## Sunday, July 23, 2017

### Solar eclipse 2017

On August 21, 2017, there will be a solar eclipse occurring coast to coast across the continental United States. The last time this happened was almost a century ago in 1918. This has two effects on states that generate a lot of solar energy: the solar energy generated decreases during an Eclipse and people turn on their lights as the sky darkens. Both of these events stress the utility grid.  California's utilities are asking people not to turn on the lights during this event. I am surprised such a short event (and California is not even in the path of totality) can cause concerns for power utilities operators. I wonder if something similar happens when there is no wind in the summer and more people turn on their fans to cool off?

## Monday, July 17, 2017

### Happy Yellow Pig Day!

Today, July 17 is Yellow Pig Day. This "holiday" was created by Princeton math students Michael Spivak and David C. Kelly when they were analyzing the number 17 and wanted to have a day to celebrate this number. This year (2017) is extra special since the year also ends in the digits 17.
Michael Spivak is the author of the wonderful differential geometry texts (published by Publish-or-Perish Press which he founded) which I fondly remember reading in graduate school. He also wrote a popular typesetting book on TeX which I did not read since Leslie Lamport's book on LaTeX makes typesetting papers using LaTeX so easy that I have skipped learning TeX and went straight to using LaTeX (which sits on top of TeX).
If I remember correctly, the typesetting software I used prior to LaTeX was ChiWriter, which is quite easy to use as it is WYSIWYG, so moving to a typesetting paradigm where you relinquish control on the formatting, page breaks, etc. to the software is quite unnerving, but I soon learn to appreciate it as I can focus on the content of the document, rather than spend time improving its appearance.

## Sunday, June 4, 2017

### Moon and Jupiter on June 3, 2017

The Moon and Jupiter are in close proximity (not physically, but as viewed from Earth) on the night of June 3, 2017.

## Sunday, May 21, 2017

### The race of Cloud Computing

I wrote earlier about a news digest app returning supplemental information which is not directly related to the article. Today, I read on the app that the horse "Cloud Computing" has won the Preakness race. And sure enough, there is a box underneath the article with a definition of cloud computing from Wikipedia. What is ironic is that this news summary is probably enabled by leveraging cloud computing. "Cloud Computing" won over "Classic Empire" by coming in from behind and won the race by a head. Perhaps this is an indication how cloud computing will overtake the classical empire of on premise computing😊.

## Thursday, May 11, 2017

### Find interesting properties of your favorite number

I created a simple web application where you can enter a positive integer < 1012 , wait a few moments and it will return a list of interesting properties about this number. Please give it a try and let me know what you think!

## Sunday, May 7, 2017

### Spring is here, pt. 2

House Finch

Chipping Sparrow
My wife says it reminds her of Red from Angry Birds

## Tuesday, May 2, 2017

### 12345

The number 12345 is interesting as it is a sphenic number (product of 3 distinct primes) and its digit reversal 54321 is also a sphenic number.

## Tuesday, April 18, 2017

### Palindromic primes which are sums of 3 consecutive primes

OEIS sequence A113846 lists palindromic primes $q$ which are the sums of 3 consecutive primes $p_1$, $p_2$, $p_3$ such that $p_2$ is also a palindromic prime.

Theorem: All terms of A113846 have an odd number of digits and the first digit (and last digit) is either 3 or 9. In addition, $p_2$ and $p_3$ have the same first digit and $p_2$ and $p_3$ have the same number of digits as $q$. The first digit of $p_2$ (and $p_3$) is equal to the first digit of $q$ divided by 3, i.e. if the first digit of $q$ is $3$, then the first digit of $p_2$ and $p_3$ is $1$.

Proof: First note that all palindromes with an even number of digits is divisible by 11. Let $p_1$, $p_2$, $p_3$ be 3 consecutive primes such that $p_2$ is a palindrome and $q = p_1+p_2+p_3$ is a palindromic prime.  It is clear that $q, p_2 > 11$ and have a odd number of digits since the first term of A113846 is 31513.
The first digit of $p_2$ cannot be 2 since $p_2$ is odd. Next we show that the first digit of $p_2$ is either 1 or 3.
Let $k$ be an even integer such that $p_2$ has $k+1$ digits. Suppose the first digit of $p_2$ is 4 or larger, i.e. $4\cdot 10^k \leq p_2 < 10^{k+1}$. Then by Bertrand's postulate (or Bertrand-Chebyshev theorem), $p_1\geq \frac{p_2}{2}$ and $p_3 \leq 2p_2$. This implies that $10^{k+1} > p_1 \geq 2\cdot 10^k$, $4\cdot 10^k < p3 \leq 2\cdot 10^{k+1}$ and thus $10^{k+1} < q < 4\cdot 10^{k+1}$, i.e. $q$ has an even number of digits, a contradiction.
Next we show that the first digit of $q$ is 3 or 9. To do this, we need a stronger result than Bertrand's postulate for the prime gap. Since $p_2 > 647$, Rohrback and Weis's 1964 result [1] shows that $p_1 \geq \frac{12}{13}p_2$ and $p_3 \leq \frac{14}{13}p_2$.
This shows that $2\frac{12}{13} p_2 \leq q \leq 3\frac{1}{13} p_2$. Since $q$ is prime, it cannot start with the digit 5. If $p_2$ start with the digit 1, then $2\cdot 10^k \leq q \leq 6 \frac{2}{13} 10^k$ and thus $q$ must start with the digit 3.
If $p_2$ start with the digit 3, then $8\frac{10}{13}10^k \leq q \leq 12\frac{4}{13}10^k$. Since $q < 10^{k+1}$, this means that $q$ must start with the digit 9.
Clearly $p_3$ must have the same number of digits as $p_2$ and as $q$.
Finally, we show that $p_3$ has the same starting digit as $p_2$. Suppose $p_2$ starts with the digit 1. If $p_3$ has a different starting digit, this must mean that $p_3 \geq 2\cdot 10^k$, i.e.
$p_2 \geq 1\frac{11}{13}10^k$, $p_1\geq 1\frac{119}{169} 10^k$ and $q \geq 4\cdot 10^k$, a contradiction.
Suppose $p_2$ start with the digit 3. If $p_3$ has a different starting digit, this must mean that $p_3 \geq 4\cdot 10^k$, i.e.
$p_2 \geq 3\frac{9}{13} 10^k$, $p_1 \geq 3\frac{69}{169}10^k$ and $q\geq 10^{k+1}$, a contradiction. QED

This argument also shows that to search for terms, for each $k$, one only needs to consider for $p_2$ the palindromic primes in the ranges $(10^k, 1\frac{7}{19}10^k)$ and $(3\cdot 10^k, 3 \frac{8}{19}\cdot 10^k)$ which amount to approximately $0.79\cdot 10^{\frac{k}{2}}$ palindromes to consider rather than a full search of $10\cdot 10^{\frac{k}{2}}$ palindromes of length $k+1$.

Conjecture: For all terms in A113846, the corresponding primes $p_1$ has the same number of digits and the same first digit as $p_2$ and $p_3$.

References:
1. H. Rohrback and J. Weis, "Zum finiten Fall des Bertrandschen Postulats", J. Reine Angew. Math., 214/215 (1964), 432–440.

## Thursday, March 30, 2017

### Spring is here

It is amazing how a couple of days of sunshine and warmer weather erased most of the evidence that we had a major winter storm 2 weeks ago.