\n
Let $a_n$ be the smallest integer $n$ such that $f(a_n) = n$. We can calculate the first few values manually.<\/p>\n
\n
$a_1 = 10$. This must be the case since every one-digit number has $f(n)=0$, and $10$ is the smallest two-digit number and $f(10)=1$.\n<\/li>
$a_2 = 19$. We seek the smallest number such that the sum of its digits needs to be summed once more to get to a one-digit number. In other words, we want the smallest number whose digits sum to $10$. This requires at least two digits and can be done in two ways ($19$ and $91$). Using more than two digits leads to larger numbers so we can exclude these possibilities.\n<\/li>
$a_3 = 199$. Now, we seek the smallest number whose digits sum to $19$. This is impossible with two digits (smallest sum is $9+9=18$), so we try with three. This can be done with a $1$ as the first digit ($1+9+9$), so this must be the answer.\n<\/li><\/ul>\n
The pattern is now clear. For all $n\\geq 1$, $a_{n+1}$ is the smallest integer whose digits sum to $a_n$. Based on the pattern, let\u2019s suppose that
\n\\[
\na_n = 1\\,\\underbrace{9\\,9\\,\\cdots\\,9}_{k_n\\text{ times}}
\n\\qquad\\text{for }n=2,3,\\dots
\n\\]where $k_2=1$ and $k_3=2$. Now $a_{n+1}$ is the smallest number whose digits sum to $a_n$. How many digits must $a_n$ have? Rewrite $a_n$ in a particular way:
\n\\begin{align}
\na_n &= 1\\,\\underbrace{9\\,9\\,\\cdots\\,9}_{k_n\\text{ times}}\\\\
\n&= 2\\cdot 10^{k_n}-1 \\\\
\n&= 2\\left( 10^{k_n}-1\\right)+1 \\\\
\n&= 2\\cdot 9\\cdot \\left( \\frac{10^{k_n}-1}{10-1}\\right)+1\\\\
\n&= 2\\cdot 9\\cdot \\left( 1+10+10^2+\\cdots+10^{k_n-1}\\right) +1 \\\\
\n&= 9 \\cdot \\bigl(\\, \\underbrace{2\\,2\\,\\cdots\\,2}_{k_n\\text{ times}}\\,\\bigr) + 1
\n\\end{align}where we used the formula for a finite geometric series<\/a> in the third step. Based on this representation, it is clear that $a_{n+1}$ requires more than $\\underbrace{2\\,2\\,\\cdots\\,2}_{k_n\\text{ times}}$ digits, since even if we made all the digits $9$ it would not be enough. W should add the extra $+1$ digit at the front of the number to get the smallest number possible. This leads to:
\n\\[
\na_{n+1} = 1\\underbrace{9\\,9\\,\\cdots\\,9}_{\\underbrace{2\\,2\\,\\cdots\\,2}_{k_n\\text{ times}}\\text{ times}}
\n\\]So $a_{n+1}$ is of the same form as $a_n$, and we find:
\n\\[
\nk_{n+1} = \\underbrace{2\\,2\\,\\cdots\\,2}_{k_n\\text{ times}}\\qquad\\text{for }k=2,3,\\dots.
\n\\]So although the sequence $f(n)$ seemed to grow at a moderate pace, for $n=0,1,2,3$, we find that
\n\\[
\nf(4) = 19,\\!999,\\!999,\\!999,\\!999,\\!999,\\!999,\\!999
\n\\]That escalated quickly!<\/p>\n
In general, we can write the full recursion as:
\n\\begin{align}
\na_1 &= 10,\\\\
\na_2 &= 19,\\\\
\na_{n+1} &= 2\\cdot 10^{(a_n-1)\/9}-1\\qquad\\text{for }n=2,3,\\dots
\n\\end{align}<\/p>\n
Extra credit<\/h3>\n
The most straightforward way to count how many $n$ between $1$ and $10,\\!000$ satisfy $f(n)=3$ is to write a short script. Here is a two-line script in Julia that does the job:<\/p>\n
\r\nf(n) = n < 10 ? 0 : 1 + f(sum(digits(n)))\r\nlength( [n for n in 1:10000 if f(n) == 3] )\r\n<\/pre>\n
and the answer is 945.\n<\/p><\/div>\n<\/body>","protected":false},"excerpt":{"rendered":"
This week\u2019s Fiddler is a puzzle about adding digits over and over again. For any positive, base-10 integer $n$, define $f(n)$ as the number of times you have to add up its digits until you get a one-digit number. For example, $f(23) = 1$ because $2+3 = 5$, a one-digit number. Meanwhile, $f(888) = 2$, … Continue reading “How many times can you add up the digits?”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"_uf_show_specific_survey":0,"_uf_disable_surveys":false,"footnotes":""},"categories":[42],"tags":[37,43,17,15],"class_list":["post-3964","post","type-post","status-publish","format-standard","hentry","category-the-fiddler","tag-coding","tag-fiddler","tag-induction","tag-recursion"],"aioseo_notices":[],"aioseo_head":"\n\t\t\n\t\n\t\n\t\n\t\n\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t