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Conjecture 1: If 3^n is good, then either 3^(n+21), or 3^(n+23) is also good.
Assume 3^n is good. Then its leading digits are between (900..and 999..). 3^21 is 1.0460353203 *10^10. 1.0460353203x=1 gives x~0.95599066360;1.0460353203x=0.9 gives x~0.860392 (which is less than 9). So from 3^n=900... to 95599.., 3^(n+21) is good. If 3^n is just a little greater that 95599... then 3^(n+21) is just a little greater than 100..., and just a little less than 10461..
Now 3^(n+23)= 9*3^(n+21). If 3^(n+21) is just little greater than 100...,and just a little less than 10461.., then 9 times that is just a little greater than 900...
So the conjecture is TRUE.
So far, it seems that the leading digit of 3^n is a nice regular, predictable quantity. But...
Conjecture 2: If 3^n is good, then 3^(m+n) is also good, for some constant, m, and n greater than 2.
Start with the values in A060956. The 'standard' pattern of the first digits is {1, 3, 9, 2, 8, 2, 7, 2, 6, 1, 5, 1, 5, 1, 4, 1, 4, 1, 3, 1, 3} but this breaks down in the third repeat which adds an extra {1,3} at the end.
Now consider just those cases where 3^n begins with a 9. {2,23,44,67,88,111....}. If we take the difference of the powers (44-23, 67-44 etc) then this new series runs {0,2,-2,2,-2,0,2,-2,2,-2,2,-2,0..} with a repeat after 153 powers. And it's true for every power up to 3^832 that if 3^n begins with 9 then so does 3^(n+153). (832+153=985, the last number given in A060956).
But we probably shouldn't take the 'rule of 153' for granted, so we try x=3^(23+153n), adding some more significant digits:
{(n=10 x=931...),(n=20 x=922...),(n=30 x=912...),(n=40 x=903...),(n=43 x=900...),(n=44 x=899...)} when it can be seen that there is a slow drift downwards below 9..., and finally at 3^(23+153*44)=6755 there is a break in the pattern. 3^6755 is ~9*10^3222; so the conjecture, statistically, holds good up to an enormous value of 3^n before failing at last.
But as soon as this 'drift' is witnessed with m=153, it is at once obvious that the same must apply to any m we choose; eventually 3^(m+n) will begin with every single first digit from 1 to 9.
So the conjecture is FALSE.
Note: When I was looking at '86 at most', I came across this:
http://mathworld.wolfram.com/GelfandsQuestion.html. Question 3 is: Will a row of all the same digit occur? No such example occurs for n less than 10^5. Some of the spadework for this question was already done in Perplexus in the past (for the numbers 2 and 5). So 3 is the obvious next place for review. Unfortunately 3^n, despite initial appearances, is not as cooperative as might have been hoped!
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