Generator-offset property: Difference between revisions

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We must have gcd(''k'', ''n'') = 1. If not, since ''n'' is odd, gcd(''k'', ''n'') is an odd number at least 3, and the ''k''-steps must form more than 2 parallel chains.

By modular arithmetic we have ''rk'' mod ''n'' = ''k''/2 iff ''r'' = ceil(''n''/2) mod ''n''. (Since gcd(2, ''n'') = 1, 2 is multiplicatively invertible mod ''n'', and we can multiply both sides by 2 to check this: we get (''n'' + 1)k = ''nk + k = k'' (mod ''n''), which is true.) This proves that the offset, which must be reached after ceil(''n''/2) generator steps, is a ''k''/2-step. (If the offset wasn't reached in ceil(n/2) steps, the two generator chains either wouldn't be disjoint or wouldn't have the assumed lengths.) <math>\square</math>

By modular arithmetic we have ''rk'' mod ''n'' = ''k''/2 iff ''r'' = ceil(''n''/2) mod ''n''. (Since gcd(2, ''n'') = 1, 2 is multiplicatively invertible mod ''n'', and we can multiply both sides by 2 to check this: we get (''n'' + 1)k = ''nk + k = k'' (mod ''n''), which is true.) This proves that the offset, which must be reached after ceil(''n''/2) generator steps, is a ''k''/2-step. (If the offset wasn't reached in ceil(''n''/2) steps, the two generator chains either wouldn't be disjoint or wouldn't have the assumed lengths.) <math>\square</math>

== Open conjectures ==

@@ Line 74: / Line 74: @@
 We must have gcd(''k'', ''n'') = 1. If not, since ''n'' is odd, gcd(''k'', ''n'') is an odd number at least 3, and the ''k''-steps must form more than 2 parallel chains.
-By modular arithmetic we have ''rk'' mod ''n'' = ''k''/2 iff ''r'' = ceil(''n''/2) mod ''n''. (Since gcd(2, ''n'') = 1, 2 is multiplicatively invertible mod ''n'', and we can multiply both sides by 2 to check this: we get (''n'' + 1)k = ''nk + k = k'' (mod ''n''), which is true.) This proves that the offset, which must be reached after ceil(''n''/2) generator steps, is a ''k''/2-step. (If the offset wasn't reached in ceil(n/2) steps, the two generator chains either wouldn't be disjoint or wouldn't have the assumed lengths.) <math>\square</math>
+By modular arithmetic we have ''rk'' mod ''n'' = ''k''/2 iff ''r'' = ceil(''n''/2) mod ''n''. (Since gcd(2, ''n'') = 1, 2 is multiplicatively invertible mod ''n'', and we can multiply both sides by 2 to check this: we get (''n'' + 1)k = ''nk + k = k'' (mod ''n''), which is true.) This proves that the offset, which must be reached after ceil(''n''/2) generator steps, is a ''k''/2-step. (If the offset wasn't reached in ceil(''n''/2) steps, the two generator chains either wouldn't be disjoint or wouldn't have the assumed lengths.) <math>\square</math>
 == Open conjectures ==