The forward equation is quite straight forward . When dealing with the odd numbers only, but when we look in the opposite direction every odd number not divisible by three has an infinite number of immediate precursors. This page explores identifying those numbers.
Looking at "1", we see that it is preceded by the infinite sequence of powers of 2. On this site, though we are focusing on the odd numbers, and we will see below, every odd number that isn't a multiple of 3 has an infinite number of odd precursors.
To find the precursors of 1 we have to find powers of two that fit the equation
2^k=3x+1{IMP}x=(2^k-1)/3
Where x is an integer. Only even values of k lead to integer solutions, giving;
| k | (2^k-1)/3 | k | (2^k-1)/3 | |
|---|---|---|---|---|
| 2 | (4-1)/3 = 1 | 10 | (1024-1)/3 = 341 | |
| 4 | (16-1)/3 = 5 | 12 | (4096-1)/3 = 1365 | |
| 6 | (64-1)/3 = 21 | 14 | (16384-1)/3 = 5461 | |
| 8 | (256-1)/3 = 85 | 16 | (65536-1)/3 = 21845 |
For numbers other than 1, say 'A', we get the similar equation
A.2^k=3x+1{IMP}x=(A.2^k-1)/3
Calculating x for a few values of A and k we can see a pattern;
'A' is 6n+1, 'k' is odd:
| x(A,k) | k = 2 | k = 4 | k = 6 |
|---|---|---|---|
| A = 1 | 1 | 5 | 21 |
| A = 7 | 9 | 37 | 149 |
| A = 13 | 17 | 69 | 277 |
'A' is 6n+5, 'k' is even:
| x(A,k) | k = 1 | k = 3 | k = 5 |
|---|---|---|---|
| A = 5 | 3 | 13 | 53 |
| A = 11 | 7 | 29 | 117 |
| A = 17 | 11 | 45 | 181 |
When A is 3n we get no predecessors, as a multiple of three can't be of the form 3x+1. The smallest precursors are therefore:
6n+1{RA}8n+1, 6n+5{RA}4n+3.
When reading down the columns the values increase arithmetically in steps of x{RA}x+2^[(k+1)], whereas across the table they increase geometrically as x{RA}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 |
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 |
Of the 18 possible remainders when dividing by 18, 9 are even and three (3,9,15) are divisible by 3. The rest divide into two sets: 6n+1 (1,7,13) and 6n+5 (5,11,17). These last 6 are the ones shown in the table.
(c) John Whitehouse 2011 - 2023