A New Take On The Sum Of Digits

Clearly only \(0\) and \(1\) satisfy the equation \(3n=3^n\) but what about the following:$$SOD(3n)=SOD(3^n)$$where SOD stands for the sum of a number's digits. The values of \(n\) that do satisfy are members of OEIS A260906:

A260906: numbers \(n\) such that \(3n\) and \(n^3\) have the same digit sum.

Here are the members of this sequence up to one million:

3, 6, 30, 60, 63, 126, 171, 252, 300, 324, 543, 585, 600, 630, 1260, 1281, 1710, 2520, 2925, 3000, 3240, 5430, 5850, 5946, 6000, 6300, 12600, 12606, 12633, 12810, 14631, 16263, 17100, 21618, 22308, 22971, 24663, 25200, 27633, 28845, 28887, 28965, 29241, 29250, 29625, 29628, 30000, 31752, 32400, 49533, 52308, 54300, 58500, 59460, 60000, 63000, 82962, 89325, 126000, 126060, 126294, 126330, 128100, 132633, 146310, 162630, 171000, 216180, 216558, 223080, 223086, 225333, 229653, 229710, 231633, 233325, 246630, 252000, 258333, 276330, 282333, 283233, 285333, 288450, 288870, 289650, 289656, 292410, 292500, 296250, 296280, 300000, 315333, 317520, 319266, 323328, 324000, 466623, 466653, 493284, 495330, 522333, 523080, 531333, 543000, 546291, 558333, 559665, 585000, 585333, 585972, 594600, 594666, 597333, 600000, 625008, 630000, 636618, 663156, 665325, 666558, 798981, 829620, 888333, 893250

Let's take the last member of the above numbers, 893250:$$ \begin{align} 3 \times 893250 &=2679750 \rightarrow 36 \\ 893250^3 &= 712720211203125000 \rightarrow 36 \end{align}$$One could ask the question as to what values of \(k\) in$$ \text{SOD}(k n) = \text{ SOD}(n^k) $$yield numbers \(n\) that satisfy in the range up to one million? For \(k=2\), there are many and they comprise OEIS A049343:

A049343: numbers \(n\) such that \(2n\) and \(n^2\) have same digit sum.

The initial members are:

0, 2, 9, 11, 18, 20, 29, 38, 45, 47, 90, 99, 101, 110, 119, 144, 146, 180, 182, 189, 198, 200, 245, 290, 299, 335, 344, 351, 362, 369, 380, 398, 450, 452, 459, 461, 468, 470, 479, 488, 495, 497, 639, 729, 794, 839, 848, 900, 929, 954, 990, 999

For example:$$ \begin{align} 2 \times 999 &= 1998 \rightarrow 27\\ 999^2 &=998001 \rightarrow 27 \end{align} $$However for \(k=4\) there are only two values that satisfy (21249 and 212490):$$ \begin{align} 4 \times 21249 &= 84996 \rightarrow 36 \\21249^4 &= 203870311303040001 \rightarrow 36 \\ 4 \times 212490 &= 849960 \rightarrow 36 \\ 212490^4 &= 2038703113030400010000 \rightarrow 36 \end{align} $$In the range up to one million (and probably beyond), no higher values of k satisfy. Oddly for \(k=1/2\), there are 31 values that satisfy in the range up to one million:

4, 324, 400, 1444, 8100, 8464, 26244, 32400, 39204, 40000, 82944, 84100, 142884, 144400, 158404, 202500, 204304, 219024, 220900, 238144, 334084, 422500, 544644, 602176, 627264, 630436, 810000, 842724, 846400, 980100, 984064

For example:$$ \begin{align} \frac{1}{2} \times 984064 &= 492032 \rightarrow 20\\ 984064^{1/2} &= 992 \rightarrow 20 \end{align} $$For \(k=1/3\), there are only three values that satisfy in the range up one million (729, 729000 and 970299). For example:$$ \begin{align} \frac{1}{3} \times 970299 &= 323433 \rightarrow 18\\ 970299^{1/3} &= 99 \rightarrow 18 \end{align} $$So there are fractional values of \(k\), namely 1/2 and 1/3, that satisfy the following as well as the integer values of 1 (trivial), 2, 3 and 4:$$ \text{SOD}(kn) = \text{ SOD}(n^k) $$There are probably other fractional values as well that satisfy but that will do for now.