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For non-negative Integers n,k with n>1

Let $S_{k}(n) = n^{k+1}-(1^k+2^k+...+n^k)$

Converting $S_{k}(n)$ in base $n$ as

$S_{k}(n) = (r_k r_{k-1} ... r_1 r_0)_n$

Definition . For $k ≥ 0$, the $k$th power-sum denominator is the smallest positive integer $d_k$ such that$ d_k · (1^k + 2^k + · · · + x^ k) $is a polynomial in $x$ with integer coefficients. The first few values of $d_k$ (see , Sequence A064538) are $d_k = 1, 2, 6, 4, 30, 12, 42, 24, 90, 20, 66, 24, 2730, 420, 90, 48, 510, . . . .$

Let $r\in \mathbb{N}$ and $0\leq j<d_k$

Define $n = rd_k+j$

$S_k(rd_k+j)=(r_k r_{k-1} ... r_1 r_0)_{rd_k+j}$

Let $r'= r+1$ and $r''= r'+1$

$S_k(r'd_k+j)=(r'_k r'_{k-1} ... r'_1 r'_0)_{r'd_k+j}$

$S_k(r''d_k+j)=(r''_k r''_{k-1} ... r''_1 r''_0)_{r''d_k+j}$

Where $r_i,r'_i,r''_i $ are digits for $0\leq i \leq k$

Show that $r''_i-r'_i= r'_i-r_i \forall r,k\in\mathbb{N}$

Example

Let $k=1, j=0 $

$$S_1( 2)=2^2-(1+2)=1=(0 ,1)_2$$

$$S_1( 4)=4^2-(1+2+3+4)=10=(1, 2)_4$$

$$S_1( 6)=6^2-(1+...+6)=15=(2 ,3)_6$$

$$S_1( 8)=8^2-(1+...+8)=28=(3, 4)_8$$

$$S_1( 10)=10^2-(1+...+10)=45=(4, 5)_{10}$$

$$S_1( 12)=12^2-(1+...+12)=66=(5, 6)_{12}$$

$$\vdots$$

Now$k=1, j=1 $

$$S_1( 3)=3^2-(1+2+3)=3=(1,0)_3$$

$$S_1( 5)=10=( 2,0)_5$$

$$S_1( 7)=21=(3,0)_7$$

$$S_1( 9)=36=(4,0)_9$$

$$S_1( 11)=55=(5,0)_{11}$$

$$\vdots$$

We can see Digit progression in arithmetic

Example2

Similarly when $k=2,j=2$

$$S_2( 8)=308=(4, 6,4)_8$$

$$S_2( 14)=(8,11,7)_{14}$$

$$S_2( 20)=(12,16,10)_{20}$$

$$S_2( 26)=(16,21,13)_{26}$$

$$S_2( 32)=(20,26,16)_{32}$$

$$\vdots$$

Now $k=4$ and $j=15$

$$S_4( 15)=(11,7,2,7,8)_{15}$$

$$S_4( 45)=(35,22,7,22,24)_{45}$$

$$S_4( 75)=(59,37,12,37,40)_{75}$$

$$S_4( 105)=(83,52,17,52,56)_{105}$$

$$\vdots$$

For motivation

If $n> 2$ and $n_k\ne 0$

then

$$n_k = n - \left\lceil\frac{\sum_{i=1}^n i^k}{n^k}\right\rceil $$

Proof

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  • 1
    $\begingroup$ Your question is not clear. Maybe you can be more explicit to show 1. What is it you already have proved, 2. What is your actual question. $\endgroup$ – Daniel Aug 23 at 20:12
  • $\begingroup$ Also, a little motivation would be nice $\endgroup$ – Daniel Aug 23 at 20:12
  • $\begingroup$ Ok, now I add some examples. I want to generalized this pattern in arithmetic progression of digits for each $k$ and it's help to understand my problem on prime number :math.stackexchange.com/q/3271631/647719 $\endgroup$ – Pruthviraj Aug 23 at 20:29
  • $\begingroup$ @Daniel I proved this type of arithmetic progression for $k=1$ only and I observed this progression have some pattern for each $k$. But not able to catch exactly. I want prove this arithmetic progression exist in $k=2$ and further $k$s. $\endgroup$ – Pruthviraj Aug 23 at 20:52

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