(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