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 Where 2^n leads, 3^n cannot be far behind... (Posted on 2012-08-24)

Carrying on in the spirit of '86 at most' here is a pair of 'statistical' conjectures about 3^n: (see A060956 in Sloane, particularly the table, for the values up to 3^1000)

To start with a definition; if the first digit of 3^n is a 9 ( e.g. 3^23 = 94,143,178,827) then we say that 3^n is 'good'.

Conjecture 1: If 3^n is good, then either 3^(n+21), or 3^(n+23) is also good.

Conjecture 2: If 3^n is good, then 3^(m+n) is also good, for some constant, m, and n greater than 2.

True or false?

 See The Solution Submitted by broll No Rating

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 solution | Comment 1 of 2

Part 1 is easy:

3^21 =  10460353203
3^23 =  94143178827

The former, when multiplied by a good number whose significand is lower than 9.5599... will produce another good number. The latter, when multiplied by a good number whose significand is greater than that same number will be good. Only if the significand of the number were exactly this (10^11 / 3^21) would the result fail to be good. But such an original number would not be a power of 3, as it would be a rational number with some power of 10 in the numerator.

BTW, the significand is the a in the scientific notation a x 10^b, and was previously called the mantissa, but the name was changed to avoid conflict of the use of "mantissa" as the fractional part of a common logarithm.

So, yes, conjecture 1 holds.

For Part 2:

If the common log of 3 is taken to be a normal number, with any sequence of digits equally likely for any count of digits anywhere in the expansion, then, again using common logs:

log(3^n)=n*log(3)

The fractional part of the resulting logarithm, called the mantissa, will then be expected to be some "random", that is, normal, number as well, equally likely to be anywhere on the unit number line.

The antilog of this mantissa is called the significand of the scientific notation for the given power of 3.

If we continue to higher and higher powers of 3 we would expect to get numbers ever closer, on both sides, to having a significand of 9 (only 3^2 has one exactly 9), as demonstrated by:

10   point 255
20   L3=log(3)/log(10):L9=2*L3:Lp=2*L3:Lpmin=99
30   for Pwr=3 to 10000000
40      Lp=Lp+L3
50      Lp=Lp-int(Lp)
60      if Lp-L9>=0 then if Lp-L9<Lpmin then Lpmin=Lp-L9:gosub *Report
70      if L9-Lp>=0 then if Lp>Lpmax then Lpmax=Lp:gosub *Report
80   next
90   end
100   *Report
110     print Pwr;tab(20);using(3,15),10^Lp
120   return

`n                    significand of 3^n`
`3                    2.7000000000000004                    8.10000000000000023                   9.41431788270000025                   8.47288609443000046                   8.86293811965250167                   9.270946314789784111                  9.129758166511361155                  8.990720186353585264                  9.120344560464466417                  9.110940660696444570                  9.101546457199243723                  9.092161939975133876                  9.0827870990366891029                 9.0734219244067871182                 9.0640664061185881335                 9.0547205342155321488                 9.0453842987513231641                 9.0360576897899211794                 9.0267406974055321947                 9.0174333116825952100                 9.0081355227157752253                 8.9988473206099474351                 9.0069818013646826602                 9.0058282277770138853                 9.00467480193384511104                9.00352152381625413355                9.00236839340532015606                9.00121541068212717857                9.00006257562775820108                8.99890988822330137963                8.99897245627167855818                8.99903502475508173673                8.99909759367351391528                8.999160163026977109383               8.999222732815476127238               8.999285303039014145093               8.999347873697593162948               8.999410444791216180803               8.999473016319886198658               8.999535588283607216513               8.999598160682381216513               8.999598160682381234368               8.999660733516211252223               8.999723306785101270078               8.999785880489053287933               8.999848454628070305788               8.999911029202156323643               8.999973604211313341498               9.000036179655545665139               9.000009783760749988780               8.9999833879433681653917              8.9999931716860582319054              9.0000029554393843972969              8.9999961271231996292021              8.9999990825613118611073              9.000002038000393`

It seems that whenever we put x digits into an exponent for 3, we can get at least x-1 signficant digits of closeness to 9 in the significand, on both the high side and the low side.

Any given single value for m will have its power of 3 have either a significand some finite amount above 9 or below 9.  Either way, it will destroy the goodness of some number that's sufficiently close to 9 in the same way, and this sufficient degree of closeness is always achievable.

So conjecture 2 would be disconfirmed.

 Posted by Charlie on 2012-08-24 15:33:51

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