Hi! I'm Tanner Helton. I'm an entrepreneur and developer based in Kansas City. My current focus is on CarePilot, a healthcare AI company that helps providers automate clinical documentation (charting) by using text-to-speech and LLM technology.
Before starting CarePilot, I worked as a Product Management Intern at BacklotCars. I got to work with some amazing people and learned a lot about startups and product management.
I attended the University of Kansas from 2020-2024, where I earned a B.S. in Computer Science, minors in Business and Mathematics, and a certificate in Cybersecurity. Prior to that, I attended Rockhurst High School in Kansas City, MO.
Technologies used: TypeScript, React, Node.js, Express.js, Docker, PostgreSQL, PHP, Laravel, AWS, Azure, OpenAI, LangChain, Python, Git, REST APIs
I co-founded CarePilot with Joseph, Samar, Adam, and John with a vision to reduce the administrative burden on healthcare providers using AI. Our full-stack solution transforms spoken conversations into structured SOAP notes in real time, saving clinicians valuable time. CarePilot now serves providers in over 10 states.
Technologies used: AI, Machine Learning, IoT, Home Security
Secured AI is a venture focused on reinventing home security by leveraging advanced AI and machine learning. It delivers effective, affordable, and convenient security solutions to modernize home protection.
I have always loved math and would love to chat about it.
Here are three classic mathematical problems I enjoy:
The recursive solution can be expressed mathematically as:
\[ T(n) = \begin{cases} 1 & \text{if } n = 1 \\ 2T(n-1) + 1 & \text{if } n > 1 \end{cases} \]
Which simplifies to: \[ T(n) = 2^n - 1 \]
For n people numbered from 1 to n and counting out every k-th person:
\[ J(n,k) = \begin{cases} 1 & \text{if } n = 1 \\ ((J(n-1,k) + k-1) \bmod n) + 1 & \text{if } n > 1 \end{cases} \]
For the special case where k = 2, there's a beautiful binary solution:
\[ J(n,2) = 2L + 1 \]
where L is the number left after writing n in binary and removing the leftmost 1.
A classic impartial combinatorial game involves n piles of objects (e.g., stones), with the i-th pile having \( x_i \) stones. On a player's turn, they may remove any positive number of stones from exactly one pile. The goal is typically to be the player who makes the last move.
The fundamental solution to Nim is expressed through the bitwise xor (also called the "Nim-sum") of the pile sizes:
\[ x_1 \oplus x_2 \oplus \cdots \oplus x_n \]
where \(\oplus\) denotes the bitwise xor operation. A position is a losing position if and only if this Nim-sum is zero. Otherwise, it's a winning position, meaning there is a winning move for the current player.