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   "source": [
    "# [Counting Fractions](https://projecteuler.net/problem=72)\n",
    "\n",
    "Like [problem 71](https://projecteuler.net/problem=71), we're looking at [Farey sequences](https://en.wikipedia.org/wiki/Farey_sequence). This time we're interested in the cardinality of $F_{1000000}$.\n",
    "\n",
    "To begin, first note that $F_1 = \\{0, 1\\}$, so $|F_1| = 2$ (this problem isn't counting 0 and 1 in its totals - we'll handle that at the end). Then consider that for any Farey sequence $F_n$, the next sequence $F_{n+1}$ will contain all the terms from $F_n$, along with all irreducible fractions $\\frac{k}{n+1}$, (since any *reducible* fraction would already be in $F_n$).\n",
    "\n",
    "How many new fractions does this get us? Well, the fraction only reduces if $k$ and $n+1$ have a common factor - in other words, if $k$ and $n+1$ are coprime, the fraction will not reduce. How many number less than $n+1$ are coprime to $n+1$? The [totient function](https://en.wikipedia.org/wiki/Euler%27s_totient_function) will tell us! So the number of irreducible fractions with denominator $n+1$ is simply $\\phi(n+1)$. This gives us\n",
    "$$|F_{n+1}| = |F_n| + \\phi(n+1)$$\n",
    "From this, we can derive a non-recursive formula:\n",
    "$$|F_n| = 1 + \\sum_{k=1}^n \\phi(k)$$\n",
    "\n",
    "The [sum of totients](https://en.wikipedia.org/wiki/Totient_summatory_function) is denoted $\\Phi(n)$.\n",
    "As mentioned before, we'll actually subtract two from this total, since the problem isn't counting 0 or 1. So this problem boils down to computing $\\Phi(1000000) - 1$.\n",
    "\n",
    "## Using SageMath\n",
    "\n",
    "Since SageMath implements a decently fast totient function, we could opt for the most straightforward approach."
   ]
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  {
   "cell_type": "code",
   "execution_count": 1,
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       "303963552391"
      ]
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     "execution_count": 1,
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   ],
   "source": [
    "sum(euler_phi(n) for n in range(1, 1000001)) - 1"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "8e268820",
   "metadata": {},
   "source": [
    "If you try to implement the totient function yourself, you might find it difficult to make this approach fast enough. An alternative is to \"sieve\" the values of totient - see [problem 70](https://projecteuler.net/problem=70). However, there's a few more *very* interesting methods to compute totient sums.\n",
    "\n",
    "## Recursive totient sum\n",
    "\n",
    "You might think a fast totient function is key to solving this problem quickly, but it turns out there's a crazy recursive formula for $\\Phi(n)$ that does not require implementing the totient function at all! Since it's a little hard to believe it works, here's a derivation of it.\n",
    "\n",
    "We start with the following identity, proven by Gauss:\n",
    "$$\\sum_{d|k}\\phi(d) = k$$\n",
    "\n",
    "Then, we can apply the formula for [triangular numbers](https://en.wikipedia.org/wiki/Triangular_number):\n",
    "$$\\sum_{k=1}^n \\sum_{d|k}\\phi(d) = \\sum_{k=1}^n k = \\frac{n(n+1)}{2}$$\n",
    "\n",
    "Now we'll rewrite the bounds.\n",
    "$$\\sum_{1 \\leq k \\leq n} \\sum_{xy=k}\\phi(x) = \\frac{n(n+1)}{2}$$\n",
    "\n",
    "This makes it easier to see that we can condense the two nested summations into one.\n",
    "$$\\sum_{1 \\leq xy \\leq n} \\phi(x) = \\frac{n(n+1)}{2}$$\n",
    "\n",
    "We can then split this single summation into a sum-of-summations, letting $y$ vary from 1 to $n$.\n",
    "$$\\sum_{1 \\leq x \\leq n} \\phi(x) + \\sum_{1 \\leq 2x \\leq n} \\phi(x) + \\sum_{1 \\leq 3x \\leq n} \\phi(x) + \\cdots + \\sum_{1 \\leq xn \\leq n} \\phi(x) = \\frac{n(n+1)}{2}$$\n",
    "\n",
    "If we apply the definition of $\\Phi(n)$, we get\n",
    "$$\\Phi\\left(\\left\\lfloor\\frac{n}{1}\\right\\rfloor\\right) + \\Phi\\left(\\left\\lfloor\\frac{n}{2}\\right\\rfloor\\right) + \\Phi\\left(\\left\\lfloor\\frac{n}{3}\\right\\rfloor\\right) + \\cdots + \\Phi\\left(\\left\\lfloor\\frac{n}{n}\\right\\rfloor\\right) = \\frac{n(n+1)}{2}$$\n",
    "\n",
    "Then we can rearrange and simplify to get\n",
    "$$\\Phi(n) = \\frac{1}{2}n(n+1) - \\sum_{d=2}^n \\Phi\\left(\\left\\lfloor \\frac{n}{d} \\right\\rfloor\\right)$$\n",
    "\n",
    "Not a super intuitive definition for $\\Phi(n)$, but pretty simple to implement, and surprisingly quick!"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "id": "858a64d2",
   "metadata": {},
   "outputs": [],
   "source": [
    "from functools import cache\n",
    "\n",
    "@cache\n",
    "def totient_sum(n):\n",
    "    if n == 1:\n",
    "        return 1\n",
    "    \n",
    "    return n*(n+1)/2 - sum(totient_sum(n//d) for d in range(2, n + 1))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "id": "3f21311a",
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    {
     "data": {
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       "303963552391"
      ]
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     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "totient_sum(1000000) - 1"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "6f1085a3",
   "metadata": {},
   "source": [
    "Both of the above strategies are simple and fast enough to get the job done on a modern computer, but it turns out we can go even deeper. But first, we need to become familiar with some important formulas.\n",
    "\n",
    "## Dirichlet convolution\n",
    "\n",
    "The [Dirichlet convolution](https://en.wikipedia.org/wiki/Dirichlet_convolution) of two [arithmetic functions](https://en.wikipedia.org/wiki/Arithmetic_function) $f$ and $g$ is\n",
    "$$(f * g)(n) = \\sum_{xy=n} f(x) g(y)$$\n",
    "\n",
    "Several important mathematical functions have some representation as a Dirichlet convolution - see the above page for a list. An important one for our purposes is $\\phi = \\mathrm{Id} * \\mu$, where $\\mathrm{Id(n) = n}$ is the [identity function](https://en.wikipedia.org/wiki/Identity_function) and $\\mu$ is the [Möbius function](https://en.wikipedia.org/wiki/M%C3%B6bius_function) - if you've never heard of the latter, don't worry about it yet.\n",
    "\n",
    "## Dirichlet's hyperbola method\n",
    "By applying the above Dirichlet convolution to\n",
    "$$\\Phi(n) = \\sum_{k=1}^n \\phi(n)$$\n",
    "we can derive an equivalent summation:\n",
    "$$\\Phi(n) = \\sum_{k=1}^n \\sum_{xy=k} x \\mu(y)$$\n",
    "\n",
    "We're going to transform this summation by applying \n",
    "[Dirichlet's hyperbola method](https://en.wikipedia.org/wiki/Dirichlet_hyperbola_method). Let $1 < a < n$, and let $b = n / a$. Then\n",
    "$$\\Phi(n) = \\sum_{x=1}^a \\sum_{y=1}^{n/x} x \\mu(y) + \\sum_{y=1}^b \\sum_{x=1}^{n/y} x \\mu(y) - \\sum_{x=1}^a \\sum_{y=1}^b x \\mu(y)$$\n",
    "\n",
    "Then with some algebra, we can transform this into\n",
    "$$\\Phi(n) = \\sum_{x=1}^a x M\\left(\\left\\lfloor \\frac{n}{x}\\right\\rfloor\\right) + \\sum_{y=1}^b \\mu(y) \\frac{\\left\\lfloor \\frac{n}{y} \\right\\rfloor \\left(\\left\\lfloor \\frac{n}{y}\\right\\rfloor + 1\\right)}{2} - M(\\lfloor b \\rfloor) \\frac{\\lfloor a \\rfloor(\\lfloor a\\rfloor +1)}{2}$$\n",
    "where $M(n)$ is the [Mertens function](https://en.wikipedia.org/wiki/Mertens_function), which is just the partial sums of the Möbius function:\n",
    "$$M(n) = \\sum_{k=1}^n \\mu(k)$$\n",
    "\n",
    "By using this formula for $\\Phi(n)$, instead of calculating $n$ different totients, our sums only run to $a$ and $b$.\n",
    "\n",
    "However, for this formula to be effective, we also need an efficient way to calculate $M(n)$. Fortunately, we can just apply the hyperbola method again, with the convolution $\\epsilon = \\mu * 1$, where $\\epsilon$ is the [unit identity](https://en.wikipedia.org/wiki/Unit_function). Once again, we choose $1 < a < n$ and let $b = n/a$.\n",
    "\n",
    "$$\\sum_{k=1}^n \\epsilon(k) = \\sum_{k=1}^n \\sum_{xy=k} \\mu(x)$$\n",
    "\n",
    "$$1 = \\sum_{x=1}^a \\sum_{y=1}^{n/x} \\mu(x) + \\sum_{y=1}^b \\sum_{x=1}^{n/y} \\mu(x) - \\sum_{x=1}^a \\sum_{y=1}^b \\mu(x)$$\n",
    "\n",
    "$$1 = \\sum_{x=1}^a  \\mu(x)\\left\\lfloor \\frac{n}{x}\\right\\rfloor + \\sum_{y=1}^b M\\left(\\left\\lfloor \\frac{n}{y}\\right\\rfloor\\right) - \\lfloor b \\rfloor M(\\lfloor a\\rfloor)$$\n",
    "\n",
    "$$M(n) = 1 + \\lfloor b\\rfloor M(\\lfloor a\\rfloor) - \\sum_{x=1}^a  \\mu(x)\\left\\lfloor \\frac{n}{x}\\right\\rfloor - \\sum_{y=2}^b M\\left(\\left\\lfloor \\frac{n}{y}\\right\\rfloor\\right)$$\n",
    "\n",
    "Now we have a recursive implementation of $M(n)$. All that's left is to calculate the values of $\\mu(n)$ that we need. By taking advantage of $\\mu$ being [multiplicative](https://en.wikipedia.org/wiki/Multiplicative_function), we can compute values using a similar strategy to the sieve of Eratosthenes - see [problem 10](https://projecteuler.net/problem=10)."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "id": "48c9275b",
   "metadata": {},
   "outputs": [],
   "source": [
    "def mobius_range(limit):\n",
    "    ms = [n for n in range(0, limit)]\n",
    "\n",
    "    for n in range(0, limit):\n",
    "        if n == 0 or n == 1 or ms[n] != n:\n",
    "            yield ms[n]\n",
    "            continue\n",
    "           \n",
    "        yield -1\n",
    "        \n",
    "        for k in range(2 * n, limit, n):\n",
    "            if k % n ^ 2 == 0:\n",
    "                ms[k] = 0\n",
    "\n",
    "            ms[k] //= -n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "f12a8b30",
   "metadata": {},
   "source": [
    "We'll use $a = \\sqrt{1000000} = 1000$ as our upper bound."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "id": "c2071ba1",
   "metadata": {},
   "outputs": [],
   "source": [
    "from math import isqrt\n",
    "mu = list(mobius_range(isqrt(1000000) + 1))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a8db5c70",
   "metadata": {},
   "source": [
    "Now we can implement $M(n)$ with our recursive formula:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "id": "f04ed69a",
   "metadata": {},
   "outputs": [],
   "source": [
    "@cache\n",
    "def M(n):\n",
    "    if n == 0 or n == 1:\n",
    "        return n\n",
    "    \n",
    "    a = sqrt(n)\n",
    "    b = n / a\n",
    "    \n",
    "    total = 1 + floor(b) * M(floor(a))\n",
    "    total -= sum(mu[x] * floor(n/x) for x in range(1, floor(a) + 1))\n",
    "    total -= sum(M(floor(n/y)) for y in range(2, floor(b) + 1))\n",
    "    \n",
    "    return total"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "132dad8d",
   "metadata": {},
   "source": [
    "And finally, yet another implementation of $\\Phi(n)$:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "id": "394d604a",
   "metadata": {
    "scrolled": true
   },
   "outputs": [],
   "source": [
    "def totient_sum(n):\n",
    "    a = sqrt(n)\n",
    "    b = n / a\n",
    "\n",
    "    total = sum(x * M(floor(n/x)) for x in range(1, floor(a) + 1))\n",
    "    total += sum(polygonal_number(3, floor(n/y)) * mu[y] for y in range(1, floor(b) + 1))\n",
    "    total -= M(floor(b)) * polygonal_number(3, floor(a))\n",
    "\n",
    "    return total"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "e59a4350",
   "metadata": {},
   "source": [
    "This approach is significantly faster than the others."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "id": "55f0719b",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "303963552391"
      ]
     },
     "execution_count": 8,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "totient_sum(1000000) - 1"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "1c67d8da",
   "metadata": {},
   "source": [
    "## Relevant sequences\n",
    "* Cardinalities of Farey sequences: [A005728](https://oeis.org/A005728)\n",
    "* Partial sums of totient function: [A002088](https://oeis.org/A002088)\n",
    "\n",
    "#### Copyright (C) 2025 filifa\n",
    "\n",
    "This work is licensed under the [Creative Commons Attribution-ShareAlike 4.0 International license](https://creativecommons.org/licenses/by-sa/4.0/) and the [BSD Zero Clause license](https://spdx.org/licenses/0BSD.html)."
   ]
  }
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