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User talk:K. Spage

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A different way to look at A243115, is to make it start with 1,2,3, with offset 1.

 1 + i * 2^ 2 ->  1 + i * 3^ 1 (1,4,2,1; does 1,5,9,...)
 2 + i * 2^ 1 ->  1 + i * 3^ 0 (see 1: 2->1; does all evens)
 3 + i * 2^ 4 ->  2 + i * 3^ 2 (3,5,16,8,4,2)
...
 

From 3 on, only 4 mod 3 terms, in a self-referential way:

| 3 + 2^2 =  7|             |             |             |
| 7 + 2^2 = 11| 3 + 2^3 = 11|             |             |
|11 + 2^2 = 15| 7 + 2^3 = 15|             |             |
|             |15 + 2^3 = 23| 7 + 2^4 = 23|             |
|23 + 2^2 = 27|             |11 + 2^4 = 27|             |
|27 + 2^2 = 31|23 + 2^3 = 31|15 + 2^4 = 31|             |
|             |31 + 2^3 = 39|23 + 2^4 = 39| 7 + 2^5 = 39|
...
 
| 1 + 2^1 =  3|
|-1 + 2^2 =  3|

| 3 + 2^2 =  7|
|-1 + 2^3 =  7|

| 7 + 2^2 = 11|
| 3 + 2^3 = 11|

|11 + 2^2 = 15|
| 7 + 2^3 = 15|
|-1 + 2^4 = 15|

|15 + 2^3 = 23|
| 7 + 2^4 = 23|

|23 + 2^2 = 27|
|11 + 2^4 = 27|

|27 + 2^2 = 31|
|23 + 2^3 = 31|
|15 + 2^4 = 31|
|-1 + 2^5 = 31|

|31 + 2^3 = 39|
|23 + 2^4 = 39|
| 7 + 2^5 = 39|
...