This file was last updated:  11/17/2018 1:33:40 AM


         Experimental Mathematics on Wisteria Tables

                        Abstract

  "Experimental mathematics is an approach to mathematics in
which computation is used to investigate mathematical objects and
to identify properties and patterns."  Eric W. Weisstein, quoted
in en.wikipedia.org/wiki/Experimental_mathematics

  The Online Encyclopedia of Integer Sequences (OEIS) contains 
over 300,000 integer sequences.  To browse it is to rekindle a 
childhood wonder at how fascinating mathematics can be.  A math 
problem provided by Marist College Prof. Joseph Kirtland a year 
ago led me to arrange 30 of the known OEIS sequences into a table,
which was published at OEIS.org/A299741.  I found multiple distinct 
algorithms which generate the table.  The wisteria name is meant 
to recall the interconnectedness of the plant.

                  ------------------------

  Here is Table 5_14_18 = OEIS.org/A299741, unbounded below and to the right.

i\j| 0  1   2    3     4      5       6        7         8          9
---+-----------------------------------------------------------------
  0| 2  2   2    2     2      2       2        2         2          2
  1| 2  3   7  _18_   47    123     322      843      2207       5778
  2| 2  4 _14_  52   194    724    2702    10084     37634     140452
  3| 2 _5_ 23  110   527   2525   12098    57965    277727    1330670
  4| 2  6  34  198  1154   6726   39202   228486   1331714    7761798
  5| 2  7  47  322  2207  15127  103682   710647   4870847   33385282
  6| 2  8  62  488  3842  30248  238142  1874888  14760962  116212808
  7| 2  9  79  702  6239  55449  492802  4379769  38925119  345946302
  8| 2 10  98  970  9602  95050  940898  9313930  92198402  912670090
  9| 2 11 119 1298 14159 154451 1684802 18378371 200477279 2186871698

  Named for the middle values of the fifth upward diagonal.

  It first appeared on OEIS on Jan 24, 2018, at oeis.org/A298675
and shortly later on Feb 18, 2018, at oeis.org/A299741.

  The table is generated by three very distinct algorithms:  A0,
A1, and A2 (to be defined) and possibly by four other algorithms
A3, A4, A5, and A6 (to be described).

  The experimental mathematics in this work occurs in the
creation of examples.  See, in particular, Section D.  But, to
expand the meaning of the phrase, also in the search, via the
web, for examples.  In the latter endeavor OEIS has been very,
very helpful.

  This material is meant to be readable and self-contained.  It
is aimed for a freshman or sophomore undergraduate mathematics
major.

========================================================== p.  2

  Outline of tonight's talk.

  Section A.  History of developing Table 5_14_18.
  
  Starting from a simple problem from Dr. Joseph Kirtland a year ago.

  This is the hardest part of the talk.
  
  Prove that if x + 1/x is an integer, then so are all these
values:  x^2 + 1/x^2,  x^3 + 1/x^3,  x^4 + 1/x^4,,,.
  
  1.  Begin with computing the values r^n + 1/r^n, n=0,1,2,...,
where r is a root of x + 1/x = k, k=1,2,3,....  Create Table
5_14_18.  Call this algorithm A2.  Yes, this is confusing.

  2.  Discover recurrence relations among the elements of each
row.  Realize that they also generate Table 5_14_18.  Call this
algorithm A1.  Forget about A2.

  3.  Look for polynomials creating the columns.  They must
exist.  Call them rule A_ ("A blank").

  4.  Discover that the polynomials are embedded in the (1,2)
Pascal triangle.

  5.  See that the polynomials are defined recursively.  Call this
algorithm A0.  Same polynomials.  Deeper understanding of their
structure.  Forget about A_.

  6.  See that the definitions of A0 and A1 are syntactically
identical, but semantically distinct.
  
  So far everything has been done with just Table 5_14_18 in mind.
Many more examples of these behaviors can be found at More tables

 Many more examples of these behaviors can be found at Table 2_4_5
 

  Section B.  Detour:  OEIS - Online Encyclopedia of Integer Sequences (OEIS)

    What it offers, how to use it, how much fun it can be.
    How helpful it can be in seeking patterns among numbers.
  
  Section C.  Detour:  scavenger hunt in OEIS for more algorithms
to generate Table 5_14_18.

    at least one:  A3.
    maybe three more: A4, A5, A6
  
  Section D.  Question.  Are there other tables which are
generated by rules of type A0?
  
  Answer:  Yes, every (a,b) Pascal triangle generates at least
three such tables ((1) using all the diagonals; (2) using the
even diagonals; and (3) using the odd diagonals).
      
  View 30 examples at  More tables
  
  Question.  Are there other wisteria tables which are not
associated with Pascal triangles?  Answer:  yes.
      
  Section E.  Conclusion.
  
  Definition.  A table T of integers is wisterian if there exists
a sequence A0 of recursively defined polynomials p[i](n) such
that polynomial i generates column i of table T when evaluated
for n = 0,1,2,....
  
  Conjecture.  For every wisteria table T with generating
polynomials A0 there is a sequence of recursion relations A1 such
that the i-th recursion relation defines the integers in row i of
T.  Further, A0 and A1 are syntactically identical (though
semantically distinct since one defines polynomials and the other
defines recurrence relations).

========================================================== p.  3

  Background info

------------------------------------------------

  The roots r1 and r2 of the quadratic equation

    a * x^2 + b * x + c = 0

are

          -b + sqrt(b^2 - 4*a*c)
    r1 = ------------------------
                 2 * a

and
          -b - sqrt(b^2 - 4*a*c)
    r2 = ------------------------
                 2 * a

------------------------------------------------ p.  4

  The best known example of a recurrence relations is the
Fibonacci sequence

    1, 1, 2, 3, 5, 8, 13, 21, 34, . . .

  which is defined by

    a(0) = 1; a(1) = 1; a(i) = a(i-1) + a(i-2), i>1.

------------------------------------------------ p.  5

  Let polynomial p(n) = a * n^3 + b * n^2 + c * n + d.

  We want to find a,b,c,d such that:

    p(0) = u
    p(1) = v
    p(2) = x
    p(3) = y

  So solve these simultaneous equations for a,b,c,d:

    p(0) =  0 * a + 0 * b + 0 * c + d = u
    p(1) =  1 * a + 1 * b + 1 * c + d = v
    p(2) =  8 * a + 4 * b + 2 * c + d = x
    p(3) = 27 * a + 9 * b + 3 * c + d = y

------------------------------------------------ p.  6

  Pascal's triangle is

                        1

                   1         1

               1        2        1

           1       3         3       1

       1       4        6        4       1

  Each number is the sum of the two numbers above it.

------------------------------------------------ p.  7

  The (1,2) Pascal triangle is:


                   1         2

               1        3        2

           1       4         5       2

       1       5        9        7       2

  Each number is the sum of the two numbers above it.
  
  AKA Lucas triangle, AKA Vieta's array (oeis.org/A029635)
  
------------------------------------------------ p.  8

  The (a,b) Pascal triangle is:

                     a              b  
          
                 a         a+b          b  
                  
             a        2a+b     a+2b        b  
               
         a      3a+b      3a+3b    a+3b       b  

      a    4a+b     6a+4b     4a+6b     a+4b     b  
         
==========================================================
========================================================== p.  9

  Section A.  History of developing Table 5_14_18

  Algorithm A2

  Last September Dr. Joseph Kirtland showed me this problem:

  Prove that if x + 1/x is an integer >= 2, then x^n + 1/x^n is
also an integer.  (^ denotes exponentiation).

  I could not believe it.  I chose instead to look for a
counterexample.

  I decided to try:

    Case 1.  Solve x + 1/x =  1
    Case 2.  Solve x + 1/x =  2
    Case 3.  Solve x + 1/x =  3
    Case 4.  Solve x + 1/x =  4
    Case 5.  Solve x + 1/x =  5
    Case 6.  Solve x + 1/x =  6, etc.

and for each case to compute the values

  x + 1/x, x^2 + 1/x^2, x^3 + 1/x^3, . . .

------------------------------------------------ p. 10

  Case 1.  If x + 1/x = 1

then x^2 - x + 1 = 0.

  The roots are:

       1 + sqrt(1 - 4)
  r1 = ---------------.
              2

       1 - sqrt(1 - 4)
  r2 = ---------------.
              2

  This gets messy, so ignore Case 1.

--------------

  Case 2.  If x + 1/x = 2, then x^2 - 2*x + 1 = 0.

  The roots are:

       2 + sqrt(4 - 4)
  r1 = --------------- = 1
              2

       2 - sqrt(4 - 4)
  r2 = --------------- = 1
              2

  so r1^n + 1/r1^n = 2, for n >= 0, and
     r2^n + 1/r2^n = 2, for n >= 0.

------------------------------------------------ p. 11

  Case 3.  If x + 1/x = 3, then x^2 - 3*x + 1 = 0.

  The roots are:

        3 + sqrt(9 - 4)     3 + sqrt(5)
  r1 = ----------------- = ------------- =  2.618034
              2                  2

        3 - sqrt(9 - 4)     3 - sqrt(5)
  r2 = ----------------- = ------------- =  0.381966
              2                  2

  Using the value 2.618034 yields

   n    0  1         2          3          4           5          6         7           8
  x^n   1  2.618034  6.854102  17.944212  46.978775  122.99187  321.9969  842.99884  2206.9996
 1/x^n  1  0.381966  0.145898   0.055728   0.021286    0.00813    0.0031    0.00118     0.0004
  sum   2  3.000000  7.000000  17.999940  47.000061  123.00000  322.0000  843.00002  2207.0000
        2  3         7         18         47         123        322       843        2207

  Using the value 0.381966 yields

   n    0  1         2          3          4           5          6         7           8
  x^n   1  0.381966  0.145898   0.055728   0.021286    0.00813    0.0031    0.00118     0.0004
 1/x^n  1  2.618034  6.854102  17.944212  46.978775  122.99187  321.9969  842.99884  2206.9996
  sum   2  3.000000  7.000000  17.999940  47.000061  123.00000  322.0000  843.00002  2207.0000
        2  3         7         18         47         123        322       843        2207

------------------------------------------------ p. 12

  One can look at the equation x^n +1/x^n = k and observe that it is
obvious that if r is a root of the equation, then so is 1/r.  Or one
can do the algebra.

  Theorem.  Let r1 and r2 be the roots of the equation
x^2 - k*x + 1 = 0.  Then r1 = 1/r2.

Proof.  We know

        k + sqrt(k^2 - 4)             k - sqrt(k^2 - 4)
  r1 = ------------------- and  r2 = -------------------
               2                             2

             k + sqrt(k^2 - 4)     k - sqrt(k^2 - 4)
  r1 * r2 = ------------------- * -------------------
                    2                     2

             k^2 - (sqrt((k^2 - 4))^2
          = --------------------------
                         4

             k^2 - k^2 + 4     4
          = --------------- = --- = 1
                   4           4

------------------------------------------------ p. 13

  Case 4.  If x + 1/x = 4, then x^2 - 4*x + 1 = 0.

  The roots are:

        4 + sqrt(16 - 4)
  r1 = ------------------ =  2 + sqrt(3) = 3.7320508
               2

        4 - sqrt(16 - 4)
  r2 = ------------------ =  2 - sqrt(3) = 0.267949
               2

   n    0  1           2          3            4          5           6           7
  x^n   1  3.7320508  13.928203  51.980762   193.99484  723.99967  2701.9996  10084.000
 1/x^n  1  0.2679492   0.0717968  0.0192319    0.00515    0.00139     0.0003      0.000
  sum   2  4.000000   14.00000   52.00000    193.99999  723.00106  2701.9999  10084.000
        2  4          14         52          194        724        2702       10084

-------------------------------------------------

  Case 5.  r1 = ( 5 + sqrt(21) ) / 2 = 4.7912818

  2     5             23        110          527       2525       12098

-------------------------------------------------

  Case 6.  r1 = ( 6 + sqrt(32) ) / 2 = 5.8284277

  2     6             34        198         1154       6726       39202

========================================================== p. 14

  These efforts lead to:

  Table 5_14_18.

i\j| 0  1   2    3     4      5       6        7         8          9
---+-----------------------------------------------------------------
  0| 2  2   2    2     2      2       2        2         2          2
  1| 2  3   7   18    47    123     322      843      2207       5778
  2| 2  4  14   52   194    724    2702    10084     37634     140452
  3| 2  5  23  110   527   2525   12098    57965    277727    1330670
  4| 2  6  34  198  1154   6726   39202   228486   1331714    7761798
  5| 2  7  47  322  2207  15127  103682   710647   4870847   33385282
  6| 2  8  62  488  3842  30248  238142  1874888  14760962  116212808
  7| 2  9  79  702  6239  55449  492802  4379769  38925119  345946302
  8| 2 10  98  970  9602  95050  940898  9313930  92198402  912670090
  9| 2 11 119 1298 14159 154451 1684802 18378371 200477279 2186871698

  Call the algorithm which led to the creation of this table A2.

========================================================== p. 15

  A digression on naming tables
  
  Conceivably "Table 5_14_18" could be too short.  If so, then
call this table Table 5_14_18_23_52_110.  Named for the values at
(3,1), (2,2), (1,3), (3,2), (2,3), (3,3).

i\j| 0  1   2    3     4      5       6        7         8          9
---+-----------------------------------------------------------------
  0| 2  2   2    2     2      2       2        2         2          2
  1| 2  3   7  _18_   47    123     322      843      2207       5778
  2| 2  4 _14_ _52_  194    724    2702    10084     37634     140452
  3| 2 _5__23__110_  527   2525   12098    57965    277727    1330670
  4| 2  6  34  198  1154   6726   39202   228486   1331714    7761798
  5| 2  7  47  322  2207  15127  103682   710647   4870847   33385282
  6| 2  8  62  488  3842  30248  238142  1874888  14760962  116212808
  7| 2  9  79  702  6239  55449  492802  4379769  38925119  345946302
  8| 2 10  98  970  9602  95050  940898  9313930  92198402  912670090
  9| 2 11 119 1298 14159 154451 1684802 18378371 200477279 2186871698

  The trouble with this convention is that, if you somehow compute this table

i\j|  0   1    2     3      4       5        6         7          8
---+---------------------------------------------------------------
  0|  2   2    2     2      2       2        2         2          2
  1|  3   7   18   _47_   123     322      843      2207       5778
  2|  4  14  _52_  194    724    2702    10084     37634     140452
  3|  5 _23_ 110   527   2525   12098    57965    277727    1330670
  4|  6  34  198  1154   6726   39202   228486   1331714    7761798
  5|  7  47  322  2207  15127  103682   710647   4870847   33385282
  6|  8  62  488  3842  30248  238142  1874888  14760962  116212808
  7|  9  79  702  6239  55449  492802  4379769  38925119  345946302
  8| 10  98  970  9602  95050  940898  9313930  92198402  912670090
  9| 11 119 1298 14159 154451 1684802 18378371 200477279 2186871698

you think its name is Table 23_52_47, but then you see that the tables
are the same, except for column 0 being omitted from the second.

  I am not sure how to name tables.

========================================================== p. 16

  Algorithm A1

  As I created each row, I looked it up on OEIS.

  There I found each row was defined as a recurrence relation.

    a(0) = 2;  a(1) = 2;   a(n) = 2 * a(n-1) - a(n-2), n >= 2.

       2      2      2      2      2       2      2     2       2

    a(0) = 2;  a(1) = 3;   a(n) = 3 * a(n-1) - a(n-2), n >= 2.

       2      3      7     18     47     123    322   843    2207

    a(0) = 2;  a(1) = 4;   a(n) = 4 * a(n-1) - a(n-2), n >= 2.

       2      4     14     52    194     724   2702 10084   37634

which generalizes to

   a(i,0) = 2, i >= 0.
   a(i,1) = i + 2, i >= 0.
   a(i,j) = (i + 2) * a(i,j-1) - a(i,j-2) for i >= 0, j >= 2.

  Call this algorithm A1.
  
i\j | 0  1   2    3     4     5      6       7        8        9
----+-----------------------------------------------------------
   0| 2  2   2    2     2     2      2       2        2        2
   1| 2  3   7   18    47   123    322     843     2207     5778
   2| 2  4  14   52   194   724   2702   10084    37634   140452
   3| 2  5  23  110   527  2525  12098   57965   277727  1330670
   4| 2  6  34  198  1154  6726  39202  228486  1331714  7761798

  Theorem.  Algorithms A1 and A2 generate the same table.

  Proof.  (by Dr. Joseph Kirtland).  See Proof_A1_is_A2

  To me it is just astonishing that two such very different
algorithms generate exactly the same numbers.

========================================================== p. 17

  Algorithm A0

  I noticed that the elements of column 1 differed by zero, the
elements of column 1 differed by i, the elements of column 2
differed by 5, 7, 9, 11, ....  The natural question was:

  What structure exists in the columns of Table 5_14_18?  We know
that for any sequence of numbers, a0, a1, a2, ..., there is a
polynomial p(n) such that p(0)=a0, p(1)=a1, p(2)=a2, ....

  Table 5_14_18.

i\j| 0  1   2    3     4      5       6        7         8          9
---+-----------------------------------------------------------------
  0| 2  2   2    2     2      2       2        2         2          2
  1| 2  3   7   18    47    123     322      843      2207       5778
  2| 2  4  14   52   194    724    2702    10084     37634     140452
  3| 2  5  23  110   527   2525   12098    57965    277727    1330670
  4| 2  6  34  198  1154   6726   39202   228486   1331714    7761798
  5| 2  7  47  322  2207  15127  103682   710647   4870847   33385282
  6| 2  8  62  488  3842  30248  238142  1874888  14760962  116212808
  7| 2  9  79  702  6239  55449  492802  4379769  38925119  345946302
  8| 2 10  98  970  9602  95050  940898  9313930  92198402  912670090
  9| 2 11 119 1298 14159 154451 1684802 18378371 200477279 2186871698

  I took successive differences for the values in each column.

  The results were:

i\j| 0 1  i\j|  0 1 2  i\j|   0  1 2 3  i\j|    0   1  2 3 4  i\j|     0    1    2   3  4 5
---+----  ---+-------  ---+-----------  ---+----------------  ---+-------------------------
  0| 2 0    0|  2 1 0    0|   2  5 2 0    0|    2  16 18 6 0    0|     2   45  102  84 24 0
  1| 2 0    1|  3 1 0    1|   7  7 2 0    1|   18  34 24 6 0    1|    47  147  186 108 24 0
  2| 2 0    2|  4 1 0    2|  14  9 2 0    2|   52  58 30 6 0    2|   194  333  294 132 24 0
  3| 2 0    3|  5 1 0    3|  23 11 2 0    3|  110  88 36 6 0    3|   527  627  426 156 24 0
  4| 2 0    4|  6 1 0    4|  34 13 2 0    4|  198 124 42 6 0    4|  1154 1053  582 180 24 0
  5| 2 0    5|  7 1 0    5|  47 15 2 0    5|  322 166 48 6 0    5|  2207 1635  762 204 24 0
  6| 2 0    6|  8 1 0    6|  62 17 2 0    6|  488 214 54 6 0    6|  3842 2397  966 228  0 0
  7| 2 0    7|  9 1 0    7|  79 19 2 0    7|  702 268 60 0 0    7|  6239 3363 1194   0  0 0
  8| 2 0    8| 10 1 0    8|  98 21 0 0    8|  970 328  0 0 0    8|  9602 4557    0   0  0 0
  9| 2 0    9| 11 0 0    9| 119  0 0 0    9| 1298   0  0 0 0    9| 14159    0    0   0  0 0

i\j|      0     1     2    3    4   5 6  i\j|       0      1      2      3     4    5   6 7
---+-----------------------------------  ---+----------------------------------------------
  0|      2   121   480  720  480 120 0    0|       2    320   2060   4956  5736 3240 720 0
  1|    123   601  1200 1200  600 120 0    1|     322   2380   7016  10692  8976 3960 720 0
  2|    724  1801  2400 1800  720 120 0    2|    2702   9396  17708  19668 12936 4680 720 0
  3|   2525  4201  4200 2520  840 120 0    3|   12098  27104  37376  32604 17616 5400 720 0
  4|   6726  8401  6720 3360  960 120 0    4|   39202  64480  69980  50220 23016 6120   0 0
  5|  15127 15121 10080 4320 1080   0 0    5|  103682 134460 120200  73236 29136    0   0 0
  6|  30248 25201 14400 5400    0   0 0    6|  238142 254660 193436 102372     0    0   0 0
  7|  55449 39601 19800    0    0   0 0    7|  492802 448096 295808      0     0    0   0 0
  8|  95050 59401     0    0    0   0 0    8|  940898 743904      0      0     0    0   0 0
  9| 154451     0     0    0    0   0 0    9| 1684802      0      0      0     0    0   0 0


  This indicates each column is generated by a well-behaved polynomial.

========================================================== p. 18

  We want to find polynomials such that

    polynomial p[0](n) generates column 0,
    polynomial p[1](n) generates column 1,
    polynomial p[2](n) generates column 2,
    polynomial p[3](n) generates column 3, etc.

where [i] shows what would normally be written as a subscript.

  It is clear that

    p[0](n) = 2, n >= 0.

    p[1](n) = n + 2, n >= 0.

    p[2](n) = n^2 + 4 * n + 2.

      p[2](0) = 0^2  +  4 * 0  +  2  =   2
      p[2](1) = 1^2  +  4 * 1  +  2  =   7
      p[2](2) = 2^2  +  4 * 2  +  2  =  14
      p[2](3) = 3^2  +  4 * 3  +  2  =  23
      p[2](4) = 4^2  +  4 * 4  +  2  =  34
      p[2](5) = 5^2  +  4 * 5  +  2  =  47
      p[2](6) = 6^2  +  4 * 6  +  2  =  62

  Table 5_14_18.

i\j| 0  1   2    3     4      5       6        7         8          9
---+-----------------------------------------------------------------
  0| 2  2   2    2     2      2       2        2         2          2
  1| 2  3   7   18    47    123     322      843      2207       5778
  2| 2  4  14   52   194    724    2702    10084     37634     140452
  3| 2  5  23  110   527   2525   12098    57965    277727    1330670
  4| 2  6  34  198  1154   6726   39202   228486   1331714    7761798
  5| 2  7  47  322  2207  15127  103682   710647   4870847   33385282
  6| 2  8  62  488  3842  30248  238142  1874888  14760962  116212808
  7| 2  9  79  702  6239  55449  492802  4379769  38925119  345946302
  8| 2 10  98  970  9602  95050  940898  9313930  92198402  912670090
  9| 2 11 119 1298 14159 154451 1684802 18378371 200477279 2186871698
  
========================================================== p. 19

  What is p[3](n) ?

  We want to find integers a, b, c, d such that

  p[3](n) = a * n^3 + b * n^2 + c * n + d = a(n,3).

  p[3](0) =                                d =    2
  p[3](1) = a       +  b      +  c      +  d =   18
  p[3](2) = a *  8  +  b * 4  +  c * 2  +  d =   52
  p[3](3) = a * 27  +  b * 9  +  c * 3  +  d =  110

  This means solving this set of simultaneous equations:

                                     d =   2
               a  +    b  +    c  +  d =  18
             8*a  +  4*b  +  2*c  +  d =  52
            27*a  +  9*b  +  3*c  +  d = 110

  Many mistakes later I got a=1, b=6, c=9, d=2 and so

      p[3](n) = n^3  +  6 * n^2  +  9 * n  +  2

  To summarize, the first four polynomials to generate the
first four columns of Table 5_14_18 are

    p[0](n) =                               2
    p[1](n) =                     1 * n  +  2
    p[2](n) =         1 * n^2  +  4 * n  +  2
    p[3](n) = n^3  +  6 * n^2  +  9 * n  +  2

    Looking just at the coefficients, we get this table:

                                            2
                                  1         2
                      1           4         2
            1         6           9         2

========================================================== p. 20

  Hoping for a miracle in order to avoid solving larger sets
of simultaneous equations, I went to OEIS, entered "1, 6, 9, 2",
and was led to the (1,2)-Pascal triangle. (This did
not work when I tried it again more recently.  Try it yourself
and see all the unbelievable places "1, 6, 9, 2" occurs.)

  The (1,2) Pascal Triangle in table form:

i\j| 0  1  2   3   4   5   6  7  8 9
---+--------------------------------
  0| 2  0  0   0   0   0   0  0  0 0
  1| 1  2  0   0   0   0   0  0  0 0
  2| 1  3  2   0   0   0   0  0  0 0
  3| 1  4  5  _2_  0   0   0  0  0 0
  4| 1  5 _9_  7   2   0   0  0  0 0
  5| 1 _6_14  16   9   2   0  0  0 0
  6|_1_ 7 20  30  25  11   2  0  0 0
  7| 1  8 27  50  55  36  13  2  0 0
  8| 1  9 35  77 105  91  49 15  2 0
  9| 1 10 44 112 182 196 140 64 17 2

  If you look at just the upward diagonals, you see:

i\j| 0 1  2  3 4 5
---+--------------
  0| 2 0  0  0 0 0
  1| 1 0  0  0 0 0
  2| 1 2  0  0 0 0
  3| 1 3  0  0 0 0
  4| 1 4  2  0 0 0
  5| 1 5  5  0 0 0
  6|_1_6__9__2_0 0
  7| 1 7 14  7 0 0
  8| 1 8 20 16 2 0
  9| 1 9 27 30 9 0

  If you look only at the even numbered upward diagonals, you
see:

i\j| 0  1  2  3  4 5
---+----------------
  0| 2  0  0  0  0 0
  2| 1  2  0  0  0 0
  4| 1  4  2  0  0 0
  6| 1  6  9  2  0 0
  8| 1  8 20 16  2 0
 10| 1 10 35 50 25 2

  But the first four lines contain exactly the coefficients
we computed to create the first four columns of Table 5_14_18.

                2
              1 2
           1  4 2
        1  6  9 2
     1  8 20 16 2
  1 10 35 50 25 2

  The polynomials are:

  p[0](n) =                                                 2
  p[1](n) =                                             n + 2
  p[2](n) =                                  n^2 +  4 * n + 2
  p[3](n) =                       n^3 +  6 * n^2 +  9 * n + 2
  p[4](n) =            n^4 +  8 * n^3 + 20 * n^2 + 16 * n + 2
  p[5](n) = n^5 + 10 * n^4 + 35 * n^3 + 50 * n^2 + 25 * n + 2

  I wrote a program to calculate the columns of a table, using
the even diagonals of the (1,2) Pascal triangle to define the
coefficients of the polynomials.

  Call this algorithm A_ ("A blank").

  Algorithm A_ generates the same table as algorithms A1 and A2.

  By construction.

========================================================== p. 21

  Evaluating p[0](n), p[0](n), p[0](n), p[0](n), p[0](n), p[0](n)
for n = 0, 1, 2, 3, 4, 5 to show how they calculate the columns
of Table 5_14_18.

  p[0](n) =                                                 2

  p[0](0) =                                                 2 =     2
  p[0](1) =                                                 2 =     2
  p[0](2) =                                                 2 =     2
  p[0](3) =                                                 2 =     2
  p[0](4) =                                                 2 =     2
  p[0](5) =                                                 2 =     2

  p[1](n) =                                             n + 2

  p[1](0) =                                             0 + 2 =     2
  p[1](1) =                                             1 + 2 =     3
  p[1](2) =                                             2 + 2 =     4
  p[1](3) =                                             3 + 2 =     5
  p[1](4) =                                             4 + 2 =     6
  p[1](5) =                                             5 + 2 =     7

  p[2](n) =                                  n^2 +  4 * n + 2

  p[2](0) =                                  0^2 +  4 * 0 + 2 =     2
  p[2](1) =                                  1^2 +  4 * 1 + 2 =     7
  p[2](2) =                                  2^2 +  4 * 2 + 2 =    14
  p[2](3) =                                  3^2 +  4 * 3 + 2 =    23
  p[2](4) =                                  4^2 +  4 * 4 + 2 =    34
  p[2](5) =                                  5^2 +  4 * 5 + 2 =    47

  p[3](n) =                       n^3 +  6 * n^2 +  9 * n + 2

  p[3](0) =                       0^3 +  6 * 0^2 +  9 * 0 + 2 =     2
  p[3](1) =                       1^3 +  6 * 1^2 +  9 * 1 + 2 =    18
  p[3](2) =                       2^3 +  6 * 2^2 +  9 * 2 + 2 =    52
  p[3](3) =                       3^3 +  6 * 3^2 +  9 * 3 + 2 =   110
  p[3](4) =                       4^3 +  6 * 4^2 +  9 * 4 + 2 =   198
  p[3](5) =                       5^3 +  6 * 5^2 +  9 * 5 + 2 =   322

  p[4](n) =            n^4 +  8 * n^3 + 20 * n^2 + 16 * n + 2

  p[4](0) =            0^4 +  8 * 0^3 + 20 * 0^2 + 16 * 0 + 2 =     2
  p[4](1) =            1^4 +  8 * 1^3 + 20 * 1^2 + 16 * 1 + 2 =    47
  p[4](2) =            2^4 +  8 * 2^3 + 20 * 2^2 + 16 * 2 + 2 =   194
  p[4](3) =            3^4 +  8 * 3^3 + 20 * 3^2 + 16 * 3 + 2 =   527
  p[4](4) =            4^4 +  8 * 4^3 + 20 * 4^2 + 16 * 4 + 2 =  1154
  p[4](5) =            5^4 +  8 * 5^3 + 20 * 5^2 + 16 * 5 + 2 =  2207

  p[5](n) = n^5 + 10 * n^4 + 35 * n^3 + 50 * n^2 + 25 * n + 2

  p[5](0) = 0^5 + 10 * 0^4 + 35 * 0^3 + 50 * 0^2 + 25 * 0 + 2 =     2
  p[5](1) = 1^5 + 10 * 1^4 + 35 * 1^3 + 50 * 1^2 + 25 * 1 + 2 =   123
  p[5](2) = 2^5 + 10 * 2^4 + 35 * 2^3 + 50 * 2^2 + 25 * 2 + 2 =   724
  p[5](3) = 3^5 + 10 * 3^4 + 35 * 3^3 + 50 * 3^2 + 25 * 3 + 2 =  2525
  p[5](4) = 4^5 + 10 * 4^4 + 35 * 4^3 + 50 * 4^2 + 25 * 4 + 2 =  6726
  p[5](5) = 5^5 + 10 * 5^4 + 35 * 5^3 + 50 * 5^2 + 25 * 5 + 2 = 15127

========================================================== p. 22

  Encapsulate all of the above information in this table:

  5  4  3  2  1 0  |i|  0 1  2   3    4     5
  -----------------+-+-----------------------
  0  0  0  0  0 2  |0|  2 2  2   2    2     2
  0  0  0  0  1 2  |1|  2 3  7  18   47   123
  0  0  0  1  4 2  |2|  2 4 14  52  194   724
  0  0  1  6  9 2  |3|  2 5 23 110  527  2525
  0  1  8 20 16 2  |4|  2 6 34 198 1154  6726
  1 10 35 50 25 2  |5|  2 7 47 322 2207 15127

    Take line i from the LHS
    Construct a polynomial p[i](n)
      using the values in line i
    Evaluate p[i](n) for n = 0,1,2,...
      to construct column i on the RHS.

========================================================== p. 23

  It is not surprising that algorithm A_ exists.  The
introduction showed that given any sequence of integers

  n0, n1, n2, ...,

  a polynomial p(n) can be created such that

    p(0) = n0
    p(1) = n1
    p(2) = n2, etc.

========================================================== p. 24

  Here is what is surprising.

  The polynomials, p[0](n), p[1](n), p[2](n), ....,
can be defined recursively.

  p[0](n) =                                                 2
  p[1](n) =                                             n + 2
  p[2](n) =                                  n^2 +  4 * n + 2
  p[3](n) =                       n^3 +  6 * n^2 +  9 * n + 2
  p[4](n) =            n^4 +  8 * n^3 + 20 * n^2 + 18 * n + 2
  p[5](n) = n^5 + 10 * n^4 + 35 * n^3 + 50 * n^2 + 25 * n + 2

  p[0](n) = 2

  p[1](n) = n + 2

  p[2](n) = (n+2) * p[1](n) - p[0](n)
          = n^2 + 4*n + 4 - 2
          = n^2 + 4*n + 2

  p[3](n) = (n+2) * p[2](n) - p[1](n)
          = n*(n^2 + 4*n + 2) + 2*(n^2 + 4*n + 2) - n - 2
          = n^3 + 6*n^2 + 9*n + 2

  p[4](n) = (n+2) * p[3](n) - p[2](n)
          = n*(n^3 + 6*n^2 + 9*n + 2) + 2*(n^3 + 6*n^2 + 9*n + 2) - (n^2 + 4*n + 2)
          = n^4 + 6*n^3 + 9*n^2 + n*2 + 2*n^3 + 12*n^2 + 18*n + 4 - n^2 - 4*n -2
          = n^4 + 8*n^3 + 20*n^2 + 16*n + 2

  p[5](n) = (n+2) * p[4](n) - p[3](n)
          = n*(n^4 +  8 * n^3 + 20 * n^2 + 16 * n + 2) + 2*(n^4 +  8 * n^3 + 20 * n^2 + 16 * n + 2) - (n^3 +  6 * n^2 +  9 * n + 2)
          = n^5 + 8*n^4 + 20*n^3 + 16*n^2 + 2*n + 2*n^4 + 16*n^3 + 40*n^2 + 32*n + 4 - n^3 - 6*n^2 - 9*n - 2
          = n^5 + 10 * n^4 + 35 * n^3 + 50 * n^2 + 25 * n + 2

  In general,

  p[0](n) = 2; p[1](n) = n + 2; p[j](n) = (n+2) * p[j-1](n) - p[j-2](n)

  Call this algorithm A0.

========================================================== p. 25

  Here is another way of seeing the same result.

  The coefficients for the first 6 polynomials are:

  0.                     2 = p[0](n)
  1.                  1  2 = p[1](n)
  2               1   4  2 = p[2](n)
  3.          1   6   9  2 = p[3](n)
  4.      1   8  20  16  2 = p[4](n)
  5.  1  10  35  50  25  2 = p[5](n)

  The procedure for deriving them recursively is:

                           1     2   = p[1](n) = n + 2
                     x     1     2   times n + 2
                    ---------------
                           2     4
                     1     2
                    ---------------
                     1     4     4
                     -           2   minus p[0](n)
                    --------------
                     1     4     2   = p[2](n) = (n+2)*p[1](n)-p[0](n)
                     x     1     2   times n + 2
                    ---------------
                     2     8     4
               1     4     2
              ---------------------
               1     6    10     4
               -           1     2   minus p[1](n)
              --------------------
               1     6     9     2   = p[3](n) = (n+2)*p[2](n)-p[1](n)
                     x     1     2   times n + 2
              ---------------------
               2    12    18     4
         1     6     9     2
        ---------------------------
         1     8    21    20     4
         -           1     4     2   minus p[2](n)
        ---------------------------
         1     8    20    16     2   = p[4](n) = (n+2)*p[3](n)-p[2](n)
                     x     1     2   times n + 2
        ---------------------------
         2    16    40    32     4
   1     8    20    16     2
  ---------------------------------
   1    10    36    56    34     4
   -           1     6     9     2   minus p[3](n)
  ---------------------------------
   1    10    35    50    25     2   = p[5](n) = (n+2)*p[4](n-p[3](n)

========================================================== p. 26

  Conclusion of Section A

  Table 5_14_18.  The even diagonals from the (1,2) Pascal Triangle.  Source 1,2,0,2.

9  8   7   6    5    4    3   2  1 0  |i|  0  1   2    3     4      5       6        7         8          9
--------------------------------------+-+------------------------------------------------------------------
0  0   0   0    0    0    0   0  0 2  |0|  2  2   2    2     2      2       2        2         2          2
0  0   0   0    0    0    0   0  1 2  |1|  2  3   7   18    47    123     322      843      2207       5778
0  0   0   0    0    0    0   1  4 2  |2|  2  4  14   52   194    724    2702    10084     37634     140452
0  0   0   0    0    0    1   6  9 2  |3|  2  5  23  110   527   2525   12098    57965    277727    1330670
0  0   0   0    0    1    8  20 16 2  |4|  2  6  34  198  1154   6726   39202   228486   1331714    7761798
0  0   0   0    1   10   35  50 25 2  |5|  2  7  47  322  2207  15127  103682   710647   4870847   33385282
0  0   0   1   12   54  112 105 36 2  |6|  2  8  62  488  3842  30248  238142  1874888  14760962  116212808
0  0   1  14   77  210  294 196 49 2  |7|  2  9  79  702  6239  55449  492802  4379769  38925119  345946302
0  1  16 104  352  660  672 336 64 2  |8|  2 10  98  970  9602  95050  940898  9313930  92198402  912670090
1 18 135 546 1287 1782 1386 540 81 2  |9|  2 11 119 1298 14159 154451 1684802 18378371 200477279 2186871698

A0:  p[0](n) = 2, for n >= 0.
     p[1](n) = n + 2, for n >= 0.
     p[j](n) = (n + 2) * p[j-1](n) - p[j-2](n), for j > 1, n >= 0.

A1:  a(i,0) = 2, for i >= 0.
     a(i,1) = i + 2, for i >= 0.
     a(i,j) = (i + 2) * a(i,j-1) - a(i,j-2) for j > 1, i >= 0.

  Note the symmetry between A0 and A1.
  
  Note the syntax for A0 and A1 are (almost) identical.
  
  Note that the semantics for A0 and A1 are very different!  A1
defines calculations among integers.  A0 describes a recursive
procedure for generating polynomials.
 
    a(2,3) = 52 = (2 + 2) * 14 - 4 = 56 - 4 = 52
    
    p[3](n) = (n + 2) * (n^2 + 4*n + 2) - (n + 2)
            = n^3 + 4*n^2 + 2*n + 2*n^2 + 8*n + 4 - n - 2
            = n^3 + 6*n^2 + 9*n + 2
            
so  p[3](2) = 8   + 6*4   + 9*2 + 2
            = 52
            
  
  Question.  Is Table 5_14_18 one of a kind?  Or are there 
other examples?

  Answer:  There are lots more.  See Section D.

==========================================================

  From nothing, beauty.

==========================================================
========================================================== p. 27

  Section B.  Online Encyclopedia of Integer Sequences (OEIS)

  History

    Founded by Neil Sloane
      He began collecting sequences in 1964
      Published sequences in two books, in 1973 and 1995
      Created OEIS in 1996 with 10,000 sequences
      OEIS now contains 315,000 sequences
      CV for Sloane is at Sloane_CV
      48,611 citations on Google Scholar as of June 29, 2016

  Using OEIS.

      To look up a sequence or an A-number
        oeis.org

        and enter the sequence or A-number in the box

      Go to www.oeis.org/Submit.html
        1.  To create an account
              You have to create an account before you can submit something to OEIS.
        2.  To login
        3.  To submit new work or a comment on the work of others

      How to log in
        Go to https://oeis.org/
        Click 'login' in the upper right corner

      How to edit
        First log in.  Then go to the array entry.
        There will be the title line; then entries in the array;
        then a line beginning "list; graph; . . .".  Click on "edit;"
        in this sequence.

      For more info on submitting
        https://oeis.org/wiki/Main_Page

      Contributor's License
        https://oeis.org/wiki/The_OEIS_Contributor%27s_License_Agreement

        You yield total editorial control to OEIS
          when you create an account.

      If you submit something, you will probably not do it perfectly.
        The editors may tidy up trivial format problems.
        They will NOT fix mathematical problems.

      Communication between contributor and editor(s) is rigid.
        You submit.  They email a response, such as "this is not clear".
        You have to read their minds in order to figure out what
          they think is a problem and then fix it.

      You edit your contribution in order to fix mistakes.
        You wait for a response from the editors.  Nothing happens.
        Finally you figure out that you have to click

          "The changes are ready for review by an OEIS Editor."

        at the end of your submission.  This tells the editors
        you (think you) have finished correcting your errors.

      There are no email addresses for editors or contributors.

      All communication with OEIS is done in this constrained format.
        OEIS has two reasons for this
        OEIS is not going to do contributors' work for them
        OEIS thereby maintains a complete record of communication
          between contributors and editors

========================================================== p. 28

  Fun Stuff

       OEIS can be fascinating to browse.  For example...
       oeis.org/A001835		a(n) = 4*a(n-1) - a(n-2), with a(0)=1, a(1)=1.
       1, 1, 3, 11, 41, 153, 571, 2131, 7953, 29681, 110771, 413403, 1542841,...
       sqrt(3) = 2/2 + 2/3 + 2/(3*11) + 2/(11*41) + 2/(41*153) + 2/(153*571)...
         - Gary W. Adamson, Dec 18 2007
       But OEIS only states results.  It does not explain.  Frustrating.

       Someplace OEIS says there is an example of two distinct sequences
         which agree for the first 14,000+ entries, but then disagree.  
         If you find that reference, please let me know.

       What is the smallest positive integer NOT on OEIS?
       I accidentally stumbled onto 1044486.
       Here is an idea for a sequence: all of the integers
         which do not otherwise occur on OEIS.
       XKCD already beat me to this; see xkcd.com/2016/

       The index to OEIS is at http://oeis.org/wiki/Index_to_OEIS.  Great fun to
       browse through.  Try going here and clicking on "humble numbers".
       http://oeis.org/wiki/Index_to_OEIS:_Section_Ho

==========================================================
========================================================== p. 29

  Section C.  More algorithms.

  We use OEIS to discover more patterns.

  We showed above that Table 5_14_18 could be generated by
algorithms A0, A1, and A2.

  Here we go through all the entries on OEIS for lines in Table
5_14_18 with the goal of finding more algorithms which generate
the table.

  More algorithms

==========================================================
========================================================== p. 30

  Section D.  How to produce wisterian tables.
    
    Method 1.
    
      1.  Create an (a,b) Pascal triangle.
      
                        a              b  
             
                    a         a+b          b  
                     
                a        2a+b     a+2b        b  
                  
            a      3a+b      3a+3b    a+3b       b  
   
         a    4a+b     6a+4b     4a+6b     a+4b     b  
      
      
      2.  Extract the upward diagonals.
    
      2.  Select subsequences of the diagonals.
    
            all, even, or odd.
    
      3.  Create polynomials p[i](n), using the values
          in the diagonals as coefficients.
    
      4.  Create column i of a table by evaluating
          polynomial p[i](n) for n = 0,1,2,3,....
    
      Many examples are shown at More tables

    Method 2.  Seek a way to generate wisteria tables
               independently of Pascal triangles.
      
      Look at the A0 rules for generating tables by Method 1.
      
      Discover they all fit in a tight mold.
      
      Deliberately create an A0 rule which does not fit the mold.
      
      Create the table.  Find that it is wisterian.
      
      Wonder:  is it really outside the realm of tables stemming
      from Pascal triangles?
      
==========================================================
========================================================== p. 31

  Section E.  Conclusion.

  Definition.  A table T of integers is wisterian if there exists
a sequence A0 of recursively defined polynomials p[i](n) such
that polynomial p[i](n) generates column i of table T when
evaluated for n = 0,1,2,....
  
  Conjecture.  For every wisteria table T with generating
polynomials A0 there is a sequence of recursion relations A1 such
that the i-th recursion relation defines the integers in row i of
T.  Further, A0 and A1 are syntactically identical (though
semantically distinct since one defines polynomials and the other
defines recurrence relations).
    
  What have we accomplished?
  
  1.  Claim.  Wisteria tables are objects of mathematical
interest in and of themselves.

  2.  Claim.  Arranging OEIS entries in tables, wisteria or
otherwise, will benefit OEIS in three ways.
 
    a.  It will reveal gaps where a rule is defined for some
entries, but not others, and should be defined for all.  This
will encourage the gaps to be filled.
  
    b.  It will create deeper understanding of connections
between entries within a table.

    c.  It could raise questions about finding connections 
between different tables.

========================================================== p. 32

  Research

  1.  Prove the above conjecture.

  2.  Find wisteria tables which are demonstrably not derived
from Pascal triangles.

  3.  Given a table which obeys algorithms A0, A1, A2, A3, A4,
A5, and A6 (based on an infinitesimally small sample of the upper
lefthand corner of the table), prove mathematically that the
entire table is created by each of the algorithms.

  4.  Scavenge further.  For each table in Section D
identify what flavor of A2, A3, A4, A5, and A6 it may obey.

  5.  Create a program which, given a table, shows what rules
the rows and columns obey.  (This is impossibly general.)

  6.  Generate (a,b)-Pascal triangles where a and b can be negative.
See what happens.

  7.  Generate Pascal triangles where the third row consists of
arbitrary integers (a,b,c).  See what happens.

========================================================== p. 33

  Further Reading

  Experimental Mathematics.    en.wikipedia.org/wiki/Experimental_mathematics
  Interview with Neil Sloane.  https://www.quantamagazine.org/neil-sloane-connoisseur-of-number-sequences-20150806/?utm_content=buffere5b2b&utm_medium=social&utm_source=facebook.com&utm_campaign=buffer
  "The On-Line Encyclopedia of Integer Sequences" by Neil Sloane in Notices of the AMS, Volume 65, Number 9.
                               https://www.ams.org/journals/notices/201809/rnoti-p1062.pdf
  Linear Recurrence Equation.  http://mathworld.wolfram.com/LinearRecurrenceEquation.html
  Generating Function.         http://mathworld.wolfram.com/GeneratingFunction.html
  Stephen Wolfram Blog.        http://blog.stephenwolfram.com/2017/03/two-hours-of-experimental-mathematics/

  Some of the opinions of Doron Zeilberger which relate to the interests of the chapter.
  Opinion149.  Tic-Tac-Toe, Checkers, Chess, Go, and next is MATHEMATICS! (Written: March 22, 2016)
  Opinion124.  A Database is Worth a Thousand Mathematical Articles: An Ode 
               to Neil Sloane's On-line Encyclopedia of Integer Sequences (OEIS) (Written: May 22, 2012)
  Opinion37.   Guess What? Programming is Even More Fun Than Proving, and, More Importantly 
               It Gives As Much, If Not More, Insight and Understanding" (Written: April 15, 1999)
               
  Acknowledgment.  Many, many thanks to Bill Rubin for acting as
sounding board for ideas, for pushing the use of spell check and
html check, and for doing a great job as webmaster for the ACM
chapter site in general and for these foils in particular.