Re: [math-fun] two "trivia"

2 Aug 2015


      (Gack, Gmail replaced a string of fixed width spaces by skinny "sans"
spaces.)
FOO!  I've gone senile.  Years ago I sent this very List a matrix product
of the form
0 A B
C 0 D
0 0 1
As NeilB points out, these matrices grouped pairwise become triangular!
(Suggested name:  cryptotriangular?)

On Fri, Jul 31, 2015 at 12:52 AM, Bill Gosper <billgosper@gmail.com> wrote:
...
A miscellaneous nontriangular matrix product:
                                            [        sqrt(%pi)  ]
                                            [ 0  0   ---------  ]
      [             1      1    5    1  ]   [             2 3   ]
 oo   [     0       -- - -----  - - --- ]   [        gamma (-)  ]
/===\ [             16       2  4   8 j ]   [               4   ]
 | |  [                  64 j           ]   [                   ]
 | |  [                                 ] = [              2 3  ]
 | |  [  1    1  2                      ]   [       4 gamma (-) ]
j = 1 [ (- - ---)       0          1    ]   [                4  ]
      [  2   4 j                        ]   [ 0  0  ----------- ]
      [                                 ]   [            3/2    ]
      [     0           0          1    ]   [         %pi       ]
                                            [                   ]
                                            [ 0  0       1      ]
(Does anybody know how to make these archaic ASCII displays in Mathematica,
This was rendered by archaic Macsyma.
...
for our archaic math-fun server?)
So grouping pairwise and converting to standard notation:
Out[372]= {{0, 0, (1/(
   64 Sqrt[\[Pi]]
     Gamma[3/
     4]^2))(91 Sqrt[\[Pi]]
       Gamma[3/
       4]^2 HypergeometricPFQ[{1/4, 3/4, 3/4, 3/4}, {1, 3/2, 3/2}, 1/
       64] - 352 Sqrt[\[Pi]]
       Gamma[3/
       4]^2 HypergeometricPFQ[{1/4, 3/4, 3/4, 3/4, 2}, {1, 1, 3/2, 3/
        2}, 1/64] +
     336 Sqrt[\[Pi]]
       Gamma[3/
       4]^2 HypergeometricPFQ[{1/4, 3/4, 3/4, 3/4, 2, 2}, {1, 1, 1, 3/
        2, 3/2}, 1/64])}, {0, 0, (1/(
   256 Sqrt[\[Pi]]
     Gamma[1/
     4]^2))(460 Sqrt[\[Pi]]
       Gamma[1/
       4]^2 HypergeometricPFQ[{1/4, 1/4, 3/4, 5/4}, {1, 3/2, 3/2}, 1/
       64] - 9 Sqrt[\[Pi]]
       Gamma[1/
       4]^2 HypergeometricPFQ[{1/4, 1/4, 3/4, 5/4}, {3/2, 3/2, 2}, 1/
       64] - 1520 Sqrt[\[Pi]]
       Gamma[1/
       4]^2 HypergeometricPFQ[{1/4, 1/4, 3/4, 5/4, 2}, {1, 1, 3/2, 3/
        2}, 1/64] +
     1344 Sqrt[\[Pi]]
       Gamma[1/
       4]^2 HypergeometricPFQ[{1/4, 1/4, 3/4, 5/4, 2, 2}, {1, 1, 1, 3/
        2, 3/2}, 1/64])}, {0, 0, 1}} == {{0, 0, Sqrt[\[Pi]]/
   Gamma[3/4]^2}, {0, 0, (4 Gamma[3/4]^2)/\[Pi]^(3/2)}, {0, 0, 1}}

In[373]:= N[%]

Out[373]= True
Mathematica had to use multiple pFqs due to irreducible polynomials in the
term ratios.
Since they're pFq[1/64], the nice little undoubled matrix is 3 bits/term.
So why don't people use matrix products?
--rwg
...
A260747, Consolidated Dragon Curve triple points.
If D:[0,1] is a Dragon curve, then besides n, there are two other integers
p and q with
D(A(n)/(15*2^k)) = D(A(p)/(15*2^k)) = D(A(q)/(15*2^k)),
where k is any integer > lg(max(A(n),A(p),A(q))/15)
13, 21, 23, 26, 37, 39, 42, 46, 47, 52, 73, 74, 78, 81, 83, 84, 92, 94,
97, 99,
Note the gap of 73 - 52 = 21 at position 10.  There are apparently no
larger gaps, and the next gap
of 21 is at position 17651.  --rwg