\\(p\\) \u3092\u81ea\u7136\u6570\u3068\u3059\u308b.\r\n\u6b21\u306e\u95a2\u4fc2\u5f0f\u3067\u5b9a\u3081\u3089\u308c\u308b\u6570\u5217 \\(\\{ a _ n \\} , \\{ b _ n \\}\\) \u3092\u8003\u3048\u308b.\r\n\\[\r\n\\left\\{ \\begin{array}{ll} a _ 1 = p , \\ b _ 1 = p+1 & \\\\ a _ {n+1} = a _ n + p b _ n & ( n = 1, 2, 3 , \\cdots ) \\\\ b _ {n+1} = p a _ n + (p+1) b _ n & ( n = 1, 2, 3, \\cdots ) \\end{array} \\right.\r\n\\]\r\n
(1)<\/strong>\u3000\\(n = 1, 2, 3, \\cdots\\) \u306b\u5bfe\u3057, \u6b21\u306e \\(2\\) \u3064\u306e\u6570\u304c\u3068\u3082\u306b \\(p^3\\) \u3067\u5272\u308a\u5207\u308c\u308b\u3053\u3068\u3092\u793a\u305b.\r\n\\[\r\na _ n -\\dfrac{n(n-1)}{2}p^2 -np , \\ b _ n -n(n-1)p^2 -np -1\r\n\\]<\/li>\r\n (2)<\/strong>\u3000\\(p\\) \u3092 \\(3\\) \u4ee5\u4e0a\u306e\u5947\u6570\u3068\u3059\u308b. \u3053\u306e\u3068\u304d, \\(a _ p\\) \u306f \\(p^2\\) \u3067\u5272\u308a\u5207\u308c\u308b\u304c, \\(p^3\\) \u3067\u306f\u5272\u308a\u5207\u308c\u306a\u3044\u3053\u3068\u3092\u793a\u305b.<\/p><\/li>\r\n<\/ol>\r\n (1)<\/strong><\/p>\r\n \\(u _ n = a _ n -\\dfrac{n(n-1)}{2}p^2 -np\\) , \\(u _ n = b _ n -n(n-1)p^2 -np -1\\) \u3068\u304a\u304f.<\/p>\r\n \u3053\u308c\u3092, \u6570\u5b66\u7684\u5e30\u7d0d\u6cd5\u3092\u7528\u3044\u3066\u793a\u3059.<\/p>\r\n 1*<\/strong>\u3000\\(n = 1\\) \u306e\u3068\u304d\r\n\\[\\begin{align}\r\nu _ 1 & = p -0 \\cdot p^2 -p = 0 , \\\\\r\nv _ 1 & = p+1 -0 \\cdot p^2 -p-1 = 0\r\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066, \\(n = 1\\) \u306e\u3068\u304d [\uff0a] \u306f\u6210\u7acb\u3059\u308b.<\/p><\/li>\r\n 2*<\/strong>\u3000\\(n = k \\ ( k \\geqq 1 )\\) \u306e\u3068\u304d, [\uff0a] \u304c\u6210\u7acb\u3059\u308b\u3068\u4eee\u5b9a\u3059\u308b\u3068\r\n\\[\\begin{align}\r\nu _ {k+1} & = a _ k +p b _ k -\\dfrac{(k+1)k}{2} p^2 -(k+1) p \\\\\r\n& = a _ k -\\dfrac{k(k-1)}{2} p^2 -kp +p ( b _ k -kp -1 ) \\\\\r\n& = u _ k +p v _ k +k(k-1) p^3\r\n\\end{align}\\]\r\n\u3086\u3048\u306b, \\(u _ {k+1}\\) \u306f \\(p^3\\) \u3067\u5272\u308a\u5207\u308c\u308b.\r\n\u307e\u305f\r\n\\[\\begin{align}\r\nv _ {k+1} & = p a _ k +(p+1) b _ k -(k+1)k p^2 -(k+1) p -1 \\\\\r\n& = b _ k -k(k-1) p^2 -kp -1 +p ( a _ k +b _ k -2kp -1 ) \\\\\r\n& = v _ k +p \\left\\{ u _ k +v _ k +\\dfrac{3 k(k-1)}{2} p^2 \\right\\} \\\\\r\n& = p u _ k +(p+1) v _ k +\\dfrac{3 k(k-1)}{2} p^3\r\n\\end{align}\\]\r\n\\(\\dfrac{3 k(k-1)}{2}\\) \u306f\u6574\u6570\u306a\u306e\u3067, \\(v _ {k+1}\\) \u306f \\(p^3\\) \u3067\u5272\u308a\u5207\u308c\u308b. \u4ee5\u4e0a\u3088\u308a, \u81ea\u7136\u6570 \\(n\\) \u306b\u5bfe\u3057\u3066 \\(u _ n , v _ n\\) \u306f\u3068\u3082\u306b \\(p^3\\) \u3067\u5272\u308a\u5207\u308c\u308b.<\/p>\r\n (2)<\/strong><\/p>\r\n \\[\\begin{align}\r\na _ p & = u _ p +\\dfrac{p(p-1)}{2} p^2 -p^2 \\\\\r\n& = u _ p +p^2 \\underline{\\left( p \\cdot \\dfrac{p-1}{2} -1 \\right)} _ {[1]} \\\\\r\n\\end{align}\\]\r\n\\(p\\) \u306f\u5947\u6570\u306a\u306e\u3067 \\(\\dfrac{p-1}{2}\\) \u306f\u6574\u6570\u3067\u3042\u308a, \u4e0b\u7dda\u90e8 [1] \u306f \\(p\\) \u306e\u500d\u6570\u3067\u306f\u306a\u3044.
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h2>\r\n
\r\n
\r\n
\r\n\u3057\u305f\u304c\u3063\u3066, \\(n = k+1\\) \u306e\u3068\u304d\u3082 [\uff0a] \u304c\u6210\u7acb\u3059\u308b.<\/p><\/li>\r\n<\/ol>\r\n
\r\n(1)<\/strong> \u306e\u7d50\u679c\u3092\u5408\u308f\u305b\u308c\u3070, \\(a _ p\\) \u306f \\(p^2\\) \u3067\u5272\u308a\u5207\u308c\u308b\u304c, \\(p^3\\) \u3067\u306f\u5272\u308a\u5207\u308c\u306a\u3044.<\/p>\r\n","protected":false},"excerpt":{"rendered":"\\(p\\) \u3092\u81ea\u7136\u6570\u3068\u3059\u308b. \u6b21\u306e\u95a2\u4fc2\u5f0f\u3067\u5b9a\u3081\u3089\u308c\u308b\u6570\u5217 \\(\\{ a _ n \\} , \\{ b _ n \\}\\) \u3092\u8003\u3048\u308b. \\[ \\left\\{ \\begin{array}{ll} a _ 1 = p , \\ b … \u7d9a\u304d\u3092\u8aad\u3080