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Proof algorithm for composite cases of A269254

Posted 9 years ago
POSTED BY: Brad Klee
Posted 9 years ago

After more calculation, we can generalize the quadratic case to the next higher cubic case, as first discussed here. Code for verifying the induction hypothesis and all base cases follows:

t1 = AbsoluteTiming[
  BasisVectors[m_] := Flatten@Outer[
     Times, X[k - #] & /@ Range[2], Y[k - #] & /@ Range[m]];
  T[n_, j_] := Expand[2 ChebyshevT[n, j/2]];
  T2[n_, j_] := T[n, T[n, j]];
  cx = T[3, j];
  cy = Expand[1 + T[3, T[2, j]]];
  cz = Expand[T2[3, j]];
  rep3 = {
    Z[k] -> cz*Z[k - 1] - Z[k - 2],
    X[k] -> cx*X[k - 1] - X[k - 2],
    Y[k] -> cy*Y[k - 1] - cy*Y[k - 2] + Y[k - 3]
    };]

t2 = AbsoluteTiming[
  v012 = Outer[Coefficient, 
     NestList[Expand[# /. k -> k + 1 /. rep3] &, 
      Expand[( Z[k] - ReplaceAll[Z[k], rep3]) /. 
         Z[x_] :> X[x] Y[x] /. rep3]
      , 2], BasisVectors[3], 1];
  ]

t3 = AbsoluteTiming[Expand@Dot[{1, -cx, 1}, v012]]    (* TEST 1 *)

t4 = AbsoluteTiming[
  rec9 = Expand@
    FoldList[Plus, -1, T[#, T[3, j]] & /@ Range[0, 8]][[2 ;; -1]];
  rec6x = {1, 1, Fold[Plus, -1, T[#, j] & /@ Range[0, 2]],
    Fold[Plus, -1, T[#, j] & /@ Range[0, 3]], 1 + T[3, j],
    Fold[Plus, -1, T[#, j] & /@ Range[0, 5]]};
  rec9x  = 
   Join[rec6x, 
    Expand[Plus @@ Times[Partition[rec6x, 3], {-1, T[3, j]}]]];
  rec9y = Expand[Factor@Expand[rec9/rec9x]];
  ]

t5 = AbsoluteTiming[
  Expand@Outer[Expand@
      Dot[
       BasisVectors[3] /. {X[x_] :> rec9x[[3 (x /. k -> 4) - #1]], 
         Y[x_] :> rec9y[[3 (x /. k -> 4) - #1]]}, #2 ] &, {0, 1, 2}, 
    v012, 1]
  ]    (* TEST 2 *)

In time $t < .1s$, the algorithm verifies all requisite zero sums. At "TEST 1", the return is $6$ zeros, while $9$ zeros are returned at "TEST 2". These zero sums depend mainly on the following cubic invariants:

v012
rec9
rec9x
rec9y

outpolys

Discussion of base case will follow on seqfans soon.

POSTED BY: Brad Klee
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