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

1,3

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).

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

Example

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.

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]]*)

Prog

(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

Crossrefs

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

Sequence in context: A206913 A117767 A340836 * A296063 A127630 A267458

Adjacent sequences: A293698 A293699 A293700 * A293702 A293703 A293704

Keyword

nonn,easy

Author

V.J. Pohjola, Oct 16 2017

Status

approved