r =0, 1, 2) x = 2k (3m + r) 1 3 =2 k m + 2k r 1 3 y =3m + r 3, r =0 5
, p =0, 1, 2, 1+4+4 2 + +4 p 2 +4 p 1 +4 p = 4p+1 1 4 1 = 22p+2 1 3 1+4+4 2 + +4 p 2 +4 p 1 + 4p 3 = 4p 1 4 1 + 4p 3 = 22p+1 1 3 y 3 1, r =1 x k 4 y 1 3, 4 2 y 1 3, 4 3 y 1 3,, 4 n y 1 3,. y 3 2, r =2 2y 1 3, 4 2y 1 3, 4 2 2y 1 3,, 4 n 2y 1 3,. 3m +1, 3m +2, 3m +3 (m =0, 1, 2, ) y 6n +1, 6n +3,6n +5(n =0, 1, 2, ) 6
x, y f(x) =y x y =6n +1 2 2p+2 (6n +1) 1 3 =4 p (8n +1)+4 p 1 + +4+1 (p =0, 1, 2, ). y =6n +3. y =6n +5 2 2p+1 (6n +5) 1 3 =4 p (4n +3)+4 p 1 + +4+1 (p =0, 1, 2, ). 1.3, 1 4 2 1 n y 1,y 2,,y n 1,y n = y 0 y n = y 0 y 1 y 2 y n 1 y n n 7
Computer x f 3x +1 2 k 2 k+1 k y = 3x +1 2 k f y = f(x) 3x +1=2 k y y f f(x) 1 ; 1 ( 2 ) 1 5 7 5 17 25 37 55 41 61 91 17 8
y y = y 0 y 1 y 2 y n 1 y n y ( ) 2 p1 y 1 =3y 0 +1 (p 1 1) 2 p2 y 2 =3y 1 +1 (p 2 1). 2 pi y i =3y i 1 +1 (p i 1). 2 pn y n =3y n 1 +1 (p n 1) ( ) y y n ( ) 2 p1+p2+ +pn y n = 2 p1+p2+ +pn 1 +3 2 p1+p2+ +pn 2 + +3 i 2 p1+p2+ +pn i 1 + +3 n 2 2 p1 +3 n 1 +3 n y n y n = y n ( ) (2 p1+p2+ +pn 3 n )y =2 p1+p2+ +pn 1 +3 2 p1+p2+ +pn 2 + +3 i 2 p1+p2+ +pn i 1 + +3 n 2 2 p1 +3 n 1 9
y, p 1 1,p 2 1,,p n 1 ( ) y =1,p 1 = p 2 = = p n =2 y = 1,p 1 = p 2 = = p n =1. (1) 5 7 5 ( ) n =2 p 1 =1,p 2 =2,y= 5 ( ) (2 1+2 3 2 )( 5) = 2 1 +3 2 1. (2) 17 25 37 55 41 61 91 17. n =7, p 1 = p 2 = p 3 = p 5 = p 6 =1,p 4 =2,p 7 =4,y= 17 p 1 + p 2 + p 3 + p 4 + p 5 + p 6 + p 7 =11 2 11 3 7 = 139 2 1+1+1+2+1+1 +3 1 2 1+1+1+2+1 +3 2 2 1+1+1+2 +3 3 2 1+1+1 +3 4 2 1+1 +3 5 2 1 +3 6 = 2363. ( 139)( 17) = 2363 ( ). 10
1.4 5 7 5. 2 p 3 q = ±1 p, q ( ) n = q 1 1 2 p 3 q =1 p, q 2 p 3 q = 1 2 p 3 q =1 p>n= q 2 p 1 =2, p 2 = p 3 = = p n 1 =1,p n = p n p 1 + p 2 + + p n = p 2 p 3 n =1 y =2 p1+p2+ +pn 1 +3 2 p1+p2+ +pn 2 + +3 i 2 p1+p2+ +pn i 1 + +3 n 2 2 p1 +3 n 1 11
( ) (2 p1+p2+ +pn 3 n )y =2 p1+p2+ +pn 1 +3 2 p1+p2+ +pn 2 + +3 i 2 p1+p2+ +pn i 1 + +3 n 2 2 p1 +3 n 1. 2 p 3 q =1 p, q. 2 p 3 q = 1 p =3,q=2 Catalan x a y b =1 a =2,b=3,x=3,y=2. n =2, 3, 4 ( ) 2 3 4. 1. ( ) y, p 1,p 2,,p n 12
( ) Computer 2 10 16 Tomas Oliveira e Silva, 1999 n y n = y 0 y 1 y 2 y n 1 y n y n = y 0 > 1, y 1 > 1,,y n 1 > 1. 1.3 ( ) 2 p1+p2+ +pn y 1 y 2 y n = (3y 0 + 1)(3y 1 +1) (3y n 1 +1) = 3 n y 0 y 1 y n 1 (1 + 1 3y 0 )(1 + 1 3y 1 ) (1 + 2 p1+p2+ +pn y n =3 n y 0 (1 + 1 3y 0 )(1 + 1 3y 1 ) (1 + 2 p1+p2+ +pn =3 n (1 + 1 3y 0 )(1 + 1 3y 1 ) (1 + a (y n =)y 0, y 1,, y n 1 >a 13 1 3y n 1 ) 1 3y n 1 ) 1 3y n 1 )
10 10 a =10 10 3 n < 2 p1+p2+ +pn =3 n (1 + 1 3y 0 )(1 + 1 3y 1 ) (1 + < 3 n (1 + 1 3a )n 1 3y n 1 ) II 12500 n 12500 ( ) x 1+x 8 y=k(x) y=3x/2 6 4 2 0-2 -4-6 -8-1 0 1 2 3 4 5 14
k(x) k(x) =log 2 (1 + x) < 3x 2 III 3 n < 2 p1+p2+ +pn < 3 n (1 + 1 3a )n log 2 (1 + x) < 3x 2 (x >0) log 2 3 < p 1 + p 2 + + p n n < log 2 3+ 1 2a log 2 3=1.5849625007 a 3 1.5 < p 1 + p 2 + + p n n a =10 10 < 2.0. log 2 3=1.5849625007 > 1.584962500 = 126797 8 10 4 0 < p 1 + p 2 + + p n n < p 1 + p 2 + + p n n log 2 3 126797 8 10 4 = (p 1 + p 2 + + p n ) 8 10 4 126797 n 8 10 4 n ( ) log 3 < log 2 + 1 2 10 10 126797 8 10 4 = log 3 log 2 126797 8 10 4 + 1 2 10 10 < 8 10 10 +0.5 10 10 < 10 9. 15
(p 1 + p 2 + + p n ) 8 10 4 126797 n 8 10 4 n > 0 1 8 10 4 n (p 1 + p 2 + + p n ) 8 10 4 126797 n 8 10 4 < 10 9 n n>10 9 /(8 10 4 ) = 12500 1+x e =2.718 log(1 + x) x log(1 + x) <x 3log2=3 0.693 =2.079 > 2 x log 2 (1 + x) < 3x 2 16
2 2.1 2 10 16 HP( ) http://mathworld.wolfram.com/collatzproblem.html 10 = 3628800 159487 183 1 60 5097000814 50. BASIC C++,Mathematica Maple. 17
., Windows BASIC Discoversoft ActiveBasic, BASIC.., BASIC, JIS Full BASIC Windows, 1000 10 10 1000. Full BASIC (ISO) BASIC. 18
BASIC [CollatzTest.BAS] #structcode STARTF=1 DO PRINT CollatzTest PRINT INPUT i( 0)=, i k=0 Collatz c=i WHILE c >1 IF c MOD 2 = 0 THEN c=c/2 ELSE c=3*c+1 END IF PRINT ;c; k=k+1 WEND PRINT ;k;! PRINT INPUT <Yes(Return)/No>=,S$ IF (S$ <> AND S$ <> Y AND S$ <> y ) THEN STARTF=0 LOOP WHILE STARTF=1 END 1 ( 19
20
11 1 ( BASIC [OddCollatzTest.BAS] #structcode STARTF=1 DO PRINT OddCollatzTest PRINT INPUT i( 0)=,i k=0 Collatz c=i WHILE c>1 c=3*c+1 r=c Mod 2 WHILE r=0 c=c/2 r=c Mod 2 WEND PRINT ;c; k=k+1 WEND PRINT ;k;! PRINT INPUT <Yes(Return)/No>=,S$ IF (S$ <> AND S$ <> Y AND S$ <> y ) THEN STARTF=0 LOOP WHILE STARTF=1 END 21
2.2 1 n 1 n n k 1 j k b n = j 1 + j 2 + + j n ( = 1 n ) j n n k b n 10.41373 log n 12.56 k=1 q(n) =10.41373 log n 12.56 n n k=1 j k b n q(n) 10 67 6.7 11.418 100 3142 31.42 35.396 1000 59542 59.542 59.375 10000 849666 84.9666 83.353 100000 10753840 107.53840 107.332 1000000 131434424 131.43442 131.310 10000000 1552724831 155.27248 155.289 100000000 17923493583 179.23493 179.267 1000000000 203234783374 203.23478 203.246 22
40 159487 64 1 29 2548500407, 30 3822750611, 31 5734125917 40 10 8 C++ 1000 BASIC 10 1000 23
1 n k 1 i k a n = i 1 + i 3 + + i 2n 1 ( = 1 n ) i n n 2k 1 a n 2.406 log n log 2 2 p(n) =2.406 log n log 2 2 q(n) 3p(n) 6.56 k=1 n n k=1 i 2k 1 a n p(n) 10 14 2.8 5.992 100 679 13.58 13.985 1000 11037 22.074 21.977 10000 153336 30.6672 29.970 100000 1899202 37.98404 37.962 1000000 22996390 45.99278 45.955 10000000 269686304 53.93726 53.947 100000000 3096869298 61.93738 61.940 1000000000 34969255812 69.93851 69.933 24
n ( 10 9 ) b n q(n) =10.41373 log n 12.56 =7.21824 log n log 2 12.56 a n p(n) =2.406 log n log 2 2 3a n b n 6.56 (10 5 n 10 9 ) n 25
Pierre de Fermat, 1601-1665 x n + y n = z n n 3 0 x, y, z. 1994 Andrew Wiles. n =2 0. Christian Goldbach, 1690-1764 1742 L. Euler, 1707-1783. 4 10 14 J. Richstein 1999. - (Erdös-Strauss) n 4 n = 1 a + 1 b + 1 c a, b, c. p 4 2+3p = 1 2+3p + 1 1+p + 1 (1 + p)(2 + 3p) n 10 8 26
3 3.1 f N = {1, 2, 3, } Z = {0, ±1, ±2, } OZ = {±1, ±3, ±5, }. x 2-0 x 2 k a (k Z, a, b OZ ) b k k x 2-, e(x) - 27
GQ = { 2 k 2m +1 m, n Z, k =0, 1, 2, } 2n +1 2-0 OQ = { 2m +1 m, n Z } 2n +1-1 3 GQ 1 3 OQ - 1 x 3 y( OQ f - k x =2 k y ( GQ, y OQ) f(x) =y x ( OQ ) k =1, 2, 3x +1=2 k y (y OQ) f(x) =y f(x) GQ x OQ f(x) f GQ OQ f(x) 28
- k x = 2k r (r, s OZ ) s f(x) = r ( OQ s x = r s OQ 3r + s - k t 3r + s =2 k t 3 x +1=3 r s +1= 3r + s s = 2k t s f(x) = 3x +1 2 k = t s ( OQ 3 5 f( 3 5 )= 7 5 f( 7 13 )= 5 5 f GQ OQ f OQ OQ f(x) OQ a(x) f OQ OQ OQ a(x) =2x + 1 3 = 6x +1 3 (x OQ) f(a(x)) = f(x) (x OQ ) a(a(x)) = 4x +1 (x OQ) f(4x +1)=f(x) (x OQ ) 29
f f(x) =y x, y f(x) =y x x 2 k y 1 3 (k =1, 2, 3, ) f OQ OQ f OQ { 1 2 k 3 k =1, 2, 3, f f(x) =x. } 3.2 2- k, x = x 0,,x k = f k (x) OQ, 30
{x 0,x 1,x 2, } x f {x k } {x k } k=0. p k e(3x k 1 +1), 3x k 1 +1=2 p k x k p k, {p 1,p 2,p 4, } x f 2- {p k } k=1 f {x k } k=0 n x 0 = x n x 1 = x n+1, 2 p1+p2+ +pn 1 +3 2 p1+p2+ +pn 2 + +3 i 2 p1+p2+ +pn i 1 + +3 n 2 2 p1 +3 n 1 C(p 1,p 2,,p n 1 ) 1.3 ( ),. x = x 0,x k = f k (x) OQ (k =1, 2, 3, ) x f 2- {p k } k=1, n 2 p1+p2+ +pn x n = C(p 1,p 2,,p n 1 )+3 n x 31
,. x = x 0,x k = f k (x) OQ (k =1, 2, 3, ) x f 2- {p k } k=1, f {x k } k=0 n, (2 p1+p2+ +pn 3 n )x = C(p 1,p 2,,p n 1 ) x = C(p 1,p 2,,p n 1 ) 2 p1+p2+ +pn 3 n OQ,, f 2- f. x = x 0, x k = f k (x), y = y 0, y k = f k (y) OQ (k = 1, 2, 3, ) x y f 2-, x y., k e(3x k 1 +1)=e(3y k 1 +1) x = y 32
x = x 0,x k = f k (x) OQ (k =1, 2, 3, ) f {x k } k=0 n, x f 2- {p k } k=1 n t N k =1, 2,,n p tn+k = p k p 1,p 2,,p n (1) f OQ OQ 1 2 p 3 (p =1, 2, 3, ) (f ) 2- {p k = p} 1 (2) f 2- {p, q} 2 2 p +3 2 p+q 9 (p, q =1, 2, 3, ) 7 f 2- {2, 1} 5 7 2 2 +3 2 2+1 9 = 7 1 = 7 f 2- {1, 3} 2 1 +3 2 1+3 9 = 5 7 33
3.3 N n 2 n N<2 n+1 { N =2 n + M 0 M<2 n, M, N N =Σ n k=0 a k2 k ( a k 0 1). - - - - - - - r =2 k a b (0 k Z, a,b OZ) 34
-. r =Σ k=0 a k2 k ( a k 0 1). 1 = Σ k=0 2k 0 = 1+( 1) = 1 + Σ k=02 k 2 = Σ k=1 2k, 2=2 1 3 = 1+Σ k=22 k, 3=1+2 4 = Σ k=2 2k, 4=2 2 5 = 1+2+Σ k=32 k, 5=1+2 2 2 p 1 = Σ p 1 k=0 2k 2 p = Σ k=p2 k 1 3 2 3 1 3 2 3 1 5 1 5 1 2 p 1 = 1 1 2 2 =Σ k=02 2k = 2+Σ k=1 22k = 1+Σ k=1 22k 1 = Σ k=02 2k+1 = 3 1 2 4 =(1+2)Σ k=02 4k =Σ k=02 4k +Σ k=02 4k+1 = 1 4 5 =1+Σ k=0 24k+2 +Σ k=0 24k+3 = 1+2 2 +2 3 +2 6 +2 7 +2 10 +2 11 + = 1 1 2 p =Σ k=02 pk 35
= 1+2 p +2 2p +2 3p +2 4p + f x = 1 5 = 1 2 p 3, p =3 -. x = 1-1, 0,, 5 2x +1-1, 0,, 3x +1-1, 0,. 2x +1 - x - 1, 1. 3x +1 - x - 2x +1 -, x -. x = 1 f 5. 1 13 =1+22 +(2 6 +2 7 +2 9 +2 10 +2 11 +2 14 )Σ k=02 12k =1+2 2 +2 6 +2 7 +2 9 +2 10 +2 11 +2 14 +2 18 +2 19 +2 21 +2 22 +2 23 +2 26 + 36
f x = 1 13 = 1 2 p 3, p =4, x = 1 -, 2x +1-13, 3x +1-1, 0,. 3x +1 - x - 2x +1 -, x -. f x = 1 2 5 3 = 1 29, x = 1 2 6 3 = 1 61 -. 1 29 =1+(22 +2 4 +2 5 +2 9 +2 14 +2 15 +2 17 +2 20 +2 21 +2 22 +2 24 +2 25 +2 26 +2 27 )Σ k=02 28k =1+2 2 +2 4 +2 5 +2 9 +2 14 +2 15 +2 17 +2 20 +2 21 +2 22 +2 24 +2 25 +2 26 +2 27 +2 30 +2 32 +2 33 +2 37 +2 42 +2 43 +2 45 +2 48 +2 49 +2 50 +2 52 +2 53 +2 54 +2 55 + 37
1 61 =1+(22 +2 4 +2 8 +2 9 +2 10 +2 12 +2 15 +2 18 +2 19 +2 24 +2 30 +2 31 +2 33 +2 35 +2 36 +2 37 +2 41 +2 43 +2 44 +2 46 +2 47 +2 50 +2 51 +2 52 +2 53 +2 55 +2 56 +2 57 +2 58 +2 59 )Σ k=0 260k =1+2 2 +2 4 +2 8 +2 9 +2 10 +2 12 +2 15 +2 18 +2 19 +2 24 +2 30 +2 31 +2 33 +2 35 +2 36 +2 37 +2 41 +2 43 +2 44 +2 46 +2 47 +2 50 +2 51 +2 52 +2 53 +2 55 +2 56 +2 57 +2 58 +2 59 +2 62 +2 64 + - - - 38
4 4.1, 1 90 =radian =radius 39
t x t 180 = x π 180 x πx cos(180 x) = cosπx cos(90 x) = cos πx 2 sin(90 x) = sin πx 2 x : g(x) = x 2 cos2 πx 2 = 1 2 + 7x 4 +(3x +1)sin2 πx 2 5x +2 4 cos πx. g(x) x x x 2 3x +1 g(x) g(x) g(x) 10 <x<22. 40
70 60 y=g(x) 50 40 30 20 10 0-10 -20-30 -10-5 0 5 10 15 20 g 2 (x) = g(g(x)) g 3 (x) = g(g(g(x))) g n (x) = g(g n 1 (x)) = g(g( (g(x)) )) x g n (x) g(x), h(x) h(x) = x 2 = 1 4 cos2 πx 2 + 3x +1 2 + x 2x +1 4 cos πx. sin 2 πx 2 41
h(x) x x 2 x 3x +1 2 h(x). h(x) 10 <x<22. 35 30 y=h(x) 25 20 15 10 5 0-5 -10-15 -10-5 0 5 10 15 20 4.2 F., F (x) = 4 πx ( ) 1 2 cos2 f(2k +1) π 2 (x 2k 1) 2 1 (2k +1) 2. k=0 42
F (x), 4 π 2 πx cos2 2 n k=0 ( ) f(2k +1) 1 (x 2k 1) 2 1 (2k +1) 2 n. F (x) 10 <x<22. 30 y=f(x) 25 20 15 10 5 0-5 -10-5 0 5 10 15 20 cos πx cos πx =1+ q=1 ( 1)q π 2q x 2q (2q)! =1 π2 x 2 2! + π4 x 4 4! π6 x 6 6! π8 x 8 8! +. F (x) x, 43
F (x) = 2 (2+ π 2 ( 1) q π2q x 2q ) { ( p (2q)! q=1 p=2 m=0 { ( = 2 f(2m +1) ) ( π 2 4 (2m +1) 3 x + 6 m=0 m=0 ( f(2m +1) + 8 (2m +1) 5 f(2m +1) ) π2 (2m +1) 3 m=0 m=0 ( f(2m +1) + 10 (2m +1) 6 3π2 2!. m=0 m=0 } f(2m +1) ) (2m +1) p+1 x p 1 f(2m +1) ) (2m +1) 4 x 2 x 3 f(2m +1) ) (2m +1) 4 x 4 +, F (0) = 8 f(2n +1) π 2 (2n +1) 3 n=0 = 8 ( ) 6k +1 π 2 (4 n (8k +1)+4 n 1 + +4+1) 3. k=0 n=0 + 8 ( π 2 k=0 n=0 (=1.04 ) > 1 ) 6k +5 (4 n (4k +3)+4 n 1 + +4+1) 3 0 F 0 F z F (1) = 0 1 F 1 F z } F (x) x z F (z) 44
A B C E F (z) z F n (z). lim n F n (z) =1 z lim n F n (z) =z 0 z z 0 = 0.03 F z 0 F z n u + iv = F n (z), u>2 10 6 v>70 z C { z = x + iy 4 <x<4, 1.5 <y<4.5 } z F n (z) E { z = x + iy 2.9400 <x<3.0200, 0.5950 <y<0.6550 } z F n (z) x cos πx cos πx =1+ q=1 ( 1)q π 2q x 2q (2q)! =1 π2 x 2 2! + π4 x 4 4! π6 x 6 6! π8 x 8 8! + x z cos πz 45
C D 46
E cos πz =1+ q=1 ( 1)q π 2q z 2q (2q)! =1 π2 z 2 2! + π4 z 4 4! π6 z 6 6! z π8 z 8 8! + 47
g(z) = z 2 = 1 2 + 7z 4 cos2 πz 2 +(3z +1)sin2 πz 2 5z +2 4 cos πz z x z = x g(z) =g(x) D g(z) { z = x + iy 4 <x<4, 3 <y<3 } z g 3 (z). g 3 (z) 1 < 0.1 z u + iv = g 3 (z) u>10 20 v>70 z F (z) z lim F n (z) =1 n? (Paul Erdös, 1913-1996) 48
( Mathematics is not yet ready for such problems. ) : - - 2000 Richard K. Guy(Ed.), : 1983 Jeffrey C. Lagarias : The 3x + 1 Problem and its Generalizations, Amer. Math. Monthly 92(1985), 3-23. http://www.cecm.sfu.ca/organics/papers/lagarias/ 49
f 3,8,27,28 f GQ OQ 28 f OQ OQ 29,30,33 F 42,44 f 31 f 2-31 f(x) a(x) 29-26 Catalan 12 f(x) 4 27,28 45 27 y 2,9 3,21,24 1,22 1,19,22 1 1 30 26 39 3n +1 1 N 27 16 7 33 31,33 7 2,7,8,14 GQ 28 GQ 28 OQ 28 OQ 28 OZ 27 Z 27 14 23 28 44 28 2-31 34-34 - 34,35 34,36,38 44 48 45 4,30,33 BASIC 18 22,24 26 48 43 34 2-27 39 50
1970.. / c T. Urata 2002 / 2002 3 31 448-8542 (0566)26-2327/2315
/. AICHI UNIVERSITY OF EDUCATION