# Special properties of the number $146$

I'm a math teacher. Next week I'll give a special lecture about number theory curiosities. It will treat special properties of numbers — the famous story with Ramanujan, taxicab numbers, later numbers divisible by all their digits, etc.

I was given class number $146$ for the lecture and I think it would be fine to start with a special property of our class's number. Ramanujan would surely find something at once, but I can't. Do you see any special properties of $146$?

Here are some of my observations, but these properties are not very special:

• $146$ is a semiprime number (product of two distinct primes), while the reversal $641$ is prime.

• $146 = 4^3 + 4^3 + 3^2 + 3^2$.

Here is a very similar question, just to show what kind of question this is and what kind of answers I would like to see.

• Wikipedia has some additional properties listed. – Stefan Mesken Jun 9 '18 at 20:28
• It is untouchable. – lulu Jun 9 '18 at 20:32
• Perhaps $146=5^2+11^2=3^2+4^2+11^2=1^2+3^2+6^2+10^2$. Both $2\cdot 146+1$ and $146^2+1$ are prime – Hagen von Eitzen Jun 9 '18 at 20:35
• OK, than "The 51st zero of $\zeta(\frac12+it)$ is near $t=146$" is probably beyond reach ... – Hagen von Eitzen Jun 9 '18 at 20:40
• I regret to inform that there is no entry for $146$ in my copy of The Penguin Dictionary of Curious and Interesting Numbers. The closest is $145$, noted as the sum of the factorials of its digits. – Joffan Jun 9 '18 at 20:50

## 19 Answers

The $146$ is the smallest number that can be expressed as a sum of three squares in exactly five ways:

\begin{align} 146 &= 0^2 + 5^2 + 11^2\\ &= 1^2 + 1^2 + 12^2\\ &= 1^2 + 8^2 + 9^2\\ &= 3^2 + 4^2 + 11^2\\ &= 4^2 + 7^2 + 9^2\\ \end{align}

$146$ can be written as squares of two primes: $$146=5^2+11^2=5^2+(1+4+6)^2=(1+4)^2+(1+4+6)^2.$$

• $137$ is another example. $137=(1+3)^2+(1+3+7)^2$, but what inside are not primes. – poyea Jun 9 '18 at 20:46
• But 137 is for physics ... – Hagen von Eitzen Jun 9 '18 at 21:05
• @HagenvonEitzen "Okay, guys, let's learn some Physics today." – poyea Jun 9 '18 at 21:13
• Oh, $1-4-6-4-1"$ appears on the fifth row of Pascal's triangle! – poyea Jun 11 '18 at 17:03

The Postmaster General has decided that only three different stamp denominations shall be produced and also that one may stick at most ten stamps on an envelope. Also, it shall be possible to stick stamps of any total value from $1$ cent, $2$ cents, ..., up to $N$ cents (inclusive). Of course, the value of $N$ depends on the three stamp denominations. What is $N$ if the denominations are $1$ cent, $10$ cents, $15$ cents? Who can find a better choice of stamp denominations? What is the best choice of stamp denominations and what is $N$ for that choice?

The answer to the last question id $N=146$.

Interesting question... Here are a few facts about $146$ I just found...

• $146 = (1^3 - 1) - 4^3 + (6^3 - 6)$
• $(1^2 + 4^2 + 6^2) + (1+4+6) = 4^3$.
• $641 - 146$ is divisible by $1+4+6$
• The sum of the sum of digits of $146^1, 146^4, 146^6$ is $2^7-1$.
• The sum of the product of the digits of $146^2$ and $146^3$ is $12^2$, which is $146 - 2$.
• The digits, with repetition, of $146^2$ are all contained in the digits of $146^3$.
• $\underbrace{\color{blue}{11^2 + 44^2 + 66^2}}_{\color{red}{3\text{ terms}}} = \color{blue}{641}\color{red}{3}$
• Worth noting that the specialness of all these properties is tied to a base 10 representation , unlike many of the answers to the linked question about 50. – Eric Jun 10 '18 at 7:43
• Indeed, the second one of these is not even to do with the number $146$ at all. Nor is the last. – John Gowers Jun 10 '18 at 16:46
• @Gowers That's not true. The second one has the property(when in base $10$) that for number $\overline{abc}$, $(a^2 + b^2 + c^2) - (a+b+c) = b^3$. The last one has the obvious(when in base $10$) property that for number $\overline{abc}$, $\overline{aa}^2 + \overline{bb}^2 + \overline{cc}^2 = \overline{cba3}$ – Jeffery Opoku-Mensah Jun 10 '18 at 17:28

Two nice facts about sums of cubes:

1. We can write $146$ as $$4 \cdot 1^3 + 3 \cdot 2^3 +2 \cdot 3^3 + 1 \cdot 4^3.$$ The previous term in this sequence is $$3 \cdot 1^3 + 2 \cdot 2^3 + 1 \cdot 3^3 = 46.$$
2. $146$ is the second-largest integer that can't be written as the sum of cubes bigger than $1$. (The largest is $154$.)

Also, http://oeis.org/A134907 gives us a really weird formula for $146$: $$146 = \left\lfloor 5 e^{-\tan 5}\right\rfloor.$$

From http://oeis.org/A172877, we see that there are exactly $146$ $4 \times 2$ matrices such as $$\begin{bmatrix}1 & 4 \\ 2 & 3 \\ 3 & 2 \\ 4 & 1\end{bmatrix}$$ with nonnegative integer entries in which each row sums to $5$ and each column sums to $10$.

• $1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 3^3=46$ – James Arathoon Jun 11 '18 at 15:19
• Don't you mean $$4\times1^3+3\times2^3+2\times3^3+1\times4^3=146$$ – James Arathoon Jun 11 '18 at 15:22
• Nice catch, thanks! – Misha Lavrov Jun 11 '18 at 15:23
• 1. is a corollary of the fact that 146 is the sum of the first four triangle numbers squared as pointed out by @Hagen von Eitzen in a comment at the top. – James Arathoon Jun 11 '18 at 15:47
• plus the fact that the sum of the first n cubes is the nth triangle number squared. – James Arathoon Jun 11 '18 at 15:58

Other than $1$, $10$ and $100$ which may be considered trivial cases, $146$ is the smallest positive integer $n$ such that, ignoring sequence and repetitions, $n^2$ and $n^3$ contain exactly the same digits:

$$146^2 = 21316$$

$$146^3 = 3112136$$

This is readily verified by reviewing the values of $n^2$ and $n^3$ for $n = 2,3,...,145$.

Addendum 11 June 2018

In fact this is the only non-trivial case of a positive integer with this property for $n < 1000$.

$$1^1 + 3^4 + 2^6 = 146$$ Note that if you swap the $2$ and $3$ for an even more "pleasing" expression, you have $$1^1 + 2^4 + 3^6 = 746 = 1^7 + 2^4 + 3^6$$ Curiously the "46"ness doesn't change.

• $1!+4!+6!=746-1$..hmm, nearly – Daniel Buck Jun 11 '18 at 20:24
• Also $1!+4!+5!=146-1$..hmm, nearly – Daniel Buck Jun 11 '18 at 20:37
• @DanielBuck So $1!+4!+(6-1)! = 146-1$ – Bram28 Jun 12 '18 at 14:58
• @Bram28 Hah-nice! – Daniel Buck Jun 12 '18 at 15:00
• @DanielBuck I made it an Answer ... but feel you should get the credit for that! – Bram28 Jun 12 '18 at 15:01

Whichever properties you choose to illustrate, in the end you can conclude by a reflexion on how likely it is that one will find "amazing" properties of any given number, as the human mind is craving for entertainment and amazement and will consider any kind of property mixing up elementary operations and digit breaking as acceptable.

Then ask your students which of the previously discussed properties they feel are actually, mathematically, interesting.

• In fact: every number has some 'amazing' property, because if not, there would be a smallest number without any 'amazing' properties ... which would make it 'amazing' :) – Bram28 Jun 12 '18 at 15:06

$$(222)_8=(146)_{10}$$ $$=2\cdot8^2+2\cdot8^1+2\cdot8^0$$ $$=2\cdot64+2\cdot8+2\cdot1$$ $$=128+16+2=146$$ This is the decimal equivalent of the octal number $222$

• And on top of all that $2\times 2\times 2$ is precisely equal to $8$, while $1\times 4\times 6$ is... wait... nevermind. – Arnaud Mortier Jun 10 '18 at 0:48

146 is the hypotenuse of a Pythagorean triple.

Here is a number pyramid: \begin{align}35^1+36^1+37^1+38^1=&146\\ 35^3+36^3+37^3+38^3=&146\cdot 1336 \\ 35^5+36^5+37^5+38^5=&146\cdot 1791556 \\ 35^7+36^7+37^7+38^7=&146\cdot 2411307676 \\ \vdots \\ 35^{2n+1}+36^{2n+1}+37^{2n+1}+38^{2n+1}=&146\cdot f(2n+1)\end{align} Exercise: Find a) $f(9)$; b) $f(2n+1)$.

A reasonably simple pattern is $$1^4+4\times2^4+3^4=146$$

Also I found that $$\sum_{k=1^2}^{2^2}k+\sum_{k=1^2}^{4^2}k=146$$ and that by adding $1$ to all the exponents we have $$\sum_{k=1^3}^{2^3}k^2+\sum_{k=1^3}^{4^3}k^2=146\times614$$ which is a multiple of 146; with the multiplier $614$ being a rearrangement of $146$.

With $\frac{614}{146}=\frac{307}{073}$; the numerator $307$ being a rearrangement of $073$ (the largest prime factor of $146$), itself being the largest prime factor of $614$.

Inspired by a comment by DanielBuck:

$$1!+4!+(6-1)!=146-1$$

• I think it’s lovely! – Daniel Buck Jun 12 '18 at 15:27
• Of course also $147-2 = 1!+4! + (7-2)!$, $148-3 = 1!+4! + (8-3)!$ and $149-4 = 1!+4! + (9-4)!$ :-) – Joffan Jun 13 '18 at 20:29
• @Joffan Not forgetting $1!+4!+(5-0)!=145-0$ – Daniel Buck Jun 13 '18 at 20:33
• @DanielBuck or indeed $144+1 = 1!+4! + (4+1)!$ etc. – Joffan Jun 13 '18 at 20:35
• @Joffan Cool!! Factorial and out – Daniel Buck Jun 13 '18 at 20:41

I am surprised how this has not been included in the answers so far and interesting story about 641 from the history and pioneers of number theory.

Fermat conjectured that all numbers of the form $2^{2^n}+1$ are primes and this conjectures was found to be true for $n = 1,2,3$. For $n = 5$, $2^{2^5}+1 = 4294967297$ was too big for him to check if it was a prime. A century later, Euler proved that $2^{2^5}+1$ is divisible by 641. In fact we have a much stronger result.

641 divides all the Generalised Fermat numbers of the form $$(2^a*5^b)^{2^5} + 1$$

where $a$ is odd and $b$ is even, or a is even and b is odd.

Representations in digits $1,...,9$, ascending and descending, using only plus and minus signs:

$$123+45+67-89=146$$

$$-98-76-5+4+321=146$$

And also, with the digits not in sequence but using only multiplication, division and exponentiation:

$$\frac{876 \times 3 \times 9^{1/2}}{54}=146$$

• $831$ of the $1000$ integers between $0$ and $999$ have representations like this. So this is an example of a property that looks special, when actually not having it would be more surprising. (The smallest positive integer with no such representation is $211$.) – Misha Lavrov Jun 10 '18 at 15:46
• That's for the ascending representations. Only $27$ integers between $0$ and $999$ have neither representation, and $772$ have both. – Misha Lavrov Jun 10 '18 at 15:52
• @MishaLavrov What exactly do you mean by "like this" (for the examples with plus and minus signs)? Perhaps I'm wrong, but I considered that representations which group the 9 digits into no more than 4 integers (ascending) or 5 integers (descending) would have extra scarcity value. – Adam Bailey Jun 10 '18 at 16:07
• Indeed I included representations with any number of groups. There are 76 numbers which have both representations of at least the same "quality" as 146. Of course, the gold standard is $12 = 123 + 45 - 67 - 89 = 987-654-321$, and $201 = 12+345-67-89 = -98-76+54+321$ is one of the $22$ integers that are almost equally good, and get a silver medal. (I give $146$ a bronze.) – Misha Lavrov Jun 10 '18 at 16:28

Expanding on lulu's comment, the $146$ is the smallest Untouchable Semiprime - see https://oeis.org/A119379.

Untouchable number - Number that cannot be expressed as the sum of all the proper divisors of any positive integer (including itself).

Semiprime - Number that is the product of two primes.

146 in base 7,13,19,31...$p \equiv 1 \pmod 6$ Is prime for the first few primes

http://oeis.org/A056899

My first thought was to look at Erich Friedman's What's Special About This Number? There I find that 146 is 222 in octal. I'm sorry, I don't find that all that special.

My second thought was to look in the OEIS. I ran the search 146 keyword:nice. This gave me 211 results. I only looked at the first ten.

Aside from the ones that have already been mentioned here, the most interesting one, in my opinion, is A003714, the Fibbinary numbers. In binary, 146 is 10010010, which you could have already figured out from the octal representation.

In Fibonacci base (Zeckendorf), 146 is 10000000010 (I hope I have the right number of zeroes there). Also no consecutive ones.

The sum of the first $n$ cubes, starting from $1^3$, is given by the square of the $n$th Triangular number, $T_n$:

\begin{align} 1^3 &= & 1^2 &= & T_{1}^2\\ 1^3+2^3 &= & 3^2 &= & T_{2}^2 \\ 1^3+2^3+3^3 &= & 6^2 &= & T_{3}^2 \\ 1^3+2^3+3^3+4^3 &= & 10^2 &= & T_{4}^2 \\ \end{align}

Noting this we can write the sum of the first $n$ squares of the Triangular numbers as: \begin{align*} \sum_{i=1}^{n}T_i^2&=T_1^2+T_2^2+\dotsb+T_n^2\tag{1}\\ &=n\times1^3+(n-1)\times2^3+\dotsb+1\times n^3\\ &=\sum_{k=1}^{n}(n-k+1)k^3\tag{2} \end{align*} Note, taking $n=4$ in $(1)$ gives $$T_{1}^2+T_{2}^2+T_{3}^2+T_{4}^2=1^2+3^2+6^2+10^2=146$$ and taking $n=4$ in $(2)$ gives $$4\times1^3+3\times2^3+2\times 3^3+1\times 4^3=146$$

As a consequence of $(1)$ another nice pattern emerges: $$\frac{1}{4}\left[2^2(1^2+3^2)+4^2(3^2+5^2)\right]=146$$