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A293702
a(n) is the length of the longest palindromic subsequence in the first n terms of A293751.
11
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
1,3
EXAMPLE
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.
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]] *)
KEYWORD
nonn
AUTHOR
V.J. Pohjola, Oct 16 2017
STATUS
approved