A293702 a(n) is the length of the longest palindromic subsequence in the first n terms of A293751.
1, 1, 3, 3, 5, 5, 7, 7, 7, 7, 7, 7, 7, 8, 8, 9, 11, 11, 12, 14, 16, 18, 20, 22, 24, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 26, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 63, 63, 63, 63, 63, 63, 63
Offset: 1
Keywords
Examples
For n = 1, Roots = 18, 21; First differences = 3; Longest palindrome = 3; a(n) = 1. For n = 2, Roots = 18, 21, 40; First differences = 3, 19; Longest palindrome = 3; a(n) = 1. For n = 3, Roots = 18, 21, 40, 43; First differences = 3, 19, 3; Longest palindrome = 3, 19, 3; a(n) = 3. For n = 20, Roots = 18, 21, 40, 43, 62, 65, 84, 87, 90, 106, 109, 112, 128, 131,134, 150, 153, 156, 172, 175; First differences = 3, 19, 3, 19, 3, 19, 3, 3, 16, 3, 3, 16, 3, 3, 16, 3, 3, 16, 3, 3; Longest palindrome = 3, 3, 16, 3, 3, 16, 3, 3, 16, 3, 3, 16, 3, 3; a(n) = 14.
Links
- V.J. Pohjola, Table of n, a(n) for n = 1..10000
- V.J. Pohjola, Line plot for n=1…30
- V.J. Pohjola, Line plot for n=1...10000
Programs
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Mathematica
rootsn = Flatten[Position[Table[Floor[Tan[-i]], {i, 1, 10^4}], 1]]; difn = Differences[rootsn]; imax = 100; palsn = {}; lenpalsn = {0}; Do[diffin = difn[[1 ;; i]]; lendiffin = Length[diffin]; pmax = i - Last[lenpalsn]; t = Table[difn[[p ;; i]], {p, 1, pmax}]; sn = Flatten[Select[t, # == Reverse[#] &]]; If[sn == {}, AppendTo[palsn, Last[palsn]] && AppendTo[lenpalsn, Last[lenpalsn]], AppendTo[palsn, sn] && AppendTo[lenpalsn, Length[Flatten[sn]]]], {i, 1, imax}]; Drop[lenpalsn, 1] (* a(n)=Drop[lenpalsn, 1][[n]] *)