cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-9 of 9 results.

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

Original entry on oeis.org

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

Views

Author

V.J. Pohjola, Oct 15 2017

Keywords

Comments

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

Examples

			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.
		

Crossrefs

Programs

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

Views

Author

V.J. Pohjola, Oct 16 2017

Keywords

Comments

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)

Crossrefs

Programs

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

Views

Author

V.J. Pohjola, Oct 16 2017

Keywords

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.
		

Crossrefs

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

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

Views

Author

V.J. Pohjola, Oct 16 2017

Keywords

Examples

			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.
		

Crossrefs

Programs

  • Mathematica
    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]] *)

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

Views

Author

V.J. Pohjola, Oct 15 2017

Keywords

Crossrefs

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

Programs

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

Formula

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

Views

Author

V.J. Pohjola, Oct 16 2017

Keywords

Comments

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

Crossrefs

Programs

  • Mathematica
    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

Views

Author

V.J. Pohjola, Oct 21 2017

Keywords

Comments

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.

Examples

			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.
		

Crossrefs

Programs

  • Mathematica
    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]]*)

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

Views

Author

V.J. Pohjola, Oct 20 2017

Keywords

Comments

-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.

Examples

			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.
		

Crossrefs

Programs

  • Mathematica
    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

Views

Author

V.J. Pohjola, Oct 23 2017

Keywords

Comments

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.

Examples

			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.
		

Crossrefs

Programs

  • Mathematica
    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
Showing 1-9 of 9 results.