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 1-7 of 7 results.

A264756 An eventually quasilinear solution to Hofstadter's Q recurrence.

Original entry on oeis.org

1, 0, 3, 3, 2, 6, 3, 2, 9, 3, 2, 12, 3, 2, 15, 3, 2, 18, 3, 2, 21, 3, 2, 24, 3, 2, 27, 3, 2, 30, 3, 2, 33, 3, 2, 36, 3, 2, 39, 3, 2, 42, 3, 2, 45, 3, 2, 48, 3, 2, 51, 3, 2, 54, 3, 2, 57, 3, 2, 60, 3, 2, 63, 3, 2, 66, 3, 2, 69, 3, 2, 72, 3, 2, 75, 3, 2, 78, 3, 2, 81
Offset: 1

Views

Author

Nathan Fox, Nov 23 2015

Keywords

Comments

a(n) is the solution to the recurrence relation a(n) = a(n-a(n-1)) +a(n-a(n-2)) [Hofstadter's Q recurrence], with the initial conditions: a(n) = 0 if n <= 0; a(1) = 1, a(2) = 0, a(3) = 3, a(4) = 3, a(5) = 2.

Links

Crossrefs

Cf. A005185, A188670, A244477, A264757, A264758.

Programs

  • Magma
    [1,0] cat [2+(n-3)*(1+Floor(-n/3)+Floor(n/3))-Floor(-(n+1)/3)-Floor((n+1)/3): n in [3..100]]; // Vincenzo Librandi, Nov 25 2015
  • Mathematica
    CoefficientList[Series[-(2*x^7 + 2*x^6 - 2*x^4 - x^3 - 3*x^2 - 1)/((x - 1)^2*(x^2 + x + 1)^2), {x, 0, 100}], x] (* Wesley Ivan Hurt, Nov 24 2015 *)
Join[{1, 0}, LinearRecurrence[{0, 0, 2, 0, 0, -1}, {3, 3, 2, 6, 3, 2}, 100]] (* Vincenzo Librandi, Nov 25 2015 *)
  • PARI
    Vec(-x*(2*x^7+2*x^6-2*x^4-x^3-3*x^2-1)/((x-1)^2*(x^2+x+1)^2) + O(x^100)) \\ Colin Barker, Nov 23 2015

Formula

a(1) = 1, a(2) = 0; thereafter a(3n) = 3n, a(3n+1) = 3, a(3n+2) = 2.
From Colin Barker, Nov 23 2015: (Start)
a(n) = 2*a(n-3) - a(n-6) for n>8.
G.f.: -x*(2*x^7+2*x^6-2*x^4-x^3-3*x^2-1) / ((x-1)^2*(x^2+x+1)^2).
(End)
a(1) = 1, a(2) = 0, a(n) = 2 + (n-3)*(1 + floor(-n/3) + floor(n/3)) - floor(-(n+1)/3) - floor((n+1)/3) for n>2. - Wesley Ivan Hurt, Nov 24 2015

A264758 An eventually quasi-cubic solution to Hofstadter's Q recurrence.

Original entry on oeis.org

7, 0, 5, 9, 3, 8, 9, 2, 9, 9, 3, 14, 9, 3, 22, 9, 2, 18, 9, 3, 32, 9, 3, 54, 9, 2, 27, 9, 3, 59, 9, 3, 113, 9, 2, 36, 9, 3, 95, 9, 3, 208, 9, 2, 45, 9, 3, 140, 9, 3, 348, 9, 2, 54, 9, 3, 194, 9, 3, 542, 9, 2, 63, 9, 3, 257, 9, 3, 799, 9, 2, 72, 9, 3, 329, 9
Offset: 1

Views

Author

Nathan Fox, Nov 23 2015

Keywords

Comments

a(n) is the solution to the recurrence relation a(n) = a(n-a(n-1)) + a(n-a(n-2)) [Hofstadter's Q recurrence], with the initial conditions: a(n) = 0 if n <= 0; a(1) = 7, a(2) = 0, a(3) = 5, a(4) = 9, a(5) = 3, a(6) = 8, a(7) = 9, a(8) = 2, a(9) = 9, a(10) = 9, a(11) = 3.

Links

Crossrefs

Cf. A005185, A188670, A244477, A264756, A264757.

Programs

  • Mathematica
    CoefficientList[Series[x (7+5x^2+9x^3+3x^4+8x^5+9x^6+2x^7+9x^8- 19x^9+ 3x^10- 6x^11- 27x^12-9x^13- 10x^14- 27x^15-6x^16-18x^17+ 15x^18- 9x^19+6x^20+ 27x^21+9x^22+14x^23+ 27x^24+ 6x^25+ 9x^26-x^27+9x^28- 5x^29-9x^30-3x^31- 3x^32-9x^33-2x^34- 2x^36-3x^37)/((1-x)^4(1+x+x^2)^4 (1+x^3+x^6)^4),{x,0,100}],x] (* Harvey P. Dale, Aug 14 2021 *)
  • PARI
    a = [7,0,5,9,3,8,9,2]; for(n=1, 10, a=concat(a, [9*n, 9, 3, 9/2*n^2+9/2*n+5, 9, 3, 3/2*n^3+9/2*n^2+8*n+8, 9, 2])); a

Formula

a(1) = 7, a(2) = 0; thereafter a(9*n) = 9*n, a(9*n+1) = 9, a(9*n+2) = 3, a(9*n+3) = 9/2*n^2+9/2*n+5, a(9*n+4) = 9, a(9*n+5) = 3, a(9*n+6) = 3/2*n^3+9/2*n^2+8*n+8, a(9*n+7) = 9, a(9*n+8) = 2.
From Colin Barker, Nov 14 2016: (Start)
G.f.: x*(7 +5*x^2 +9*x^3 +3*x^4 +8*x^5 +9*x^6 +2*x^7 +9*x^8 -19*x^9 +3*x^10 -6*x^11 -27*x^12 -9*x^13 -10*x^14 -27*x^15 -6*x^16 -18*x^17 +15*x^18 -9*x^19 +6*x^20 +27*x^21 +9*x^22 +14*x^23 +27*x^24 +6*x^25 +9*x^26 -x^27 +9*x^28 -5*x^29 -9*x^30 -3*x^31 -3*x^32 -9*x^33 -2*x^34 -2*x^36 -3*x^37) / ((1 -x)^4*(1 +x +x^2)^4*(1 +x^3 +x^6)^4).
a(n) = 4*a(n-9) - 6*a(n-18) + 4*a(n-27) - a(n-36) for n>38.
(End)

A268368 An eventually quasi-quadratic sequence satisfying a Hofstadter-like recurrence.

Original entry on oeis.org

0, 1, 0, 4, 4, 4, 3, 12, 8, 4, 3, 24, 12, 4, 3, 40, 16, 4, 3, 60, 20, 4, 3, 84, 24, 4, 3, 112, 28, 4, 3, 144, 32, 4, 3, 180, 36, 4, 3, 220, 40, 4, 3, 264, 44, 4, 3, 312, 48, 4, 3, 364, 52, 4, 3, 420, 56, 4, 3, 480, 60, 4, 3, 544, 64, 4, 3, 612, 68, 4, 3, 684
Offset: 1

Views

Author

Nathan Fox, Feb 23 2016

Keywords

Comments

a(n) is the solution to the recurrence relation a(n) = a(n-a(n-1)) + a(n-a(n-2)) + a(n-a(n-3)), with the initial conditions: a(n) = 0 if n <= 0; a(1) = 0, a(2) = 1, a(3) = 0, a(4) = 4, a(5) = 4, a(6) = 4, a(7) = 3.

Links

Crossrefs

Cf. A005185, A188670, A244477, A264757, A267501.

Programs

  • PARI
    concat(0, Vec(x^2*(1 + 4*x^2 + 4*x^3 + x^4 + 3*x^5 - 4*x^7 - 5*x^8 - 6*x^9 + 3*x^12 + 3*x^13) / ((1 - x)^3*(1 + x)^3*(1 + x^2)^3) + O(x^100))) \\ Colin Barker, Jun 22 2017

Formula

a(2) = 1, a(3) = 0; otherwise a(4n) = 2n^2+2n, a(4n+1) = 4n, a(4n+2) = 4, a(4n+3) = 3.
From Colin Barker, Jun 22 2017: (Start)
G.f.: x^2*(1 + 4*x^2 + 4*x^3 + x^4 + 3*x^5 - 4*x^7 - 5*x^8 - 6*x^9 + 3*x^12 + 3*x^13) / ((1 - x)^3*(1 + x)^3*(1 + x^2)^3).
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12) for n>12.
(End)

A275153 A linear-recurrent solution to Hofstadter's Q-recurrence.

Original entry on oeis.org

9, 0, 0, 0, 7, 9, 9, 10, 4, 9, 9, 3, 9, 16, 9, 9, 20, 4, 9, 18, 3, 9, 34, 9, 9, 40, 4, 9, 27, 3, 9, 61, 9, 9, 80, 4, 9, 36, 3, 9, 97, 9, 9, 160, 4, 9, 45, 3, 9, 142, 9, 9, 320, 4, 9, 54, 3, 9, 196, 9, 9, 640, 4, 9, 63, 3, 9, 259, 9, 9, 1280, 4, 9, 72, 3, 9, 331, 9, 9, 2560
Offset: 1

Views

Author

Nathan Fox, Jul 17 2016

Keywords

Comments

a(n) is the solution to the recurrence relation a(n) = a(n-a(n-1)) + a(n-a(n-2)) [Hofstadter's Q recurrence], with the initial conditions: a(n) = 0 if n <= 0; a(1) = 9, a(2) = 0, a(3) = 0, a(4) = 0, a(5) = 7, a(6) = 9, a(7) = 9, a(8) = 10, a(9) = 4, a(10) = 9, a(11) = 9, a(12) = 3.
This sequence is an interleaving of nine simpler sequences. Six are eventually constant, one is a linear polynomial, one is a quadratic polynomial, and one is a geometric sequence.

Links

Crossrefs

Cf. A005185, A188670, A244477, A264756, A264757, A264758, A268368.

Programs

  • Mathematica
    Join[{9, 0, 0, 0}, LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 0, -9, 0, 0, 0, 0, 0, 0, 0, 0, 7, 0, 0, 0, 0, 0, 0, 0, 0, -2}, {7, 9, 9, 10, 4, 9, 9, 3, 9, 16, 9, 9, 20, 4, 9, 18, 3, 9, 34, 9, 9, 40, 4, 9, 27, 3, 9, 61, 9, 9, 80, 4, 9, 36, 3, 9}, 100]] (* Jean-François Alcover, Dec 01 2018 *)

Formula

a(3) = 0, a(4) = 0; otherwise:
a(9n) = 4, a(9n+1) = 9 a(9n+2) = 9n a(9n+3) = 3.
a(9n+4) = 9, a(9n+5) = (9*n^2 + 9*n + 14)/2, a(9n+6) = 9.
a(9n+7) = 9, a(9n+8) = 10*2^n.
a(n) = 5*a(n-9) - 9*a(n-18) + 7*a(n-27) - 2*a(n-36) for n>40.
G.f.: -(18*x^39 +6*x^38 +8*x^35 +10*x^34 +18*x^33 +18*x^32 +14*x^31 -45*x^30 -15*x^29 -18*x^28 +18*x^27 -20*x^26 -30*x^25 -45*x^24 -45*x^23 -17*x^22 +36*x^21 +12*x^20 +27*x^19 -45*x^18 +16*x^17 +30*x^16 +36*x^15 +36*x^14 +19*x^13 -9*x^12 -3*x^11 -9*x^10 +36*x^9 -4*x^8 -10*x^7 -9*x^6 -9*x^5 -7*x^4 -9)/((2*x^9-1)*(x-1)^3*(x^2+x+1)^3*(x^6+x^3+1)^3).

A275361 An eventually quasilinear solution to Hofstadter's Q-recurrence.

Original entry on oeis.org

0, 4, -40, -9, 8, -8, 7, 1, 5, 13, -24, -1, 8, 8, 8, 1, 5, 13, -8, 7, 8, 8, 23, 1, 5, 13, 8, 15, 8, 16, 31, 1, 5, 13, 24, 23, 8, 24, 39, 1, 5, 13, 40, 31, 8, 32, 47, 1, 5, 13, 56, 39, 8, 40, 55, 1, 5, 13, 72, 47, 8, 48, 63, 1, 5, 13, 88, 55, 8, 56, 71
Offset: 1

Views

Author

Nathan Fox, Jul 24 2016

Keywords

Comments

a(n) is the solution to the recurrence relation a(n) = a(n-a(n-1)) + a(n-a(n-2)) [Hofstadter's Q recurrence], with the first 45 terms as initial conditions.
This is a quasilinear sequence with quasiperiod 8. Four of the component sequences are constant, three have slope 1, and one has slope 2.

Links

Crossrefs

Cf. A005185, A188670, A244477, A264756, A264757, A264758, A268368, A275153, A275362.

Programs

  • Mathematica
    Join[{0, 4, -40, -9, 8, -8, 7, 1, 5, 13, -24, -1, 8, 8, 8}, LinearRecurrence[ {0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, -1}, {1, 5, 13, -8, 7, 8, 8, 23, 1, 5, 13, 8, 15, 8, 16, 31}, 100]] (* Jean-François Alcover, Dec 12 2018 *)

Formula

a(1) = 0, a(2) = 4, a(14) = 8, a(15) = 8; otherwise:
a(8n) = 1, a(8n+1) = 5, a(8n+2) = 13, a(8n+3) = 16n-40, a(8n+4) = 8n-9, a(8n+5) = 8, a(8n+6) = 8n-8, a(8n+7) = 8n+7.
a(n) = 2*a(n-8) - a(n-16) for n>31.
G.f.: -(7*x^30 -8*x^29 -14*x^22 +16*x^21 +9*x^17 +5*x^16 +x^15 +6*x^14 -24*x^13 +8*x^12 -17*x^11 -56*x^10 -5*x^9 -5*x^8 -x^7 -7*x^6 +8*x^5 -8*x^4 +9*x^3 +40*x^2 -4*x)/((x-1)^2*(x+1)^2*(x^2+1)^2*(x^4+1)^2).

A275362 An eventually quasilinear solution to Hofstadter's Q recurrence.

Original entry on oeis.org

-9, 2, 9, 2, 0, 7, 9, 10, 3, 0, 2, 9, 2, 9, 9, 9, 20, 3, 9, 22, 9, 2, 18, 9, 18, 30, 3, 18, 32, 9, 2, 27, 9, 27, 40, 3, 27, 42, 9, 2, 36, 9, 36, 50, 3, 36, 52, 9, 2, 45, 9, 45, 60, 3, 45, 62, 9, 2, 54, 9, 54, 70, 3, 54, 72, 9, 2, 63, 9, 63, 80, 3, 63, 82
Offset: 1

Views

Author

Nathan Fox, Jul 24 2016

Keywords

Comments

a(n) is the solution to the recurrence relation a(n) = a(n-a(n-1)) + a(n-a(n-2)) [Hofstadter's Q recurrence], with the initial conditions: a(n) = 0 if n <= 0; a(1) = -9, a(2) = 2, a(3) = 9, a(4) = 2, a(5) = 0, a(6) = 7, a(7) = 9, a(8) = 10, a(9) = 3, a(10) = 0, a(11) = 2, a(12) = 9, a(13) = 2, a(14) = 9, a(15) = 9, a(16) = 9.
This is a quasilinear sequence with quasiperiod 9. Four of the component sequences are constant, three have slope 1, and two have slope 10/9.

Links

Crossrefs

Cf. A005185, A188670, A244477, A264756, A264757, A264758, A268368, A275153, A275361.

Programs

  • Mathematica
    Join[{-9, 2, 9, 2, 0, 7, 9, 10, 3, 0, 2}, LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, -1}, {9, 2, 9, 9, 9, 20, 3, 9, 22, 9, 2, 18, 9, 18, 30, 3, 18, 32}, 100]] (* Jean-François Alcover, Dec 13 2018 *)

Formula

a(6) = 7, a(7) = 9, a(11) = 2; otherwise:
a(9n) = 3, a(9n+1) = 9n-9, a(9n+2) = 10n+2, a(9n+3) = 9, a(9n+4) = 2, a(9n+5) = 9n, a(9n+6) = 9, a(9n+7) = 9n, a(9n+8) = 10n+10.
a(n) = 2*a(n-9) - a(n-18) for n>29.
G.f.: -(10*x^28 -9*x^24 +2*x^23 -20*x^19 +3*x^17 +9*x^15 +5*x^14 -9*x^13 +2*x^12 +9*x^11 +2*x^10 -18*x^9 -3*x^8 -10*x^7 -9*x^6 -7*x^5 -2*x^3 -9*x^2 -2*x+9)/((x-1)^2*(x^2+x+1)^2*(x^6+x^3+1)^2).

A275365 a(1)=2, a(2)=2; thereafter a(n) = a(n-a(n-1)) + a(n-a(n-2)).

Original entry on oeis.org

0, 2, 2, 4, 2, 6, 2, 8, 2, 10, 2, 12, 2, 14, 2, 16, 2, 18, 2, 20, 2, 22, 2, 24, 2, 26, 2, 28, 2, 30, 2, 32, 2, 34, 2, 36, 2, 38, 2, 40, 2, 42, 2, 44, 2, 46, 2, 48, 2, 50, 2, 52, 2, 54, 2, 56, 2, 58, 2, 60, 2, 62, 2, 64, 2, 66, 2, 68, 2, 70, 2, 72, 2, 74
Offset: 0

Views

Author

Nathan Fox, Jul 24 2016

Keywords

Comments

a(n) is the solution to the recurrence relation a(n) = a(n-a(n-1)) + a(n-a(n-2)) [Hofstadter's Q recurrence], with the initial conditions: a(n) = 0 if n <= 0; a(1) = 2, a(2) = 2.
Starting with n=1, a(n) is A005843 interleaved with A007395.
This sequence is the same as A133265 with the leading 2 changed to a 0.

Links

Crossrefs

Cf. A005185, A133265, A188670, A244477, A264756, A264757, A264758, A268368, A275153, A275361, A275362.

Programs

  • Mathematica
    Join[{0}, LinearRecurrence[{0, 2, 0, -1}, {2, 2, 4, 2}, 73]] (* Jean-François Alcover, Feb 19 2019 *)

Formula

a(0) = 0; thereafter, a(2n) = 2, a(2n+1) = 2n+2.
a(n) = 2*a(n-2) - a(n-4) for n>4.
G.f.: -(2*x^3 -2*x -2)/((x-1)^2*(x+1)^2).
Showing 1-7 of 7 results.

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