Although the forward equation is quite easy to calculate, calculating the precursors of a given number is more interesting. Looking at "1", we see that it is preceded by the infinite sequence of powers of 2. We are focusing on odd predecessors, so we have to find powers of two that fit the equation
Where x is an integer. Only even values of k lead to integer solutions, giving;
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Etc. For numbers other than 1, say 'A', we get the similar equation
Calculating x for a few values of A and k we can see a pattern;
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When A is a multiple of three we get no predecessors, as a multiple of three can't be of the form 3x+1. When A is one more than a multiple of 6, k will be even, and when it is one less than a multiple of 6, k will be odd. The smallest precursors are therefore: 6n+1→8n+1 and 6n+5→4n+3.
When reading down the columns the values increase linearly in steps of 2(k+1), whereas across the table they increase geometrically as x → 4x+1.
The next two tables are extensions of the previous tables, but now the numbers are colour-coded according to their remainders when divided by 6.
| A\k | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 | |
|---|---|---|---|---|---|---|---|---|---|---|
| 1 | 1 | 5 | 21 | 85 | 341 | 1,365 | 5,461 | 21,845 | 87,381 | |
| 7 | 9 | 37 | 149 | 597 | 2,389 | 9,557 | 38,229 | 152,917 | 611,669 | |
| 13 | 17 | 69 | 277 | 1,109 | 4,437 | 17,749 | 70,997 | 283,989 | 1,135,957 | |
| 19 | 25 | 101 | 405 | 1,621 | 6,485 | 25,941 | 103,765 | 415,061 | 1,660,245 | |
| 25 | 33 | 133 | 533 | 2,133 | 8,533 | 34,133 | 136,533 | 546,133 | 2,184,533 | |
| 31 | 41 | 165 | 661 | 2,645 | 10,581 | 42,325 | 169,301 | 677,205 | 2,708,821 |
| A\k | 1 | 3 | 5 | 7 | 9 | 11 | 13 | 15 | 17 | |
|---|---|---|---|---|---|---|---|---|---|---|
| 5 | 3 | 13 | 53 | 213 | 853 | 3,413 | 13,653 | 54,613 | 21,8453 | |
| 11 | 7 | 29 | 117 | 469 | 1,877 | 7,509 | 30,037 | 120,149 | 480,597 | |
| 17 | 11 | 45 | 181 | 725 | 2,901 | 11,605 | 46,421 | 185,685 | 742,741 | |
| 23 | 15 | 61 | 245 | 981 | 3,925 | 15,701 | 62,805 | 251,221 | 1,004,885 | |
| 29 | 19 | 77 | 309 | 1,237 | 4,949 | 19,797 | 79,189 | 316,757 | 1,267,029 |
Notice that numbers of the form 6n+5 have a smaller odd predecessor. This can be used to generate arbitrary long sequences of numbers where (3x+1)/2 is odd and (3*(3x+1)/2+1) is larger than 3x+1.
The colour patterns repeat every 3 rows, as we can see from the following tables
| A\k | 2 | 4 | 6 | |
|---|---|---|---|---|
| 18n+1 | 24n+1 = 6.(4n)+1 | 96n+5 = 6.(16n)+5 | 384n+21 = 6.(64n+3)+3 | |
| 18n+7 | 24n+9 = 6.(4n+1)+3 | 96n+37 = 6.(16n+6)+1 | 384n+149 = 6.(64n+24)+5 | |
| 18n+13 | 24n+17 = 6.(4n+2)+5 | 96n+69 = 6.(16n+11)+3 | 384n+277 = 6.(64n+46)+1 |
| A\k | 1 | 3 | 5 | |
|---|---|---|---|---|
| 18n+5 | 12n+3 = 6.(2n)+3 | 48n+13 = 6.(8n+2)+1 | 192n+53 = 6.(32n+8)+5 | |
| 18n+11 | 12n+7 = 6.(2n+1)+1 | 48n+29 = 6.(8n+4)+5 | 192n+117 = 6.(128n+19)+3 | |
| 18n+17 | 12n+11 = 6.(2n+1)+5 | 48n+45 = 6.(8n+7)+3 | 192n+181 = 6.(128n+30)+1 |
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Blue Chains: We can use the fact that the smallest predecessor of 6n+5 values is less that the original value to create arbitrary long chains of increasing odd values. |
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How reverse iteration works: This page explains the reverse iteration algorithm and uses it to find those numbers that converge on 85. |