A244477 -id:A244477 - cp's OEIS frontend

A309567 a(1) = 4, a(2) = 2, a(3) = 5, a(4) = 3, a(5) = 1; a(n) = a(n-a(n-1)) + a(n-a(n-4)) for n > 5.

4, 2, 5, 3, 1, 4, 7, 5, 8, 6, 4, 12, 5, 13, 6, 9, 17, 5, 18, 6, 9, 22, 5, 23, 11, 9, 27, 5, 28, 11, 9, 32, 5, 33, 11, 14, 37, 5, 38, 11, 14, 42, 5, 43, 11, 14, 47, 5, 48, 16, 14, 52, 5, 53, 16, 14, 57, 5, 58, 16, 14, 62, 5, 63, 16, 19, 67, 5, 68, 16, 19, 72, 5, 73, 16, 19, 77, 5, 78, 16, 19, 82, 5, 83, 21, 19, 87, 5

Offset: 1

Altug Alkan and Rémy Sigrist, Aug 08 2019

A well-defined quasi-periodic solution for Hofstadter V recurrence (a(n) = a(n-a(n-1)) + a(n-a(n-4))).

Robert Israel, Table of n, a(n) for n = 1..10000
Altug Alkan, Nathan Fox, Orhan Ozgur Aybar, Zehra Akdeniz, On Some Solutions to Hofstadter's V-Recurrence, arXiv:2002.03396 [math.DS], 2020.

Cf. A063882, A244477, A296518, A309492, A309494, A309496, A309554.

f:= proc(n) local k,j;
  j:= n mod 5;
  k:= (n-j)/5;
  if j=0 then 5*floor(sqrt(k-1))+1
  elif j=1 then 5*round(sqrt(k))-1
  elif j=2 then 5*k+2
  elif j=3 then 5
  else 5*k+3
  fi
end proc:
f(1):= 4:
map(f, [$1..100]); # Robert Israel, Aug 08 2019

a[n_] := a[n] = If[n < 6, {4, 2, 5, 3, 1}[[n]], a[n - a[n-1]] + a[n - a[n-4]]]; Array[a, 88] (* Giovanni Resta, Aug 08 2019 *)

q=vector(100); q[1]=4; q[2]=2; q[3]=5; q[4]=3; q[5]=1; for(n=6, #q, q[n]=q[n-q[n-1]]+q[n-q[n-4]]); q

For k >= 1:
a(5*k)   = 5*floor(sqrt(k-1))+1,
a(5*k+1) = 5*round(sqrt(k))-1,
a(5*k+2) = 5*k+2,
a(5*k+3) = 5,
a(5*k+4) = 5*k+3.

A309496 a(1) = 1, a(2) = 3, a(3) = a(4) = a(6) = 6, a(5) = 2, a(7) = 4; a(n) = a(n-a(n-4)) + a(n-a(n-5)) for n > 7.

Original entry on oeis.org

1, 3, 6, 6, 2, 6, 4, 6, 10, 12, 6, 12, 10, 9, 16, 18, 6, 16, 18, 9, 22, 24, 6, 22, 24, 9, 28, 30, 6, 28, 30, 9, 34, 36, 6, 34, 36, 9, 40, 42, 6, 40, 42, 9, 46, 48, 6, 46, 48, 9, 52, 54, 6, 52, 54, 9, 58, 60, 6, 58, 60, 9, 64, 66, 6, 64, 66, 9, 70, 72, 6, 70, 72, 9, 76, 78, 6, 76, 78, 9, 82, 84, 6, 82, 84, 9, 88

Offset: 1

Altug Alkan, Aug 05 2019

A well-defined solution sequence for recurrence a(n) = a(n-a(n-4)) + a(n-a(n-5)).

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,1,0,0,1,0,0,-1).

Cf. A244477, A309492, A309494.

I:=[1,3,6,6,2,6,4];[n le 7 select I[n] else Self(n-Self(n-4))+Self(n-Self(n-5)): n in [1..90]]; // Marius A. Burtea, Aug 07 2019

a[n_] := a[n] = If[n < 8, {1, 3, 6, 6, 2, 6, 4}[[n]], a[n - a[n-4]] + a[n - a[n-5]]]; Array[a, 87] (* Giovanni Resta, Aug 07 2019 *)

q=vector(100); q[1]=1; q[2]=3; q[3]=q[4]=q[6]=6; q[5]=2; q[7]=4; for(n=8, #q, q[n] = q[n-q[n-4]]+q[n-q[n-5]]); q

Vec(x*(1 + 3*x + 6*x^2 + 5*x^3 - x^4 - 3*x^6 + x^7 - 2*x^8 + 3*x^9 + x^10 + 2*x^11 - x^13 - 3*x^16 - 2*x^17 + 2*x^18 + 2*x^20 - 2*x^21) / ((1 - x)^2*(1 + x)*(1 - x + x^2)*(1 + x + x^2)^2) + O(x^80)) \\ Colin Barker, Aug 15 2019

For k > 2:
  a(6*k-4) = 9,
  a(6*k-3) = 6*k-2,
  a(6*k-2) = 6*k,
  a(6*k-1) = 6,
  a(6*k)   = 6*k-2,
  a(6*k+1) = 6*k.
From Colin Barker, Aug 05 2019: (Start)
G.f.: x*(1 + 3*x + 6*x^2 + 5*x^3 - x^4 - 3*x^6 + x^7 - 2*x^8 + 3*x^9 + x^10 + 2*x^11 - x^13 - 3*x^16 - 2*x^17 + 2*x^18 + 2*x^20 - 2*x^21) / ((1 - x)^2*(1 + x)*(1 - x + x^2)*(1 + x + x^2)^2).
a(n) = a(n-3) + a(n-6) - a(n-9) for n > 22.
(End)

A269328 An eventually quasilinear solution to Hofstadter's Q recurrence.

Original entry on oeis.org

5, 2, 0, 3, 6, 5, 2, 5, 5, 12, 5, 2, 10, 5, 18, 5, 2, 15, 5, 24, 5, 2, 20, 5, 30, 5, 2, 25, 5, 36, 5, 2, 30, 5, 42, 5, 2, 35, 5, 48, 5, 2, 40, 5, 54, 5, 2, 45, 5, 60, 5, 2, 50, 5, 66, 5, 2, 55, 5, 72, 5, 2, 60, 5, 78, 5, 2, 65, 5, 84, 5, 2, 70, 5, 90

Offset: 1

Nathan Fox, Feb 23 2016

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) = 5, a(2) = 2, a(3) = 0, a(4) = 3, a(5) = 6, a(6) = 5, a(7) = 2.
Starting from n=5, this sequence consists of five interleaved linear sequences with three different slopes.
Square array read by rows: T(j,k), j>=1, 1<=k<=5, in which row j list [5, 2, 5*(j-1), 5, 6*j], except T(1,4) = 3, not 5. - Omar E. Pol, Jun 22 2016

			From _Omar E. Pol_, Jun 22 2016: (Start)
Written as a square array T(j,k) with five columns the sequence begins:
5, 2,  0, 3,  6;
5, 2,  5, 5, 12;
5, 2, 10, 5, 18;
5, 2, 15, 5, 24;
5, 2, 20, 5, 30;
5, 2, 25, 5, 36;
5, 2, 30, 5, 42;
5, 2, 35, 5, 48;
5, 2, 40, 5, 54;
5, 2, 45, 5, 60;
5, 2, 50, 5, 66;
5, 2, 55, 5, 72;
5, 2, 60, 5, 78;
5, 2, 65, 5, 84;
5, 2, 70, 5, 90;
...
Note that T(1,4) = 3, not 5. (End)

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Nathan Fox, Finding Linear-Recurrent Solutions to Hofstadter-Like Recurrences Using Symbolic Computation, arXiv:1609.06342 [math.NT], 2016.
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 2, 0, 0, 0, 0, -1).

Cf. A005185, A188670, A244477, A264756.

I:=[5,2,0,3,6,5,2,5,5,12,5,2,10,5]; [n le 14 select I[n] else 2*Self(n-5)-Self(n-10): n in [1..100]]; // Vincenzo Librandi, Dec 16 2018

Join[{5, 2, 0, 3}, LinearRecurrence[{0, 0, 0, 0, 2, 0, 0, 0, 0, -1} , {6, 5, 2, 5, 5, 12, 5, 2, 10, 5}, 80]] (* Jean-François Alcover, Dec 16 2018 *)
CoefficientList[Series[(-2 x^13 - x^8 + 5 x^7 - 2 x^6 - 5 x^5 + 6 x^4 + 3 x^3 + 2 x + 5) / (x^10 - 2 x^5 + 1), {x, 0, 100}], x] (* Vincenzo Librandi, Dec 16 2018 *)

a(4) = 3; otherwise a(5n) = 6n, a(5n+1) = 5, a(5n+2) = 2, a(5n+3) = 5n, a(5n+4) = 5.
From Chai Wah Wu, Jun 22 2016: (Start)
a(n) = 2*a(n-5) - a(n-10) for n > 14.
G.f.: x*(-2*x^13 - x^8 + 5*x^7 - 2*x^6 - 5*x^5 + 6*x^4 + 3*x^3 + 2*x + 5)/(x^10 - 2*x^5 + 1). (End)

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

Nathan Fox, Jul 24 2016

sign

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.

Nathan Fox, Table of n, a(n) for n = 1..1000
Nathan Fox, Finding Linear-Recurrent Solutions to Hofstadter-Like Recurrences Using Symbolic Computation, arXiv:1609.06342 [math.NT], 2016.
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, -1).

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

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 *)

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

Nathan Fox, Jul 24 2016

sign

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.

Nathan Fox, Table of n, a(n) for n = 1..1000
Nathan Fox, Finding Linear-Recurrent Solutions to Hofstadter-Like Recurrences Using Symbolic Computation, arXiv:1609.06342 [math.NT], 2016.
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, -1).

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

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 *)

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).

A283878 An eventually quasilinear solution to Hofstadter's Q recurrence.

Original entry on oeis.org

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

Offset: 1

Nathan Fox, Mar 19 2017

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) = 0, a(2) = 2, a(3) = 3, a(4) = 1.

Nathan Fox, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (0, 0, 2, 0, 0, -1).

Cf. A005185, A188670, A244477, A264756, A283879.

A283878:=proc(n) option remember: if n <= 0 then 0: elif n = 1 then 0: elif n = 2 then 2: elif n = 3 then 3: elif n = 4 then 1: else A283878(n-A283878(n-1)) + A283878(n-A283878(n-2)): fi: end:

G.f.: (-x*(x^6 + x^5 + x^3 - x^2 - 3*x - 2)) / ((-1 + x)^2*(1 + x + x^2)^2).
a(n) = 2*a(n-3) - a(n-6) for n > 8.
a(1) = 0, a(2) = 2; thereafter a(3*k) = 3*k, a(3*k+1) = 1, a(3*k+2) = 3.

A283879 An eventually quasilinear solution to Hofstadter's Q recurrence.

Original entry on oeis.org

4, 1, 0, 3, 3, 1, 4, 8, 7, 1, 4, 12, 11, 1, 4, 16, 15, 1, 4, 20, 19, 1, 4, 24, 23, 1, 4, 28, 27, 1, 4, 32, 31, 1, 4, 36, 35, 1, 4, 40, 39, 1, 4, 44, 43, 1, 4, 48, 47, 1, 4, 52, 51, 1, 4, 56, 55, 1, 4, 60, 59, 1, 4, 64, 63, 1, 4, 68, 67, 1, 4, 72, 71, 1, 4

Offset: 1

Nathan Fox, Mar 19 2017

nonn

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) = 4, a(2) = 1, a(3) = 0, a(4) = 3, a(5) = 3, a(6) = 1, a(7) = 4, a(8) = 8.

Nathan Fox, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1, -1, 1, 1, -1, 1, -1).

Cf. A005185, A188670, A244477, A264756, A283878.

A283879:=proc(n) option remember: if n <= 0 then 0: elif n = 1 then 4: elif n = 2 then 1: elif n = 3 then 0: elif n = 4 then 3: elif n = 5 then 3: elif n = 6 then 1: elif n = 7 then 4: elif n = 8 then 8: else A283879(n-A283879(n-1)) + A283879(n-A283879(n-2)): fi: end:

LinearRecurrence[{1,-1,1,1,-1,1,-1},{4,1,0,3,3,1,4,8,7,1,4},80] (* Harvey P. Dale, May 25 2025 *)

a(1) = 4, a(4) = 3; otherwise a(4n) = 4n, a(4n+1) = 4n-1, a(4n+2) = 1, a(4n+3) = 4.
G.f.: (-x^10-3*x^9+3*x^8+2*x^7+4*x^5-5*x^4+3*x^2-3*x+4) / ((1+x)*(-1+x)^2*(1+x^2)^2).
a(n) = a(n-1) - a(n-2) + a(n-3) + a(n-4) - a(n-5) + a(n-6) - a(n-7) for n > 11.

A309554 a(1) = a(6) = 1, a(2) = a(3) = a(8) = 2, a(4) = a(7) = 7, a(5) = 5; a(n) = a(n-a(n-1)) + a(n-a(n-3)) for n > 8.

Original entry on oeis.org

1, 2, 2, 7, 5, 1, 7, 2, 9, 3, 11, 3, 6, 4, 14, 5, 9, 16, 6, 15, 6, 10, 8, 21, 21, 21, 2, 28, 3, 30, 3, 6, 4, 33, 5, 9, 35, 6, 34, 6, 10, 8, 40, 40, 40, 2, 47, 3, 49, 3, 6, 4, 52, 5, 9, 54, 6, 53, 6, 10, 8, 59, 59, 59, 2, 66, 3, 68, 3, 6, 4, 71, 5, 9, 73, 6, 72, 6, 10, 8, 78, 78, 78, 2, 85, 3, 87, 3, 6, 4

Offset: 1

Altug Alkan and Rémy Sigrist, Aug 07 2019

A well-defined solution sequence for recurrence a(n) = a(n-a(n-1)) + a(n-a(n-3)).

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-1).

Cf. A046700, A244477, A296518, A309492, A309494, A309496.

I:=[1,2,2,7,5,1,7,2];[n le 8 select I[n] else Self(n-Self(n-1))+Self(n-Self(n-3)): n in [1..90]]; // Marius A. Burtea, Aug 08 2019

q=vector(100); q[1]=q[6]=1; q[2]=q[3]=q[8]=2; q[4]=q[7]=7; q[5]=5; for(n=9, #q, q[n]=q[n-q[n-1]]+q[n-q[n-3]]); q

Vec(x*(1 + x)*(1 + x + x^2 + 6*x^3 - x^4 + 2*x^5 + 5*x^6 - 3*x^7 + 12*x^8 - 9*x^9 + 20*x^10 - 17*x^11 + 23*x^12 - 19*x^13 + 33*x^14 - 28*x^15 + 37*x^16 - 21*x^17 + 27*x^18 - 14*x^19 + 16*x^20 - 10*x^21 + 4*x^22 + 7*x^23 + 12*x^24 - 5*x^25 + 3*x^26 + 7*x^27 - 10*x^28 + 18*x^29 - 21*x^30 + 15*x^31 - 19*x^32 + 24*x^33 - 29*x^34 + 20*x^35 - 17*x^36 + 11*x^37 - 6*x^38 + 2*x^39 - 10*x^40 + 9*x^41 - 6*x^42 + 5*x^43) / ((1 - x)^2*(1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 + x^10 + x^11 + x^12 + x^13 + x^14 + x^15 + x^16 + x^17 + x^18)^2) + O(x^90)) \\ Colin Barker, Aug 11 2019

For k >= 1:
a(19*k-11) = 2,
a(19*k-10) = 19*k-10,
a(19*k-9) = 3,
a(19*k-8) = 19*k-8,
a(19*k-7) = 3,
a(19*k-6) = 6,
a(19*k-5) = 4,
a(19*k-4) = 19*k-5,
a(19*k-3) = 5,
a(19*k-2) = 9,
a(19*k-1) = 19*k-3,
a(19*k) = 6,
a(19*k+1) = 19*k-4,
a(19*k+2) = 6,
a(19*k+3) = 10,
a(19*k+4) = 8,
a(19*k+5) = a(19*k+6) = a(19*k+7) = 19*k+2.
From Colin Barker, Aug 08 2019: (Start)
G.f.: x*(1 + x)*(1 + x + x^2 + 6*x^3 - x^4 + 2*x^5 + 5*x^6 - 3*x^7 + 12*x^8 - 9*x^9 + 20*x^10 - 17*x^11 + 23*x^12 - 19*x^13 + 33*x^14 - 28*x^15 + 37*x^16 - 21*x^17 + 27*x^18 - 14*x^19 + 16*x^20 - 10*x^21 + 4*x^22 + 7*x^23 + 12*x^24 - 5*x^25 + 3*x^26 + 7*x^27 - 10*x^28 + 18*x^29 - 21*x^30 + 15*x^31 - 19*x^32 + 24*x^33 - 29*x^34 + 20*x^35 - 17*x^36 + 11*x^37 - 6*x^38 + 2*x^39 - 10*x^40 + 9*x^41 - 6*x^42 + 5*x^43) / ((1 - x)^2*(1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 + x^10 + x^11 + x^12 + x^13 + x^14 + x^15 + x^16 + x^17 + x^18)^2).
a(n) = 2*a(n-19) - a(n-38) for n > 45.
(End)

A309636 a(1) = 3, a(2) = 1, a(3) = 4, a(4) = 2, a(5) = 5; a(6) = 3; a(n) = a(n-a(n-1)) + a(n-a(n-4)) for n > 6.

Original entry on oeis.org

3, 1, 4, 2, 5, 3, 6, 4, 7, 10, 8, 6, 9, 7, 10, 13, 6, 14, 12, 10, 18, 6, 14, 17, 10, 23, 11, 14, 22, 10, 28, 16, 14, 27, 10, 33, 16, 14, 32, 10, 38, 16, 19, 37, 10, 43, 16, 24, 42, 10, 48, 16, 24, 47, 10, 53, 16, 24, 52, 10, 58, 16, 24, 57, 10, 63, 21, 24, 62, 10, 68, 26, 24, 67, 10

Offset: 1

Altug Alkan and Nathan Fox, Aug 10 2019

A well-defined quasi-periodic solution for Hofstadter V recurrence (a(n) = a(n-a(n-1)) + a(n-a(n-4))).

Michael De Vlieger, Table of n, a(n) for n = 1..10006
Altug Alkan, Nathan Fox, Orhan Ozgur Aybar, Zehra Akdeniz, On Some Solutions to Hofstadter's V-Recurrence, arXiv:2002.03396 [math.DS], 2020.

Cf. A063882, A244477, A296518, A309492, A309494, A309496, A309554, A309567.

I:=[3,1,4,2,5,3]; [n le 6 select I[n] else  Self(n-Self(n-1)) + Self(n-Self(n-4)): n in [1..80]]; // Marius A. Burtea, Aug 11 2019

Nest[Append[#, #[[-#[[-1]] ]] + #[[-#[[-4]] ]]] &, {3, 1, 4, 2, 5, 3}, 69] (* Michael De Vlieger, May 08 2020 *)

q=vector(100); q[1]=3; q[2]=1; q[3]=4; q[4]=2; q[5]=5; q[6]=3; for(n=7, #q, q[n] = q[n-q[n-1]] + q[n-q[n-4]]); q

For k > 1:
a(5*k)   = 10,
a(5*k+1) = 5*k-2,
a(5*k+2) = 5*(floor((sqrt(2*k-1)-1)/2) + floor((sqrt(2*k-3)-1)/2)) + 6,
a(5*k+3) = 5*(floor(sqrt(k/2)) + floor(sqrt((k-1)/2))) + 4,
a(5*k+4) = 5*k-3.
Also, a(5*k+2) = 5*f(k)+1 and a(5*k+3) = 5*g(k)-1 where f(k) = g(k-g(k-1)) and g(k) = f(k-f(k))+2 with f(1) = g(1) = 1, g(2) = 2.

A275363 a(1)=3, a(2)=6, a(3)=3; thereafter a(n) = a(n-a(n-1)) + a(n-1-a(n-2)).

Original entry on oeis.org

3, 6, 3, 3, 9, 6, 3, 12, 9, 3, 15, 12, 3, 18, 15, 3, 21, 18, 3, 24, 21, 3, 27, 24, 3, 30, 27, 3, 33, 30, 3, 36, 33, 3, 39, 36, 3, 42, 39, 3, 45, 42, 3, 48, 45, 3, 51, 48, 3, 54, 51, 3, 57, 54, 3, 60, 57, 3, 63, 60, 3, 66, 63, 3, 69, 66, 3, 72, 69

Offset: 1

Nathan Fox, Jul 24 2016

nonn

Same recurrence as in A046699 but with different starting values.

This sequence is quasilinear.

Nathan Fox, Table of n, a(n) for n = 1..1000
Nathan Fox, Finding Linear-Recurrent Solutions to Hofstadter-Like Recurrences Using Symbolic Computation, arXiv:1609.06342 [math.NT], 2016.
Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).

Cf. A046699, A005185, A244477.

Flatten[Array[{3, 3*# + 6, 3*# + 3} &, 30, 0]] (* Paolo Xausa, Oct 23 2024 *)
LinearRecurrence[{0,0,2,0,0,-1},{3,6,3,3,9,6},80] (* Harvey P. Dale, Nov 27 2024 *)

a(3n) = 3n, a(3n+1) = 3, a(3n+2) = 3n+6.
a(n) = 2*a(n-3) - a(n-6) for n>6.
G.f.: -(3*x^4 +3*x^3 -3*x^2 -6*x-3)/((x-1)^2*(x^2+x+1)^2).

cp's OEIS Frontend

A309567 a(1) = 4, a(2) = 2, a(3) = 5, a(4) = 3, a(5) = 1; a(n) = a(n-a(n-1)) + a(n-a(n-4)) for n > 5.

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A309496 a(1) = 1, a(2) = 3, a(3) = a(4) = a(6) = 6, a(5) = 2, a(7) = 4; a(n) = a(n-a(n-4)) + a(n-a(n-5)) for n > 7.

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A269328 An eventually quasilinear solution to Hofstadter's Q recurrence.

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

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

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A283878 An eventually quasilinear solution to Hofstadter's Q recurrence.

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A283879 An eventually quasilinear solution to Hofstadter's Q recurrence.

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A309554 a(1) = a(6) = 1, a(2) = a(3) = a(8) = 2, a(4) = a(7) = 7, a(5) = 5; a(n) = a(n-a(n-1)) + a(n-a(n-3)) for n > 8.