From 6fe1881e82586b695e164e86cc1d3987cfe15ee6 Mon Sep 17 00:00:00 2001
From: "xw_y_am@rmbp" <xw_y_am@163.com>
Date: Wed, 13 Sep 2017 17:48:19 +0800
Subject: [PATCH] add 131 133 135 137

---
 python/120.py   |  0
 python/131.1.py | 33 +++++++++++++++++++++++++++++++
 python/131.md   | 13 +++++++++++++
 python/131.py   | 12 ++++++++++++
 python/133.1.py | 32 ++++++++++++++++++++++++++++++
 python/135.1.py |  8 ++++++++
 python/137.md   | 52 +++++++++++++++++++++++++++++++++++++++++++++++++
 python/137.py   | 19 ++++++++++++++++++
 8 files changed, 169 insertions(+)
 mode change 100755 => 100644 python/120.py
 create mode 100644 python/131.1.py
 create mode 100644 python/131.md
 create mode 100644 python/131.py
 create mode 100644 python/133.1.py
 create mode 100644 python/135.1.py
 create mode 100644 python/137.md
 create mode 100644 python/137.py

diff --git a/python/120.py b/python/120.py
old mode 100755
new mode 100644
diff --git a/python/131.1.py b/python/131.1.py
new file mode 100644
index 0000000..06fb237
--- /dev/null
+++ b/python/131.1.py
@@ -0,0 +1,33 @@
+from tools import number_theory
+import math
+
+def try_root(n):
+    root = int(n ** 0.5)
+    if root ** 2 == n:
+        return root
+    raise ValueError
+
+def solve(p, k):
+    num, mod = divmod((3 * k + try_root(k * (4 * p - 3 * k))) * k, 2 * (p - 3 * k))
+    if not mod:
+        return num
+    raise ValueError
+
+def try_n(p, offset):
+    for k in range(int(p * offset) + 1, p // 3 + 1):
+        try:
+            return solve(p, k), k / p
+        except:
+            pass
+
+def find(limit):
+    offset = 0
+    for p in number_theory.make_prime(limit):
+        try:
+            solve, offset = try_n(p, offset)
+            print(solve, offset)
+            yield solve
+        except:
+            pass
+
+print(list(find(1000000)))
diff --git a/python/131.md b/python/131.md
new file mode 100644
index 0000000..76ceb8a
--- /dev/null
+++ b/python/131.md
@@ -0,0 +1,13 @@
+$$
+m^3=n^3+n^2p
+$$
+
+let $m=n+k$ , then
+$$
+3n^2k+3nk^2+k^3=n^2p\\
+(p-3k)n^2-3k^2n-k^3=0
+$$
+witch $p>3k$ . solving this function, we get
+$$
+n = \frac{3k^2 \pm k\sqrt{4kp-3k^2}}{p-3k}
+$$
diff --git a/python/131.py b/python/131.py
new file mode 100644
index 0000000..7494e64
--- /dev/null
+++ b/python/131.py
@@ -0,0 +1,12 @@
+
+from tools import number_theory
+
+_eps = 0.0000000000001
+
+def n_peak(p):
+    for n in range(1, p):
+        right = n ** 3 + p * (n ** 2)
+        m = int(right ** (1 / 3) + _eps)
+        print(n, m, right - m ** 3)
+
+print(n_peak(19))
diff --git a/python/133.1.py b/python/133.1.py
new file mode 100644
index 0000000..6ed75c3
--- /dev/null
+++ b/python/133.1.py
@@ -0,0 +1,32 @@
+from tools import number_theory
+
+def unit(p):
+    base = 1
+    loop = 1
+    while True:
+        base %= p
+        if not base:
+            return loop
+        base = 10 * base + 1
+        loop += 1
+
+def is_dec(num):
+    while not num % 2:
+        num //= 2
+    while not num % 5:
+        num //= 5
+    return 1 == num
+
+def statistic(limit):
+    prime = number_theory.make_prime(limit)
+    next(prime)
+    next(prime)
+    next(prime)
+    for p in prime:
+        if is_dec(unit(p)):
+            yield p
+
+#print(list(statistic(100000)))
+get = [11, 17, 41, 73, 101, 137, 251, 257, 271, 353, 401, 449, 641, 751, 1201, 1409, 1601, 3541, 4001, 4801, 5051, 9091, 10753, 15361, 16001, 19841, 21001, 21401, 24001, 25601, 27961, 37501, 40961, 43201, 60101, 62501, 65537, 69857, 76001, 76801]
+
+print(sum(set(number_theory.make_prime(100000)) - set(get)))
diff --git a/python/135.1.py b/python/135.1.py
new file mode 100644
index 0000000..722829b
--- /dev/null
+++ b/python/135.1.py
@@ -0,0 +1,8 @@
+'''
+let x = y + d, z = y - d
+then x**2 - y**2 - z**2 = n can be written b**2 - 4bd + n = 0
+so b = 2d +- sqrt(4 * d**2 - n)
+let t**2 = 4 * d**2 - n, then (2d + t) * (2d - t) = n
+'''
+
+def dual_factor(num):
diff --git a/python/137.md b/python/137.md
new file mode 100644
index 0000000..a576704
--- /dev/null
+++ b/python/137.md
@@ -0,0 +1,52 @@
+$$
+A_F(x)= xF_1+x^2F_2+x^3F_3\cdots
+$$
+
+$$
+xA_F(x)= \; x^2F_1+x^3F_2+x^4F_3+\cdots
+$$
+
+$(1)+(2)$, and consider that $F_1=F_2$, then we have
+$$
+(1+x)A_F(x)=xF_2+x^2F_2+x^3F_3+\cdots
+$$
+so
+$$
+x(1+x)A_F(x)=A_F(x)-xF_1
+$$
+then
+$$
+A_F(x)=\frac{x}{1-x-x^2}
+$$
+we want that $A_F(x)$ could be a nature number, so let
+$$
+\frac{x}{1-x-x^2}=n
+$$
+then
+$$
+nx^2+(n+1)x-n=0
+$$
+solve this, get
+$$
+x=\frac{\sqrt{5n^2+2n+1}-(n+1)}{2n}
+$$
+only $5n^2+2n+1$ is square number that can make x to be a rational.
+
+let's try to analyse how $n$ increasing when $5n^2+2n+1$ is square number.
+
+| n          | increasing rate of n |
+| ---------- | -------------------- |
+| 2          |                      |
+| 15         | 7.5                  |
+| 104        | 6.933333333          |
+| 714        | 6.865384615          |
+| 4895       | 6.855742297          |
+| 33552      | 6.854341164          |
+| 229970     | 6.854136862          |
+| 1576239    | 6.854107057          |
+| 10803704   | 6.854102709          |
+| 74049690   | 6.854102075          |
+| 507544127  | 6.854101982          |
+| 3478759200 | 6.854101969          |
+
+it is obvious that the rate of $n$ is converge to $6.8541\cdots$ . with using search engein we know $\varphi^4=6.8541\cdots$ and $\varphi=\frac{\sqrt{5}-1}{2}$.
\ No newline at end of file
diff --git a/python/137.py b/python/137.py
new file mode 100644
index 0000000..17a3eb2
--- /dev/null
+++ b/python/137.py
@@ -0,0 +1,19 @@
+
+def fib_jump():
+    a, b = 13, 21
+    while True:
+        yield a
+        for i in range(4):
+            a, b = b, a + b
+
+
+def search(limit):
+    n = 2
+    for jump in fib_jump():
+        yield n
+        n += jump
+        limit -= 1
+        if not limit:
+            return
+
+print(list(search(15))[-1])