A293704 a(n) is the shift of the longest palindromic subsequence in the first n terms of A293700.
0, -1, 0, -1, 0, -1, 0, -1, 0, -1, 0, -1, 0, -1, 0, -1, 0, -1, -2, -3, -4, -5, -6, -7, -8, -9, -10, -11, -12, -13, -14, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, -1, -2, -3, -4, -5, -6, -7, -8, -9, -10, -11, -12, -13, -14, -15, -16, -17, -18, 18, 17, 16
Offset: 1
Keywords
Examples
For n = 1, first differences = 3; longest palindrome = 3; a(1) = 0 - 0 = 0. For n = 2, differences = 3, 19; longest palindrome = 3; a(2) = 0 - 1 = -1. For n = 22, differences = 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 3, 16, 3, 3, 16; longest palindrome = 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3; a(22) = 0 - 5 = -5.
Links
- V.J. Pohjola, Table of n, a(n) for n = 1..10000
- V. J. Pohjola, Line plot drawn for n=1...150
- V. J. Pohjola, Line plot drawn for n=1...2500
- V. J. Pohjola, Line plot drawn for n=1...10000
Programs
-
Mathematica
rootsp = Flatten[Position[Table[Floor[Tan[i]], {i, 1, 10^4}], 1]]; lrootsp = Length[rootsp]; difp = Differences[rootsp]; ldp = Length[difp]; kmax = 500; palsp = {}; lenpalsp = {0}; shiftp = {}; posp = {}; Do[diffip = difp[[1 ;; k]]; lendiffip = Length[diffip]; pmax = k - Last[lenpalsp]; t = Table[difp[[p ;; k]], {p, 1, pmax}]; sp = Flatten[Select[t, # == Reverse[#] &]]; If[sp == {}, AppendTo[palsp, Last[palsp]] && AppendTo[lenpalsp, Last[lenpalsp]], AppendTo[palsp, sp] && AppendTo[lenpalsp, Length[Flatten[sp]]]]; AppendTo[posp, Position[t, Last[palsp]]]; pp = Last[Flatten[posp]] - 1; qq = lendiffip - (pp + Last[lenpalsp]); AppendTo[shiftp, pp - qq], {k, 1, kmax}]; lenpalsp; shiftp (*a(n)=shiftp[[n]]*)
Comments