V.J. Pohjola - cp's OEIS frontend

A295285 Numbers n such that for positive integers i, the union of sequences n+22i contains the positive roots of floor(tan(k)) = 1 (A293698).

1, 4, 183, 538, 893, 1248, 1603, 1958, 2313, 2668, 3023, 3378, 3733, 4088, 4443, 4798, 5153, 5508, 5863, 6218, 6573, 6928, 225919, 226274, 226629, 226984, 227339, 227694, 228049, 228404, 228759, 229114, 229469, 229824

Offset: 1

V.J. Pohjola, Nov 19 2017

nonn

Each number n is the first term of the corresponding individual sequence n+22i, and the whole is union of these overlapping sequences. Due to periodicity, there is a single generating function (n-(n-22)*j)/(j-1)^2 for the sequences. However, the function does not provide predictive means for generating A293698, because also terms which are not the roots are generated. The roots appear in each n+22i as finite subsequences of given length, at given steps. There is, however slight but difficult or impossible to predict variability both in the length, which is either 37 or 36, and the step which is either 7810 or 7832. A293698 is union of these subsequences. - V.J. Pohjola, Feb 25 2018

			For n = 1, i = 0..12, the terms 1, 23, 45, .., 265 are the roots.
For n = 4, i = 0..28, the terms 4, 26, 48, .., 620 are the roots.
For n = 183, i = 0..36, the terms 183, 205, 227, .., 975 are the roots.
For n = 1, i = 331..367, the terms 7283, 7305, 7327, .., 8075 are the roots.
For n = 4, i = 347..383, the terms 7638, 7660, 7682, .., 8430 are the roots.
For n = 183, i = 355..391, the terms 7993, 8015, 8037, .., 8785 are the roots.
The subsequences have the length of either 36 or 37 beyond the initial ranges 1+22i and 4+22i which are 13 and 29, respectively.

V.J. Pohjola, Table of n, a(n) for n = 1..101
V. J. Pohjola, Line plot for n=1..3
V. J. Pohjola, Line plot for n=1..100

Cf. A293698, A000503.

posroots6 = {}; Do[If[Floor[Tan[n]] == 1, AppendTo[posroots6, n]], {n, 0, 10^6}]
bigroots = {1}; posrootsi = {{1}}; Do[jj = {};
Do[lastb = Last[bigroots];
If[MemberQ[posroots6, lastb + 22*j], AppendTo[jj, j]], {j, 0, 10^4}];
posrootsi = Flatten[AppendTo[posrootsi, Table[lastb + 22*jj[[k]], {k, 1, Length[jj]}]]];
bigroot = First[Complement[posroots6, posrootsi]];
AppendTo[bigroots, bigroot], {i, 1, 100}]; bigroots

Name edited by V.J. Pohjola, Mar 15 2018

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

A294923 Number n such that the whole sequence of the first n terms of A293700 is a palindrome.

Original entry on oeis.org

1, 3, 5, 7, 9, 11, 13, 15, 17, 46, 83, 120, 157, 194, 231, 268, 305, 342, 379, 416, 453, 490, 527, 564, 601, 638, 675, 712, 749, 786, 823, 860, 897, 934, 971, 1008, 1045, 1082, 1119, 1156, 9105, 19792, 51817, 83842, 201253, 318664, 436075, 553486

Offset: 1

V.J. Pohjola, Nov 11 2017

nonn

A293700 are the first differences of A293698 which are the positive integers i such that floor(tan(i))=1.

A293701 are the lengths of the longest palindromic subsequences in the first n terms of A293700.

			The first 3 terms of A293701 are (3,19,3) which is a palindromic sequence, so 3 is a term.
The first 4 terms of A293701 are (3,19,3,19) which is not a palindromic sequence, so 4 is not a term.
The first 17 terms of A293701 are (3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3) which is a palindromic sequence, so 17 is a term.
The first 18 terms of A293701 are (3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 3) which is not a palindromic sequence, so 18 is not a term.
The first 19 terms of A293701 are (3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 19, 3, 3, 16) which is not a palindromic sequence, so 19 is not a term.

V.J. Pohjola, Line plot for n=1..50

Cf. A293701, A293700, A293698, A294924.

rootsp7 = Flatten[Position[Table[Floor[Tan[n]], {n, 1, 10^7}], 1]];
difp7 = Differences[rootsp7];
nx = {}; Do[
If[Table[difp7[[i]], {i, 1, n}] == Reverse[Table[difp7[[i]], {i, 1, n}]],
  AppendTo[nx, n]], {n, 1, Length[difp7]}]
nx

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)).

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

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]]*)

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]]*)

A293702 a(n) is the length of the longest palindromic subsequence in the first n terms of A293751.

Original entry on oeis.org

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

V.J. Pohjola, Oct 16 2017

nonn

			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.

V.J. Pohjola, Table of n, a(n) for n = 1..10000
V.J. Pohjola, Line plot for n=1…30
V.J. Pohjola, Line plot for n=1...10000

Cf. A293698, A293751, A293700, A293701, A293705, A293704, A293699, A293703, A293706.

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]] *)

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]]*)

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

cp's OEIS Frontend

User: V.J. Pohjola

A295285 Numbers n such that for positive integers i, the union of sequences n+22i contains the positive roots of floor(tan(k)) = 1 (A293698).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Extensions

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A294923 Number n such that the whole sequence of the first n terms of A293700 is a palindrome.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

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

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Formula

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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A293702 a(n) is the length of the longest palindromic subsequence in the first n terms of A293751.

Views

Author

Keywords

Examples

Links

Crossrefs