A293702 -id:A293702 - cp's OEIS frontend

A293698 Values of positive integer i such that floor(tan(i)) = 1.

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, 315, 334, 337, 356, 359, 378, 381, 400, 403, 422, 425, 444, 447, 466, 469, 488, 491, 510, 513, 532, 535, 538, 554, 557, 560, 576, 579, 582, 598, 601, 604, 620

Offset: 1

V.J. Pohjola, Oct 15 2017

nonn

The sequence is the first result in the chain of iteration leading to the ultimate sequence A258024.
Sequence terms are also the roots of A000503(i)=1, starting from i=1.
This is a subsequence of A258024 from which this differs for the first time at n=11, where a(11) = 111, while A258024(11) = 105, the term not included in this sequence. Note that A000503(105) = 4, a term which is included in this sequence. - Antti Karttunen, Oct 30 2017
Numbers k such that Pi/4 <= k - m*Pi < arctan(2) for some m. - Robert Israel, Nov 06 2017

			The values of floor(tan(i)), starting from i=0, are given in A000503. Those i, for which floor(tan(i))=1 is true, are the roots of this equation. Thus the roots are the positions of 1 in A000503(i>0).
For n=1, i=1; a(1)=1.
For n=2, i=4; a(2)=4.
For n=3, i=23; a(3)=23.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A000503, A258024, A293751, A293700, A293701, A293704, A293699, A293702, A293705, A004112, A024814.

rootsp = Flatten[Position[Table[Floor[Tan[i]], {i, 1, 10^3}], 1]]
(* a(n) = rootsp[[n]] *)
(* Alternatively: *)
rootsp = {}; Do[If[Floor[Tan[n]] == 1, AppendTo[rootsp, n]], {n, 1, 10^3}]
rootsp (* a(n) = rootsp[[n]] *)
Select[ Range@ 622, Floor@ Tan@ # == 1 &] (* Robert G. Wilson v, Nov 06 2017 *)

isok(n) = floor(tan(n)) == 1; \\ Michel Marcus, Oct 24 2017

first(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()])} \\ David A. Corneth, Oct 24 2017

A293700 First differences of A293698.

Original entry on oeis.org

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, 3, 19, 3, 19, 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, 3, 19, 3, 19, 3, 19, 3

Offset: 1

V.J. Pohjola, Oct 16 2017

nonn

Sequence seems to be composed of only three different integers: 3, 16 and 19.
Despite its apparent simplicity, it has interesting palindromic and periodic features and may be conjectured not to be represented in a closed form.
It has a resemblance to the sequences in DNA being composed of four nucleotide bases in varying orders. These sequences, too, contain palindromic substructures having an important role for the genome.
From Robert Israel, Nov 06 2017: (Start)
The only possible values are 3, 16 and 19.
k is in A293698 iff Pi/4 <= k - m*Pi < arctan(2) for some m. We may then verify the following:
If Pi/4 <= k - m*Pi < arctan(2) - 16 + 5*Pi, then k+16 is the next term of A293698.
If arctan(2) - 16 + 5*Pi <= k - m*Pi < 5*Pi/4 - 3, then k+19 is the next term of A293698.
If 5*Pi/4 - 3 <=  k - m*Pi < arctan(2), then k+3 is the next term of A293698. (End)

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A258200, A258007, A293701, A293704, A293698, A293751, A293699, A293702, A293705.

A293698:= select(i -> floor(tan(i))=1, [$1..1000]):
A293698[2..-1]-A293698[1..-2]; # Robert Israel, Nov 06 2017

rootsp = Flatten[Position[Table[Floor[Tan[i]], {i, 1, 10^6}], 1]];
difp = Differences[rootsp]
(*a(n)=difp[[n]]*)
Differences@ Select[ Range@750, Floor@ Tan@# == 1 &] (* Robert G. Wilson v, Nov 06 2017 *)

lista(nn) = {last = 0; for (n=1, nn, if (floor(tan(n)) == 1, if (last, print1(n-last, ", ")); last = n;););} \\ Michel Marcus, Oct 24 2017

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

Original entry on oeis.org

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

A293751 Values of positive integer i such that floor(tan(-i)) = 1.

Original entry on oeis.org

18, 21, 40, 43, 62, 65, 84, 87, 90, 106, 109, 112, 128, 131, 134, 150, 153, 156, 172, 175, 178, 197, 200, 219, 222, 241, 244, 263, 266, 285, 288, 307, 310, 329, 332, 351, 354, 373, 376, 395, 398, 417, 420, 439, 442, 445, 461, 464, 467, 483, 486, 489, 505, 508, 511, 527, 530, 533, 552, 555, 574, 577, 596

Offset: 1

V.J. Pohjola, Oct 15 2017

nonn

Robert G. Wilson v, Table of n, a(n) for n = 1..1040

Cf. A000530, A293698, A293699, A293702, A293705, A293700, A293701, A293704. a(n) are also the roots of A195910(i)=1, starting from i=1.

rootsn = Flatten[Position[Table[Floor[Tan[-i]], {i, 1, 10^3}], 1]];
(*a(n) = rootsn[[n]]*)
Select[Range@600, Floor@Tan@-# == 1 &] (* Robert G. Wilson v, Nov 19 2017 *)

isok(n) = floor(tan(-n)) == 1; \\ Michel Marcus, Nov 03 2017

floor(tan(-n)) = -ceiling(tan(n)).

A293699 First differences of A293751.

Original entry on oeis.org

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, 19, 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, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3

Offset: 1

V.J. Pohjola, Oct 16 2017

nonn

Sequence seems to be composed of three different integers 3, 16 and 19. Further comments in A293700.

Cf. A293751, A293698, A293700, A293702, A293705, A293701, A293704.

rootsn = Flatten[Position[Table[Floor[Tan[-n]], {n, 1, 10^6}], 1]];
difn = Differences[rootsn];

A293704 a(n) is the shift of the longest palindromic subsequence in the first n terms of A293700.

Original entry on oeis.org

0, -1, 0, -1, 0, -1, 0, -1, 0, -1, 0, -1, 0, -1, 0, -1, 0, -1, -2, -3, -4, -5, -6, -7, -8, -9, -10, -11, -12, -13, -14, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, -1, -2, -3, -4, -5, -6, -7, -8, -9, -10, -11, -12, -13, -14, -15, -16, -17, -18, 18, 17, 16

Offset: 1

V.J. Pohjola, Oct 21 2017

sign

Shift is the measure of the position of the palindromic subsequence within the corresponding sequence of first differences, defined as the number of terms being dropped from the left end of the sequence of first differences minus those dropped from its right end. Thus, when shift is negative, the palindrome has moved leftward from its symmetric position.

			For n = 1, first differences = 3; longest palindrome = 3; a(1) = 0 - 0 = 0.
For n = 2, differences = 3, 19; longest palindrome = 3; a(2) = 0 - 1 = -1.
For n = 22, differences = 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 3, 16, 3, 3, 16; longest palindrome = 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3; a(22) = 0 - 5 = -5.

V.J. Pohjola, Table of n, a(n) for n = 1..10000
V. J. Pohjola, Line plot drawn for n=1...150
V. J. Pohjola, Line plot drawn for n=1...2500
V. J. Pohjola, Line plot drawn for n=1...10000

Cf. A293698, A293751, A293700, A293701, A293705, A293699, A293702, A293706, A293703.

rootsp = Flatten[Position[Table[Floor[Tan[i]], {i, 1, 10^4}], 1]];
lrootsp = Length[rootsp];
difp = Differences[rootsp];
ldp = Length[difp];
kmax = 500; palsp = {}; lenpalsp = {0}; shiftp = {}; posp = {};
Do[diffip = difp[[1 ;; k]]; lendiffip = Length[diffip];
  pmax = k - Last[lenpalsp];
  t = Table[difp[[p ;; k]], {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]]]];
  AppendTo[posp, Position[t, Last[palsp]]]; pp = Last[Flatten[posp]] - 1;
  qq = lendiffip - (pp + Last[lenpalsp]);
  AppendTo[shiftp, pp - qq], {k, 1, kmax}];
lenpalsp;
shiftp (*a(n)=shiftp[[n]]*)

A293705 a(n) is the shift of the longest palindromic subsequence in the first n terms of A293699.

Original entry on oeis.org

0, -1, 0, -1, 0, -1, 0, -1, -2, -3, -4, -5, -6, 6, 5, 7, 6, 5, 7, 6, 5, 4, 3, 2, 1, 0, -1, -2, -3, -4, -5, -6, -7, -8, -9, -10, -11, -12, -13, -14, -15, -16, -17, -18, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, -1, -2, -3, -4, -5

Offset: 1

V.J. Pohjola, Oct 21 2017

sign

Shift is the measure of the position of the palindromic subsequence within the corresponding sequence of first differences, defined as the number of terms being dropped from the left end of the sequence of first differences minus those dropped from its right end. When shift is a positive number, it indicates the number of steps that the palindrome has moved to the right from its symmetric position.

			For n = 1, differences = 3; longest palindrome = 3; a(1) = 0 - 0 = 0.
For n = 2, differences = 3, 19; longest palindrome = 3; a(2) = 0 - 1 = -1.
For n = 14, differences = 3, 19, 3, 19, 3, 19, 3, 3, 16, 3, 3, 16, 3, 3; longest palindrome = 3, 3, 16, 3, 3, 16, 3, 3; a(14) = 6 - 0 = 6.

V.J. Pohjola, Table of n, a(n) for n = 1..10000
V.J. Pohjola, Line plot for n=1...10000
V.J. Pohjola, Line plot for n=1...100

Cf. A293698, A293751, A293700, A293702, A293704, A293699, A293701, A293706, A293703.

rootsn = Flatten[Position[Table[Floor[Tan[-i]], {i, 1, 10^4}], 1]];
difn = Differences[rootsn];
ldn = Length[difn];
kmax = 500; palsn = {}; lenpalsn = {0}; shiftn = {}; posn = {};
Do[diffin = difn[[1 ;; k]]; lendiffin = Length[diffin];
  pmax = k - Last[lenpalsn];
  t = Table[difn[[p ;; k]], {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]]]];
  AppendTo[posn, Position[t, Last[palsn]]]; pp = Last[Flatten[posn]] - 1;
  qq = lendiffin - (pp + Last[lenpalsn]);
  AppendTo[shiftn, pp - qq], {k, 1, kmax}];
shiftn (*a(n)=shiftn[[n]]*)

A293703 a(n) is the length of the longest palindromic subsequence in the first differences of the list of the first n negative and positive roots of floor(tan(k))=1.

Original entry on oeis.org

1, 3, 5, 7, 9, 11, 13, 15, 15, 17, 17, 19, 19, 21, 21, 23, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117

Offset: 1

V.J. Pohjola, Oct 20 2017

nonn

-A293751 are the negative roots of floor(tan(k))=1.

Each increment of n increases the length of the sequence of the first differences by two, whereby the length of the palindrome increases by 0, 1 or 2.

			For n = 1, the roots are -18, 1; the first differences are 19; the longest palindrome is 19; so a(n) = 1.
For n = 2, the roots are -21, -18, 1, 4; the first differences are 3, 19, 3; the longest palindrome is 3, 19, 3; so a(n) = 3.
For n = 8, the roots are -87, -84, -65, -62, -43, -40, -21, -18, 1, 4, 23, 26, 45, 48, 67, 70; the first differences are 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3; the longest palindrome is 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3; so a(n) = 15.
For n = 9, the roots are -90, -87, -84, -65, -62, -43, -40, -21, -18, 1, 4, 23, 26, 45, 48, 67, 70, 89; first differences are 16, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19; the longest palindrome is 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3; so a(n) = 15.

V.J. Pohjola, Table of n, a(n) for n = 1..3001
V.J. Pohjola, Line plot for n=1...20
V.J. Pohjola, Line plot for n=1...200
V.J. Pohjola, Line plot for n=1...3000

Cf. A293698, A293751, A293700, A293701, A293706, A293699, A293702, A293704, A293705.

rootsA = {}; Do[
If[Floor[Tan[i]] == 1, AppendTo[rootsA, i]], {i, -10^5, 10^5}]
lenN = Length[Select[rootsA, # < 0 &]]
r = 200; roots = rootsA[[lenN - r ;; lenN + r + 1]]
diff = Differences[roots]
center = (Length[diff] + 1)/2; kmax = (Length[diff] + 1)/2 -
  1; pals = {}; lenpals = {}; lenpal = 1;
Do[diffk = diff[[center - k ;; center + k]];
lendiffk = Length[diffk]; w = 3;
lenpal = lenpal + 2; (Label[alku]; w = w - 1;
  pmax = lendiffk - lenpal - (w - 1);
  t = Table[diffk[[p ;; lenpal + w + p - 1]], {p, 1, pmax}];
  s = Select[t, # == Reverse[#] &]; If[s != {}, Goto[end], Goto[alku]];
  Label[end]); AppendTo[pals, First[s]];
AppendTo[lenpals, Length[Flatten[First[s]]]];
lenpal = Length[Flatten[First[s]]], {k, 0, kmax}]
lenpals (*a[n]=lenpals[[n]]*)

A293706 a(n) is the shift of the longest palindromic subsequence within the first differences of the concatenation of the first n negative and positive roots of floor(tan(k)) = 1.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 4, 4, 6, 6, 8, 8, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10

Offset: 1

V.J. Pohjola, Oct 23 2017

sign

Shift is the measure of the position of a palindromic subsequence within the corresponding sequence of first differences, being defined as the number of terms omitted from the left end of the sequence of first differences minus those omitted from its right end. Thus, when shift is, say, 10, the position of the palindrome is 10 steps to the right from the center of the first differences.

a(n) remains at value 10 from n=18 to 1183 after which it drops stepwise linearly to -1544.

			For n = 1, roots=-18,1; differences = 19; longest palindrome = 19; a(n) = 0.
For n = 2, roots=-21, -18, 1, 4; differences = 3,19,3; longest palindrome = 3,19,3  a(2) = 0.
For n = 9, roots=-106, -90, -87, -84, -65, -62, -43, -40, -21, -18, 1, 4, 23, 26, 45, 48, 67, 70, 89, 92; differences = 16, 3, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3; longest palindrome = 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3; a(9) = 2 - 0 = 2.

V.J. Pohjola, Table of n, a(n) for n = 1..3001
V.J. Pohjola, Line plot for n=1..30
V.J. Pohjola, Line plot for n=1..3000

Cf. A293698, A293751, A293700, A293703, A293704, A293699, A293701, A293705, A293702.

rootsA = {}; Do[
If[Floor[Tan[i]] == 1, AppendTo[rootsA, i]], {i, -10^4, 10^4}]
lenN = Length[Select[rootsA, # < 0 &]];
r = 1000; roots = rootsA[[lenN - r ;; lenN + r + 1]];
diff = Differences[roots];
center = Length[roots]/2;
pals = {}; lenpals = {}; lenpal = 1; pos = {}; shift = {};
Do[diffn = diff[[center - (n - 1) ;; center + (n - 1)]];
lendiffn = Length[diffn]; w = 3;
lenpal = lenpal + 2; (Label[alku]; w = w - 1;
  pmax = lendiffn - lenpal - (w - 1);
  t = Table[diffn[[p ;; lenpal + w + p - 1]], {p, 1, pmax}];
  s = Select[t, # == Reverse[#] &]; If[s != {}, Goto[end], Goto[alku]];
  Label[end]); AppendTo[pals, First[s]];
AppendTo[lenpals, Length[Flatten[First[s]]]];
AppendTo[pos, Flatten[Position[t, First[s]]]]; pp = Last[Flatten[pos]];
qq = lendiffn - (pp - 1 + Last[lenpals]);
AppendTo[shift, pp - 1 - qq], {n, 1, center}]
shift

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

A293698 Values of positive integer i such that floor(tan(i)) = 1.

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A293700 First differences of A293698.

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

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A293751 Values of positive integer i such that floor(tan(-i)) = 1.

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A293699 First differences of A293751.

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A293704 a(n) is the shift of the longest palindromic subsequence in the first n terms of A293700.

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A293705 a(n) is the shift of the longest palindromic subsequence in the first n terms of A293699.

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A293703 a(n) is the length of the longest palindromic subsequence in the first differences of the list of the first n negative and positive roots of floor(tan(k))=1.

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