# Sum of the digits in base $p+1$

Definition

Let $$W$$ be the function , defined as $$W(a,b)=r$$

given $$a,b\in \mathbb{Z_+}$$ and $$a>1$$

Take $$m$$ to be the integer s.t. $$a^{m+1} \ge b > a^{m}$$, i.e. $$m = \lceil \log{b}/\log{a} \rceil - 1$$.

Convert number $$a^{m+1} - b$$ in base $$a$$ and add it's digits

$$a^{m+1} - b = (r_{l} r_{l-1} ... r_{1} r_{0})_{a}$$

Where $$r=\sum_{i=0}^{l}r_{i}$$

Example

$$W(5,77)=8$$

Identity$$1$$

if $$W(a,b)=r$$ then $$b+r\equiv 1($$ mod $$a-1)$$

$$S$$ is a function defined as

$$S(a,n)=\sum_{i=1}^{a}i^{n}$$

Where $$a$$ and $$n$$ are positive integer.

Let $$p$$ is prime and $$p+1=z$$

Question

show that

If $$z>2n+2$$ Then $$W(z,W(z,S(z,2n)))=z$$

Example

Let $$n=1$$ here, choose any $$z>4$$

Let $$z=6$$

So $$W(6,W(6,S(6,2)))=W(6,W(6,91))=W(6,10)=6$$

Python programming for calculate $$W$$ function

n1=5
n2=77
rem_array = []
while n2 != 1:
mod = n2%n1
if mod != 0:
rem = n1-mod
n2 = n2 + rem
rem_array.append(round(rem))
n2=n2/n1
else:
n2 = n2/n1
rem_array.append(0)
print(rem_array[::-1])
print(sum(rem_array))



Proof for, if $$p>n+1$$ then $$p|S(p,n)$$

Formula

$$S(a,n)= \sum_{i=1}^{a} i^{n}=\sum_{b=1}^{n+1} \binom{a}b\sum_{j=0}^{b-1} (-1)^{j}(b-j)^{n}\binom{b-1}j$$

for formula

Proof

Let $$a=p(prime)>n+1$$

We can see, $$a$$ can be common out from $$\sum_{b=1}^{n+1}\binom{a}b\sum_{j=0}^{b-1} ...$$

$$\implies a|S(a,n)$$

Proof for, If $$p|S(p,2n)$$ Then $$W(z,W(z,S(z,2n)))=(z-1)r+1=pr+1$$

Proof

See $$S(z,2n)=pr_1+1$$

$$\implies W(z,W(z,S(z,2n)))$$ $$\ \ \ by\ identity1$$

$$=W(z,W(z,pr_1+1))$$

$$=W(z,pr_2)$$

$$=pr+1=(z-1)r+1$$

For some $$r,r_1,r_2\in\mathbb{Z}$$

I believe $$r$$ is always $$1$$ for all $$z>2n+2$$, that's my question.

Related questions

To count such $$p$$ which $$p\nmid S(p,2n)$$

Special observation on prime number and π(n)