From 2820d50cbec08891edc9752d2722eec3338e0d9a Mon Sep 17 00:00:00 2001
From: "xw_y_am@rmbp" <xw_y_am@163.com>
Date: Sat, 30 Sep 2017 11:03:38 +0800
Subject: [PATCH] add 139 140

---
 python/139.md | 24 +++++++++++++++++++++++
 python/139.py | 41 +++++++++++++++++++++++++++++++++++++++
 python/140.md | 53 +++++++++++++++++++++++++++++++++++++++++++++++++++
 python/140.py | 13 +++++++++++++
 4 files changed, 131 insertions(+)
 create mode 100644 python/139.md
 create mode 100644 python/139.py
 create mode 100644 python/140.md
 create mode 100644 python/140.py

diff --git a/python/139.md b/python/139.md
new file mode 100644
index 0000000..ace3a57
--- /dev/null
+++ b/python/139.md
@@ -0,0 +1,24 @@
+毕达哥拉斯数可以由如下公式得出
+
+$$
+a=2mn \\
+b=m^2-n^2 \\
+c=m^2+n^2
+$$
+显然 $m>n$ ，因此不妨设 $m=n+k$ ，则
+$$
+a=2n^2+2nk \\
+b=k^2+2nk \\
+c=2n^2+2nk+k^2
+$$
+所以 $a+b+c=4n^2+6nk+2k^2$ 。
+
+$4n^2+6n+2<N$ ，所以
+$$
+n<\frac{\sqrt{4N+1}-3}{4}
+$$
+$2k^2+6nk+4n^<N$ ，所以
+$$
+k<\frac{\sqrt{2N+n^2}-3n}{2}
+$$
+
diff --git a/python/139.py b/python/139.py
new file mode 100644
index 0000000..ef95237
--- /dev/null
+++ b/python/139.py
@@ -0,0 +1,41 @@
+from tools import number_theory
+
+def range_n(N):
+    return int(((4 * N + 1) ** 0.5 - 3) / 4) + 1
+
+def range_k(N, n):
+    return int(((2 * N + n ** 2) ** 0.5 - 3 * n) / 2) + 1
+
+def search(limit):
+    count = 0
+    for n in range(1, range_n(limit)):
+        for k in range(1, range_k(limit, n), 2):
+            if number_theory.gcd(n, k) > 1:
+                continue
+            p = 2 * n ** 2
+            q = k ** 2
+            r = 2 * n * k
+            if (p + q + r) % abs(p - q):
+                continue
+            print(n, k)
+            count += limit // (2 * p + 2 * q + 3 * r)
+    return count
+
+def perfect(limit):
+    def side(n, k):
+        return 4 * n ** 2 + 6 * n * k + 2 * k ** 2
+    def sqrt2(n):
+        m = int(n * 2 ** 0.5)
+        return m + (1 - m % 2)
+    n, k = 1, 1
+    border = 12
+    count = 0
+    while border < limit:
+        yield limit // border
+        n = n + k
+        k = sqrt2(n)
+        border = side(n, k)
+
+#print(search(100000000))
+print(sum(list(perfect(100000000))))
+
diff --git a/python/140.md b/python/140.md
new file mode 100644
index 0000000..80c584c
--- /dev/null
+++ b/python/140.md
@@ -0,0 +1,53 @@
+根据137题有
+$$
+A_G(x)=\frac{x+3x^2}{1-x-x^2}
+$$
+
+所以
+$$
+(n+3)x^2+(n+1)x-n=0
+$$
+解此方程，得
+$$
+x=\frac{\sqrt{5n^2+14n+1}-(n+1)}{2(n+3)}
+$$
+只有 $5n^2+14n+1$ 是平方数时， $x$ 才会是有理数。
+
+尝试分析 $N=5n^2+14n+1$ 是平方数时 $n$ 的变化规律
+
+| N        | increasing rate of N |
+| -------- | -------------------- |
+| 2        |                      |
+| 5        | 2.5                  |
+| 21       | 4.2                  |
+| 42       | 2.0                  |
+| 152      | 3.619047619047619    |
+| 296      | 1.9473684210526316   |
+| 1050     | 3.5472972972972974   |
+| 2037     | 1.94                 |
+| 7205     | 3.5370643102601864   |
+| 13970    | 1.9389312977099236   |
+| 49392    | 3.535576234788833    |
+| 95760    | 1.9387755102040816   |
+| 338546   | 3.535359231411863    |
+| 656357   | 1.9387527839643652   |
+| 2320437  | 3.535327573256627    |
+| 4498746  | 1.9387494683113569   |
+| 15904520 | 3.535322954441082    |
+
+观察发现 $1.9387\ldots*3.5353\ldots=\frac{\sqrt{5}+1}{2}$.
+
+进一步尝试
+$$
+5-2=3 \\
+42-21=21 \\
+296-152=144
+$$
+
+$$
+21-5=16=2*8 \\
+152-42=110=2*55 \\
+1050-296=754=2*377
+$$
+
+$3,21,144,\ldots,8,55,377\ldots$ 它们都是 Fibonacci 数。
\ No newline at end of file
diff --git a/python/140.py b/python/140.py
new file mode 100644
index 0000000..52dc9b2
--- /dev/null
+++ b/python/140.py
@@ -0,0 +1,13 @@
+
+def calc(limit):
+    a, b = 3, 5
+    n = 2
+    d = 1
+    for i in range(limit):
+        yield n
+        n += d * a
+        a, b = b, a + b
+        a, b = b, a + b
+        d = 3 - d
+
+print(sum(list(calc(30))))