**dap1** (10/21/82)

#N:ihlpb:6200001: 0:2940 ihlpb!dap1 Oct 20 13:59:00 1982 Speaking of insomnia relievers, I have one that I have been grappling with for several years now (off and on). I used to work part time for the government and was told that I could not work a "schedule" due to union rules. This didn't make much sense to me so I decided to work it out to it's logical extremes. As I see it, "working a schedule" means working on days which are in arithmetic sequence (i.e. every other day, every third day, etc.). So the question I posed was this: If I start work on day zero and then go to work on any day which doesn't form an arithmetic sequence of length N with the days I've already worked, then on which days do I work? As an example, if N=3 then I can work on days 0 and 1 but not on day 2 since 0,1,2 forms an arithmetic sequence of length 3. I can work again on day 3 and 4 but not on day 5,6,7 or 8 (0,3,5-2,4,6-1,4,7-0,4,8). The sequence starts as: 0,1,3,4,9,10,12,13,27,28,30... I've solved the problem for N prime and the answer is rather surprising. The i'th term in the sequence of days can be found as follows: 1. Expand i in base (N-1) notation. 2. Interpret the expansion in base N. 3. The resultant number is the i'th day worked. For instance, if N=3 then the 10th term is 30 since 10(base 10)=1010(base 2) and 1010(base 3)=30(base 10). The proof of this involves the fact that no prime order group has any subgroups. What's more, I don't see any reasonable way to generalize to the problem where N is composite. I have generated the sequence on the computer and for composite N it seems to be an extremely unruly beast. Some other questions also come to mind in regard to this question: 1. Is this sequence the most densely packed set of numbers which possess this property? For any M it is not hard to find examples of tighter sequences of the integers 0 through M which enjoy the NAS (non arithmetic sequence) property. However it seems that as M grows larger these tightly packed sequences will have to skip numbers that the proposed sequence includes so maybe there is still some sort of optimality property along the order of "As M approaches infinity..." 2. What about three and higher dimensional lattices? In particular, the original problem can be thought of as points on a one dimensional lattice. Could this be expanded to two and higher dimensions? In order to simulate the original problem exactly some sort of order would have to be given to the points in the lattice but a variation similar to that discussed in 1.) above would be something like "In an M by Q two dimensional lattice how many points may be placed such that no line through the lattice has N evenly space points lying on it?". 3. Etc., etc. Well I've gone on too long but if anybody wants to talk this over send mail to me. Thanks, Darrell Plank BTL-Indian Hills (312)979-4582 ...ihlpb!dap1

**dap1** (10/22/82)

#N:ihlpb:6200002: 0:2940 ihlpb!dap1 Oct 21 8:45:00 1982 Speaking of insomnia relievers, I have one that I have been grappling with for several years now (off and on). I used to work part time for the government and was told that I could not work a "schedule" due to union rules. This didn't make much sense to me so I decided to work it out to it's logical extremes. As I see it, "working a schedule" means working on days which are in arithmetic sequence (i.e. every other day, every third day, etc.). So the question I posed was this: If I start work on day zero and then go to work on any day which doesn't form an arithmetic sequence of length N with the days I've already worked, then on which days do I work? As an example, if N=3 then I can work on days 0 and 1 but not on day 2 since 0,1,2 forms an arithmetic sequence of length 3. I can work again on day 3 and 4 but not on day 5,6,7 or 8 (0,3,5-0,3,6-1,4,7-0,4,8). The sequence starts as: 0,1,3,4,9,10,12,13,27,28,30... I've solved the problem for N prime and the answer is rather surprising. The i'th term in the sequence of days can be found as follows: 1. Expand i in base (N-1) notation. 2. Interpret the expansion in base N. 3. The resultant number is the i'th day worked. For instance, if N=3 then the 10th term is 30 since 10(base 10)=1010(base 2) and 1010(base 3)=30(base 10). The proof of this involves the fact that no prime order group has any subgroups. What's more, I don't see any reasonable way to generalize to the problem where N is composite. I have generated the sequence on the computer and for composite N it seems to be an extremely unruly beast. Some other questions also come to mind in regard to this question: 1. Is this sequence the most densely packed set of numbers which possess this property? For any M it is not hard to find examples of tighter sequences of the integers 0 through M which enjoy the NAS (non arithmetic sequence) property. However it seems that as M grows larger these tightly packed sequences will have to skip numbers that the proposed sequence includes so maybe there is still some sort of optimality property along the order of "As M approaches infinity..." 2. What about three and higher dimensional lattices? In particular, the original problem can be thought of as points on a one dimensional lattice. Could this be expanded to two and higher dimensions? In order to simulate the original problem exactly some sort of order would have to be given to the points in the lattice but a variation similar to that discussed in 1.) above would be something like "In an M by Q two dimensional lattice how many points may be placed such that no line through the lattice has N evenly space points lying on it?". 3. Etc., etc. Well I've gone on too long but if anybody wants to talk this over send mail to me. Thanks, Darrell Plank BTL-Indian Hills (312)979-4582 ...ihlpb!dap1