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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| ...