Leading one bit in 3^x

General
1

The number of leading one bit in 3^x is important when placing bounds on the
length of cycles. This is the basis of the constants in Eliahou’s paper.
Playing around I’ve also confirmed this by a different method.

A lower bound on 3^x for a given number of leading one bits could be key to
the Collatz puzzle. It would be interesting if anyone that is better at math
than me wants to look into it. Programmers may want to come up with an
algorithm as well.

Exponents can be calculated using a high precision representation of the log2
f 3. The fractional part of the its product with the exponent is near 2^Ones
minus one divided by 2^Ones. Here are some examples to give you a sense of
how the numbers work out.

Log2(3) = 1.58496_25007_21156_18145_37389_43947_81650_87598_14407_69248_10604_55752_65454_10982_2779 ...


                                             2^Ones-1
Ones   Exponent      Log2(3) * Exponent      --------
                                             2^Ones

 7          147         232.98948_76060    0.99218_75000
 8          253         400.99551_26825    0.99609_37500
 9          306         484.99852_52207    0.99804_68750
10        1_636        2592.99865_11798    0.99902_34375
11        8_951       14186.99934_39551    0.99951_17188
12       12_276       19456.99965_88529    0.99975_58594
13       14_271       22618.99984_77916    0.99987_79297
14       31_202       49453.99994_75015    0.99993_89648
15       15_601       24726.99997_37508    0.99996_94824
16       47_468       75234.99998_42318    0.99998_47412
17      158_670      251485.99998_94258    0.99999_23706
18       79_335      125742.99999_47129    0.99999_61853
19    2_858_055     4529909.99999_86042    0.99999_80927
20    1_524_296     2415951.99999_92554    0.99999_90463
21      762_148     1207975.99999_96277    0.99999_95232
22      381_074      603987.99999_98138    0.99999_97616
23      190_537      301993.99999_99069    0.99999_98808
2

I think what you’re trying to achieve is get a way of finding the sequence of local maximums of 3^x/2^k<1 such that for all 1<=x’<=x 1>3^x/2^k>3^x’/2^k’ which is indeed closely related to Collatz Cycles. In this case, it can be shown that this problems boils down to finding the rationnals semi convergents of ln2/ln3=log3(2) such that they are strictly smaller than ln2/ln3 (you can find more info on that on internet) on which it is quite easy to make a program to compute them as they are quite well known and there is an explicit recursive formula to compute them. Now using this you can indeed find Eliahou’s results that are in his paper using a different method that puts a constraint on how close does 3^x/2^k has to be from 1 while being under 1, and the bigger we know which is the biggest N for which we know that all numbers below this N verifies the Collatz Conjecture the closer does 3^x/2^k has to be from 1. I was able of doing exactly that and this method can actually yield better minimum cycles lengths in function of N (not by much) than the one used by Eliahou.