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+#!/usr/bin/python3
+# https://theweeklychallenge.org/blog/perl-weekly-challenge-246/#TASK2
+#
+# Task 2: Linear Recurrence of Second Order
+# =========================================
+#
+# You are given an array @a of five integers.
+#
+# Write a script to decide whether the given integers form a linear recurrence
+# of second order with integer factors.
+#
+# A linear recurrence of second order has the form
+#
+#  a[n] = p * a[n-2] + q * a[n-1] with n > 1
+#
+# where p and q must be integers.
+#
+#
+## Example 1
+##
+## Input: @a = (1, 1, 2, 3, 5)
+## Output: true
+##
+## @a is the initial part of the Fibonacci sequence a[n] = a[n-2] + a[n-1]
+## with a[0] = 1 and a[1] = 1.
+#
+## Example 2
+##
+## Input: @a = (4, 2, 4, 5, 7)
+## Output: false
+##
+## a[1] and a[2] are even. Any linear combination of two even numbers with
+## integer factors is even, too.
+## Because a[3] is odd, the given numbers cannot form a linear recurrence of
+## second order with integer factors.
+#
+## Example 3
+##
+## Input: @a = (4, 1, 2, -3, 8)
+## Output: true
+##
+## a[n] = a[n-2] - 2 * a[n-1]
+#
+############################################################
+##
+## discussion
+##
+############################################################
+#
+# We need a solution that allows for
+# p * a[0] + q * a[1] = a[2]
+# p * a[1] + q * a[2] = a[3]
+# p * a[2] + q * a[3] = a[4]
+#
+# A linear combination of a number a in term of two other numbers b and c
+# is possible if gcd(b,c) divides a; however there might be multiple
+# such combinations possible so it is not possible from one triplet of
+# numbers to determine which (if any) linear recurrence works for all
+# 5 numbers. For example (4,2,4): 1*4+0*2=4, 0*4+2*2=4, 2*4-2*2=4,
+# 3*4-4*2=4, -1*4+8*2=4, ...
+# The gcd method however allows to check whether there is any potential
+# solution at all, so let's start with that. Since the gcd of two numbers
+# can slso be combined linearly out of the two numbers I can only assume
+# the task was meant to use that, however in the general case we might have
+# to check any of the (potentially infinitely many) linear combinations for
+# whether or not their factors are suitable for the whole chain of numbers,
+# and it doesn't even hold true for the first example.
+# So we can try to find a few linear combinations from the first triplet,
+# and if any of those works for all numbers we're good.
+# As a heuristic for the range of p and q we use +/- |a|+|b| and try more or
+# less all combinations of these; since we have multiple numbers we just
+# take the maximum of those numbers * 2 instead.
+#
+
+def true():
+    print("Output: true")
+    return True
+
+def false():
+    print("Output: false")
+    return False
+
+def absmax(elems: list):
+    max = abs(elems[0])
+    for elem in elems:
+        if abs(elem) > max:
+            max = abs(elem)
+    return max
+
+def gcd(x: int, y: int) -> int:
+    if x < 0:
+        return gcd(-x, y)
+    if y < 0:
+        return gcd(x, -y)
+    if x < y:
+        return gcd(y, x)
+    z = x % y
+    if z > 0:
+        return gcd(y, z)
+    return y
+
+def linear_recurrence_of_second_order(a: list) -> bool:
+    (i, j, k, l, m) = a
+
+    if k % gcd(i, j) > 0:
+        return false()
+    if l % gcd(j, k) > 0:
+        return false()
+    if m % gcd(k, l) > 0:
+        return false()
+
+    limit = 2 * absmax(a)
+
+    for p in range(-limit, 2*limit):
+        for q in range(-limit, 2*limit):
+            if p*i + q*j == k:
+                if p*j + q*k == l:
+                    if p*k + q*l == m:
+                        return true()
+    return false()
+
+
+linear_recurrence_of_second_order([1, 1, 2, 3, 5])
+linear_recurrence_of_second_order([4, 2, 4, 5, 7])
+linear_recurrence_of_second_order([4, 1, 2, -3, 8])