0

The number $C(n)$ , where $n$ is a positive integer is defined as follows : Write the numbers $j$ from $1$ to $n$ $j$ times. The resulting number is $C(n)$. The letter $C$ comes from the name I invented for these numbers : "Crazy Numbers"

Two examples ($C(8)$ and $C(10)$) including the PARI-routine calculating $C(n)$ :

? s=0;n=8;for(j=1,n,for(k=1,j,s=s*10^length(digits(j))+j));print(s)
122333444455555666666777777788888888
? s=0;n=10;for(j=1,n,for(k=1,j,s=s*10^length(digits(j))+j));print(s)
12233344445555566666677777778888888899999999910101010101010101010
?

Here :

How many digits does $C(n)$ have?

I asked for the number of digits of $C(n)$ (for the definition of $C(n)$, see the link in the linked question), which I finally found out. Another user asked for the digit sum of $C(n)$, and I only can say, if we denote $d(n)$ to be the digit sum of $n$, the digit sum of $C(n)$ is given by $$\sum_{j=1}^n j\cdot d(j)$$

If there would be a formula for $$\sum_{j=1}^n d(j)$$ we could derive the desired digit sum from that.

Does anyone know a concrete formula for $$\sum_{j=1}^n d(j)$$ where $d(n)$ is the digit sum of $n$ ?

Peter
  • 84,454
  • https://math.stackexchange.com/questions/2374537/are-there-prime-numbers-cn-or-rcn?noredirect=1&lq=1 – Peter Aug 03 '17 at 16:12
  • Please put all relevant definitions in your question. Don't make readers click through, not once but twice, just to be able to understand the quantity in the title of the question. – Greg Martin Aug 03 '17 at 17:01
  • https://math.stackexchange.com/questions/817038/sum-of-digits-of-number-from-1-to-n?rq=1 – Peter Aug 03 '17 at 17:45

1 Answers1

-1

Consider the digit sum of $C(8) = 122333444455555666666777777788888888$.

We add one copy of $1$ to two copies of $2$ to three copies of $3$ and so on. . . .

In other words $1 + 2^2 + 3^2 + 4^2 + \ldots + 8^2 = \displaystyle \sum_{m=1}^8 m^2$.

In general $C(n)$ has digit sum $\displaystyle \sum_{m=1}^n m^2$.

And there's a formula for that.

The digit sum of $C(n)$ is equal to $\displaystyle \frac{n(n+1)(2n+1)}{6}$

Daron
  • 10,300