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### Problem Description

This program will generate "Magic Sequences".  .A Magic Sequence of length N has the property that each value in the series represents the number of times that its rank (position number) relative to zero, appears in the sequence. If the elements are represented by Xi, 0<=i<=N-1 then Xi = the number of occurrences of "i" in the sequence.

So for sequences of length 5, one (and only) example is [2,1,2,0,0], magic because x0 = 2 and there are 2 zeroes in the sequence; X1=1 and there is only a single 1 in the sequence; x2=2 and there are two "2"s, X3=X4=0 and there are no "3"s or "4"s.

### Background & Techniques

A viewer wrote recently asking about a program to generate "Magic Sequences". He included code that calculated the sequence of length 7 by nesting loops 7 levels deep, each level generating a digit position.  Clearly he felt that there must be a better.  Probably so.

There does not seem to be much literature on sequences with this property and they appear to degenerate into a predictable pattern rather soon so are perhaps not so interesting from a mathematical standpoint but did make a good programming exercise.

I generalized the code to produce three additional ways to calculate sequences for a given N. Each method is faster than the preceding version.

Method 1 simply counts in base N and checks each to see if it is "magic". [00001, 00002, 00003, 00004, 00010 ,..., 21200, ..., 44444, 44445] for N=5, for example.

The next two methods take advantage of the fact that the sum of the non-zero elements of a sequence form an partition of N. (Each value in the sequence represents the count of a particular value and there are N of these values
altogether, so the sum must equal N.) So we might as well start with the integer partitions of N.

Method 2 generates integer partitions of N, checks that one of the values equals the number of zeros in the sequence and then expands the sequence with zeros and permutes the last N-1 digits looking for magic sequences. For N=7, the partition {1,1,2,3} is a potential solution since it contains 4 numbers so there must be 3 zeros in the sequence and we have a 3 in the partition that can occupy the 1st position.    Method 2 then permutes the remaining 6 sequence members, {1,1,2,0,0,0}, checking for than would complete a valid magic sequence.   This requires checking 6!=720 possibilities.

Method 3 improves on Method 2 by permuting the potential position where the digits of each partition might appear in the sequence.  so in the N=7  case,  we  only check the number of places that {2,1,1} can be inserted into the last 6 positions.  This requires that only  (6!/3!)=6*5*4=120 patterns be checked.

#### Notes for programmers

 From the programming perspective, our TIntPartition class in unit UIntPartition2 and TComboset class in the UComboV2 library unit  makes generating the partitions and permutations quite easy.   Each of these units initializes a default instance of the class (DefPartition for TIntpartition and  Combos for TComboSet) which avoids creating our own. There is a commented version of a procedure for Method 1 (Button1Click) which  works as literally described above, counting in base N.   The equivalent, but faster, procedure actually used is a version of the TComboset  NextLexRepRPermute function which generates permutations with repeats.  For permuting N of N items, this is equivalent to counting in Base N. Run times for each method are displayed, making the it easy to determine the effect of changes on speed.

### Suggestions for Further Explorations

Other enhancements to increase speed?

For the math types, it appears that for N>6, there is only a single sequence with four non-zero members and they are: X0=N-4,  X1=2,  X2=1 and XN-4=1.  How about proving this?

 Original Date: May 1, 2007 Modified: July 29, 2017