pe2 and css

static/posts/pe-p2/post.typ

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+#let post_slug = "pe-p2"
+#let post_preview_image = "image.png"
+#let post_summary = "Solution to Project Euler Problem 2"
+#let post_date = "2026-07-29"
+
+#set math.mat(delim: "[")
+
+= Project Euler Problem 2
+
+== Statement
+Each new term in the Fibonacci sequence is generated by adding the previous two terms.
+
+By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
+
+== Exploration
+
+
+The even terms in the sequence are:
+#text(fill: white, html.frame(
+$ cal(F) = {0, cancel(1), cancel(1), 2, cancel(3), cancel(5), 8, cancel(13), cancel(21), 34, cancel(55), ...} $
+))
+
+This can be computed quickly with code, some Fibonacci sequence equations would be interesting to derive.
+
+Notice every 3rd number is even, as the previous two numbers are odd, which sums to even. A similar statement can be said about the odd numbers.
+
+The Fibonacci sequence can be generated with the equation:
+#text(fill: white, html.frame(
+$ mat(cal(F)_i; cal(F)_(i+1)) = mat(0, 1; 1, 1) mat(cal(F)_(i-1); cal(F)_i) $
+))
+Then every third number can be defined with the equation:
+#text(fill: white, html.frame(
+$
+&mat(cal(F)_(i+2); cal(F)_(i+3)) = mat(0, 1; 1, 1)^3 mat(cal(F)_(i-1); cal(F)_i)\
+=>&mat(cal(F)_(i+2); cal(F)_(i+3)) = mat(1, 2; 2, 3) mat(cal(F)_(i-1); cal(F)_i)
+$
+))
+Also something interesting I noticed for #text(fill: white, html.frame($cal(F) = {0, 1, 1, 2, 3, 5, 8, ...} (cal(F)_0 = 0, i>= 1)$)) is:
+#text(fill: white, html.frame(
+$
+mat(0, 1; 1, 1)^i = mat(cal(F)_(i-1), cal(F)_(i); cal(F)_(i), cal(F)_(i+1))
+$
+))
+
+Playing with that leads to the similar equation
+
+#text(fill: white, html.frame(
+[$cal(F)_(i+j) = cal(F)_(i) cal(F)_(j+1) + cal(F)_(i-1) cal(F)_(j)$,\ thus $cal(F)_(i+3) = 3cal(F)_i + 2cal(F)_(i-1)$]
+))
+
+An equation to represent this problem is as follows, however this would need a computer to finish in reasonable time. 
+#text(fill: white, html.frame(
+$
+  sum_(i=1)^(n: cal(F)_(3n)<4 times 10^6) cal(F)_(3i)
+$
+))
+
+== Solution
+
+Since every third number is even, take each third element of the sequence until four million. Three step jumps can be taken as defined by the matrix above.
+
+```py
+>>> class Pair:
+...   def __init__(self, fi_m1, fi):
+...     self.fi_m1 = fi_m1
+...     self.fi = fi
+>>> def f_3(pair):
+...   return Pair(
+...     pair.fi_m1 + (2 * pair.fi),
+...     (2 * pair.fi_m1) + (3 * pair.fi)
+...   )
+>>> x = Pair(1, 2)
+>>> s = 0
+>>> while x.fi < 4e6:
+...   s += x.fi
+...   x = f_3(x)
+>>> print(s) 
+4613732
+```