A293701 a(n) is the length of the longest palindromic subsequence in the first n terms of A293700.

1, 1, 3, 3, 5, 5, 7, 7, 9, 9, 11, 11, 13, 13, 15, 15, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 47, 49

Offset: 1

Author

V.J. Pohjola, Oct 16 2017

Keywords

• nonn
• easy

Comments

If a(n + 1) > a(n) for some n, then A293700(n + 1) is in the longest palindrome. So to find a(n + 1) it suffices to check if A293700 is in the palindrome, which must be at least of length a(n). - David A. Corneth, Nov 25 2017

At points where a(n) = n, the whole sequence is a palindrome. For example at n=9105, the length of the longest palindrome a(9105) is 9105 (see A294923).

Examples

			For n = 1, roots = 1, 4; first differences = 3; longest palindrome = 3; a(n) = 1.
For n = 2, roots = 1, 4, 23; first differences = 3, 19; longest palindrome = 3; a(n) = 1.
For n = 3, roots = 1, 4, 23, 26; first differences = 3, 19, 3; longest palindrome = 3, 19, 3; a(n) = 3.
For n = 33, roots = 1, 4, 23, 26, 45, 48, 67, 70, 89, 92, 111, 114, 133, 136, 155, 158, 177, 180, 183, 199, 202, 205, 221, 224, 227, 243, 246, 249, 265, 268, 271, 290, 293, 312; first differences = 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 3, 16, 3, 3, 16, 3, 3, 16, 3, 3, 16, 3, 3, 19, 3, 19; longest palindrome = 19, 3, 19, 3, 3, 16, 3, 3, 16, 3, 3, 16, 3, 3, 16, 3, 3, 19, 3, 19; a(n) = 20.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10000 (computed with the given Scheme-program from the b-file of A293700 provided by Robert Israel)
• V.J. Pohjola, Line plot of n, a(n) for n=1...100
• V.J. Pohjola, Line plot of n, a(n) for n=1...1400
• V.J. Pohjola, Line plot of n, a(n) for n=1...10000

Crossrefs

Cf. A293698, A293699, A293700, A293702, A293703, A293704, A293705, A293706, A293751.

Programs

Mathematica

rootsp = Flatten[Position[Table[Floor[Tan[i]], {i, 1, 10^4}], 1]];
lrootsp = Length[rootsp];
difp = Differences[rootsp];
ldp = Length[difp];
nmax = 200; palsp = {}; lenpalsp = {0};
Do[diffip = difp[[1 ;; n]]; lendiffip = Length[diffip];
  pmax = n - Last[lenpalsp];
  t = Table[difp[[p ;; n]], {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]]]], {n, 1, nmax}];
Drop[lenpalsp,1](*a(n)=Drop[lenpalsp,1][[n]]*)

PARI

firstsols(n) = {my(res = vector(n), i = 0, pi = [Pi, Pi], sols = [atan(1), atan(2)]); while(1, for(j = ceil(sols[1]), floor(sols[2]), i++; if(i>n, return(res)); res[i] = j); sols+=[Pi(), Pi()])} \\ A293698
diff(v) = vector(#v-1,i,v[i+1]-v[i]);
first(n) = {my(res = vector(n), m = 0, check = diff(firstsols(n+1))); for(i=1, n, for(j = 1, i - m, if(ispalindrome(check, j, i), m = i - j + 1; next(1))); res[i] = m); res}
ispalindrome(v, {llim = 1}, {ulim = #v}) = {for(i=0, (ulim - llim) \ 2, if(v[llim + i]!=v[ulim - i], return(0))); 1} \\ David A. Corneth, Nov 25 2017

Scheme

;; This uses memoization-macro definec and assumes also that A293700 is available:
(definec (A293701 n) (if (= 1 n) n (let outloop ((k n)) (cond ((<= k (A293701 (- n 1))) (A293701 (- n 1))) (else (let inloop ((i n)) (let ((low-ind (+ 1 (- n k) (- n i)))) (cond ((< i low-ind) (max k (A293701 (- n 1)))) ((not (= (A293700 i) (A293700 low-ind))) (outloop (- k 1))) (else (inloop (- i 1)))))))))))
;; Antti Karttunen, Nov 25 2017