A294923 -id:A294923 - cp's OEIS frontend

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

V.J. Pohjola, Oct 16 2017

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).

			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.

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

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

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]]*)

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

;; 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

A294924 Numbers n such that the whole sequence of the first n terms of A293699 is a palindrome.

Original entry on oeis.org

1, 3, 5, 7, 26, 63, 100, 137, 174, 211, 248, 285, 322, 359, 396, 433, 470, 507, 544, 581, 618, 655, 692, 729, 766, 803, 840, 877, 914, 951, 988, 1025, 1062, 1099, 1136, 1173, 1210, 1247, 1284, 1321, 1358, 1395, 1432, 1469, 1506, 1543, 1580, 1617, 1654, 1691, 1728, 1765, 1802, 1839, 1876, 1913, 1950, 1987, 2024, 2061, 2098, 2135

Offset: 1

V.J. Pohjola, Nov 11 2017

nonn

A293699 are the first differences of A293751 which are the positive integers i such that floor(tan(-i))=1.

A293702 are the lengths of the longest palindromic subsequences in the first n terms of A293699.

			The first 7 terms of A293699 are (3, 19, 3, 19, 3, 19, 3) which is a palindromic sequence, so 7 is a term.
The first 8 terms of A293699 are (3, 19, 3, 19, 3, 19, 3, 3) which is not a palindromic sequence, so 8 is not a term.
The first 9 terms of A293699 are (3, 19, 3, 19, 3, 19, 3, 3, 16) which is not a palindromic sequence, so 9 is not a term.
The first 25 terms of A293699 are (3, 19, 3, 19, 3, 19, 3, 3, 16, 3, 3, 16, 3, 3, 16, 3, 3, 16, 3, 3, 19, 3, 19, 3, 19) which is not a palindromic sequence, so 25 is not a term.
The first 26 terms of A293699 are (3, 19, 3, 19, 3, 19, 3, 3, 16, 3, 3, 16, 3, 3, 16, 3, 3, 16, 3, 3, 19, 3, 19, 3, 19, 3) which is a palindromic sequence, so 26 is a term.

V.J. Pohjola, Line plot for n=1..7
V.J. Pohjola, Line plot for n=1..85

Cf. A293702, A293699, A293751, A294923.

rootsn7 = Flatten[Position[Table[Floor[Tan[-n]], {n, 1, 10^7}], 1]];
difn7 = Differences[rootsn7];
ny = {}; Do[
If[Table[difn7[[i]], {i, 1, n}] == Reverse[Table[difn7[[i]], {i, 1, n}]],
  AppendTo[ny, n]], {n, 1, Length[difn7]}]
ny

cp's OEIS Frontend

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

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