A349407 -id:A349407 - cp's OEIS frontend

A350279 Irregular triangle T(n,k) read by rows in which row n lists the iterates of the Farkas map (A349407) from 2*n - 1 to 1.

1, 3, 1, 5, 3, 1, 7, 11, 17, 9, 3, 1, 9, 3, 1, 11, 17, 9, 3, 1, 13, 7, 11, 17, 9, 3, 1, 15, 5, 3, 1, 17, 9, 3, 1, 19, 29, 15, 5, 3, 1, 21, 7, 11, 17, 9, 3, 1, 23, 35, 53, 27, 9, 3, 1, 25, 13, 7, 11, 17, 9, 3, 1, 27, 9, 3, 1, 29, 15, 5, 3, 1

Offset: 1

Paolo Xausa, Dec 22 2021

			Written as an irregular triangle, the sequence begins:
  n\k   1   2   3   4   5   6   7
  -------------------------------
   1:   1
   2:   3   1
   3:   5   3   1
   4:   7  11  17   9   3   1
   5:   9   3   1
   6:  11  17   9   3   1
   7:  13   7  11  17   9   3   1
   8:  15   5   3   1
   9:  17   9   3   1
  10:  19  29  15   5   3   1
  11:  21   7  11  17   9   3   1
  12:  23  35  53  27   9   3   1

Paolo Xausa, Table of n, a(n) for n = 1..12301 (rows 1..1000 of triangle, flattened).
H. M. Farkas, "Variants of the 3N+1 Conjecture and Multiplicative Semigroups", in Entov, Pinchover and Sageev, Geometry, Spectral Theory, Groups, and Dynamics, Contemporary Mathematics, vol. 387, American Mathematical Society, 2005, p. 121.
J. C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, American Mathematical Society, 2010, p. 74.

Cf. A349407, A375909 (# of iterations), A375910 (row sums), A375911 (row maxs).

Cf. A070165.

FarkasStep[x_] := Which[Divisible[x, 3], x/3, Mod[x, 4] == 3, (3*x + 1)/2, True, (x + 1)/2];
Array[Most[FixedPointList[FarkasStep, 2*# - 1]] &, 15] (* Paolo Xausa, Sep 03 2024 *)

T(n,1) = 2*n-1; T(n,k) = A349407((T(n,k-1)+1)/2), where n >= 1 and k >= 2.

A375909 Number of iterations of the Farkas map (A349407) to reach 1 starting from 2*n - 1.

Original entry on oeis.org

0, 1, 2, 5, 2, 4, 6, 3, 3, 5, 6, 6, 7, 3, 4, 9, 5, 5, 6, 7, 7, 7, 4, 8, 8, 4, 4, 10, 6, 6, 10, 7, 6, 6, 7, 7, 7, 8, 8, 9, 4, 9, 8, 5, 5, 9, 10, 10, 9, 6, 5, 11, 6, 6, 11, 7, 7, 7, 8, 8, 11, 8, 8, 13, 8, 8, 7, 5, 8, 8, 9, 9, 8, 9, 9, 10, 5, 10, 10, 5, 5, 10, 11, 11, 9

Offset: 1

Paolo Xausa, Sep 02 2024

			a(10) = 5 because the trajectory 19 -> 29 -> 15 -> 5 -> 3 -> 1 takes 5 steps.

Paolo Xausa, Table of n, a(n) for n = 1..10000

(Row lengths of A350279) - 1.

Cf. A006577, A006666, A349407, A375267, A375911.

FarkasStep[x_] := Which[Divisible[x, 3], x/3, Mod[x, 4] == 3, (3*x + 1)/2, True, (x + 1)/2];
Array[Length[FixedPointList[FarkasStep, 2*# - 1]] - 2 &, 100]

A375911 Largest value in the trajectory of 2*n - 1 in the Farkas map (A349407).

Original entry on oeis.org

1, 3, 5, 17, 9, 17, 17, 15, 17, 29, 21, 53, 25, 27, 29, 161, 33, 53, 37, 39, 41, 65, 45, 161, 49, 51, 53, 125, 57, 89, 161, 63, 65, 101, 69, 161, 73, 75, 77, 269, 81, 125, 85, 87, 89, 137, 161, 485, 97, 99, 101, 233, 105, 161, 125, 111, 113, 173, 117, 269, 161

Offset: 1

Paolo Xausa, Sep 02 2024

			a(10) = 29 because 29 is the largest value in the trajectory 19 -> 29 -> 15 -> 5 -> 3 -> 1.

Paolo Xausa, Table of n, a(n) for n = 1..10000

Cf. A349407, A350279, A375909.

Cf. A025586, A365478, A375280.

FarkasStep[x_] := Which[Divisible[x, 3], x/3, Mod[x, 4] == 3, (3*x + 1)/2, True, (x + 1)/2];
Array[Max[FixedPointList[FarkasStep, 2*# - 1]] &, 100]

a(n) = max{A350279(n,k) for 1 <= k <= A375909(n) + 1}.

A375265 a(n) = n/3 if n mod 3 = 0; otherwise a(n) = n/2 if n mod 2 = 0; otherwise a(n) = 3*n + 1.

Original entry on oeis.org

4, 1, 1, 2, 16, 2, 22, 4, 3, 5, 34, 4, 40, 7, 5, 8, 52, 6, 58, 10, 7, 11, 70, 8, 76, 13, 9, 14, 88, 10, 94, 16, 11, 17, 106, 12, 112, 19, 13, 20, 124, 14, 130, 22, 15, 23, 142, 16, 148, 25, 17, 26, 160, 18, 166, 28, 19, 29, 178, 20, 184, 31, 21, 32, 196, 22, 202, 34, 23

Offset: 1

Paolo Xausa, Aug 08 2024

Anderson (1987) reformulates the 3x+1 conjecture using this function.

Paolo Xausa, Table of n, a(n) for n = 1..10000
Stuart Anderson, Struggling with the 3x+1 problem, The Mathematical Gazette, Vol. 71, Issue 458, December 1987, p. 273 (see A005186 for a scanned copy).
Jeffrey C. Lagarias, The 3x + 1 Problem: An Annotated Bibliography (1963-1999), arXiv:math/0309224 [math.NT], 2011, p. 6.
Index entries for sequences related to 3x+1 (or Collatz) problem.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,2,0,0,0,0,0,-1).

Cf. A375266 (trajectories).

Cf. A005186, A006370, A014682, A185452, A349407, A350515.

a := n -> ifelse(irem(n, 3) = 0, iquo(n, 3), ifelse(irem(n, 2) = 0, iquo(n, 2), 3*n + 1)): seq(a(n), n = 1..69);  # Peter Luschny, Aug 14 2024

A375265[n_] := Which[Divisible[n, 3], n/3, Divisible[n, 2], n/2, True,3*n + 1];
Array[A375265, 100]

A350515 a(n) = (n-1)/3 if n mod 3 = 1; a(n) = n/2 if n mod 6 = 0 or n mod 6 = 2; a(n) = (3n+1)/2 if n mod 6 = 3 or n mod 6 = 5.

Original entry on oeis.org

0, 0, 1, 5, 1, 8, 3, 2, 4, 14, 3, 17, 6, 4, 7, 23, 5, 26, 9, 6, 10, 32, 7, 35, 12, 8, 13, 41, 9, 44, 15, 10, 16, 50, 11, 53, 18, 12, 19, 59, 13, 62, 21, 14, 22, 68, 15, 71, 24, 16, 25, 77, 17, 80, 27, 18, 28, 86, 19, 89, 30, 20, 31, 95, 21, 98, 33, 22, 34, 104

Offset: 0

Paolo Xausa, Jan 02 2022

This is a variant of the Farkas map (A349407).
Yolcu, Aaronson and Heule prove that the trajectory of the iterates of the map starting from any nonnegative integer always reaches 0.
If displayed as a rectangular array with six columns, the columns are A008585, A005843, A016777, A017221, A005408, A017257 (see example). - Omar E. Pol, Jan 02 2022

			From _Omar E. Pol_, Jan 02 2022: (Start)
Written as a rectangular array with six columns read by rows the sequence begins:
   0,  0,  1,  5,  1,  8;
   3,  2,  4, 14,  3, 17;
   6,  4,  7, 23,  5, 26;
   9,  6, 10, 32,  7, 35;
  12,  8, 13, 41,  9, 44;
  15, 10, 16, 50, 11, 53;
  18, 12, 19, 59, 13, 62;
  21, 14, 22, 68, 15, 71;
  24, 16, 25, 77, 17, 80;
  27, 18, 28, 86, 19, 89;
  30, 20, 31, 95, 21, 98;
...
(End)

H. M. Farkas, "Variants of the 3N+1 Conjecture and Multiplicative Semigroups", in Entov, Pinchover and Sageev, Geometry, Spectral Theory, Groups, and Dynamics, Contemporary Mathematics, vol. 387, American Mathematical Society, 2005, p. 121.
Emre Yolcu, Scott Aaronson and Marijn J. H. Heule, An Automated Approach to the Collatz Conjecture, arXiv:2105.14697 [cs.LO], 2021, pp. 21-25.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,2,0,0,0,0,0,-1).

Cf. A005408, A005843, A006370, A014682, A016777, A017221, A017257, A349407, A350279.

nterms=100;Table[If[Mod[n,3]==1,(n-1)/3,If[Mod[n,6]==0||Mod[n,6]==2,n/2,(3n+1)/2]],{n,0,nterms-1}]
(* Second program *)
nterms=100;LinearRecurrence[{0,0,0,0,0,2,0,0,0,0,0,-1},{0,0,1,5,1,8,3,2,4,14,3,17},nterms]

def a(n):
    r = n%6
    if r == 1 or r == 4: return (n-1)//3
    if r == 0 or r == 2: return n//2
    if r == 3 or r == 5: return (3*n+1)//2
print([a(n) for n in range(70)]) # Michael S. Branicky, Jan 02 2022

a(n) = (A349407(n+1)-1)/2.

a(n) = 2*a(n-6)-a(n-12). - Wesley Ivan Hurt, Jan 03 2022

A375910 Row sums of A350279.

Original entry on oeis.org

1, 4, 9, 48, 13, 41, 61, 24, 30, 72, 69, 151, 86, 40, 53, 538, 74, 128, 109, 100, 110, 182, 69, 507, 135, 81, 93, 395, 129, 217, 599, 132, 139, 249, 220, 460, 182, 161, 177, 850, 121, 340, 267, 140, 158, 448, 631, 1625, 232, 173, 182, 708, 233, 389, 504, 220, 242, 428

Offset: 1

Paolo Xausa, Sep 02 2024

1

Paolo Xausa, Table of n, a(n) for n = 1..10000

Cf. A349407, A350279, A375909, A375911.

Cf. A033493, A285098.

FarkasStep[x_] := Which[Divisible[x, 3], x/3, Mod[x, 4] == 3, (3*x + 1)/2, True, (x + 1)/2];
Array[Total[FixedPointList[FarkasStep, 2*# - 1]] - 1 &, 100]

a(n) = Sum_{k = 1..A375909(n) + 1} A350279(n,k).

A350548 Irregular triangle T(n,k) read by rows in which row n lists the iterates of the A350515 map from n to 0.

Original entry on oeis.org

0, 1, 0, 2, 1, 0, 3, 5, 8, 4, 1, 0, 4, 1, 0, 5, 8, 4, 1, 0, 6, 3, 5, 8, 4, 1, 0, 7, 2, 1, 0, 8, 4, 1, 0, 9, 14, 7, 2, 1, 0, 10, 3, 5, 8, 4, 1, 0, 11, 17, 26, 13, 4, 1, 0, 12, 6, 3, 5, 8, 4, 1, 0, 13, 4, 1, 0, 14, 7, 2, 1, 0, 15, 23, 35, 53, 80, 40, 13, 4, 1, 0

Offset: 0

Paolo Xausa, Jan 04 2022

			Written as an irregular triangle, the sequence begins:
  n\k   0   1   2   3   4   5   6
  -------------------------------
   0:   0
   1:   1   0
   2:   2   1   0
   3:   3   5   8   4   1   0
   4:   4   1   0
   5:   5   8   4   1   0
   6:   6   3   5   8   4   1   0
   7:   7   2   1   0
   8:   8   4   1   0
   9:   9  14   7   2   1   0
  10:  10   3   5   8   4   1   0
  11:  11  17  26  13   4   1   0
  ...

Emre Yolcu, Scott Aaronson and Marijn J. H. Heule, An Automated Approach to the Collatz Conjecture, arXiv:2105.14697 [cs.LO], 2021, pp. 21-25.
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A070165, A349407, A350279, A350515.

A350515[n_]:=If[Mod[n,3]==1,(n-1)/3,If[Mod[n,6]==0||Mod[n,6]==2,n/2,(3n+1)/2]];
nrows=20;Table[NestWhileList[A350515,n,#>0&],{n,0, nrows-1}]

T(n,0) = n; T(n,k) = A350515(T(n,k-1)), where n >= 0 and k >= 1.

T(n,k) = (A350279(n+1,k+1)-1)/2, where n >= 0 and k >= 0.

cp's OEIS Frontend

A350279 Irregular triangle T(n,k) read by rows in which row n lists the iterates of the Farkas map (A349407) from 2*n - 1 to 1.

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A375909 Number of iterations of the Farkas map (A349407) to reach 1 starting from 2*n - 1.

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A375911 Largest value in the trajectory of 2*n - 1 in the Farkas map (A349407).

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A375265 a(n) = n/3 if n mod 3 = 0; otherwise a(n) = n/2 if n mod 2 = 0; otherwise a(n) = 3*n + 1.

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A350515 a(n) = (n-1)/3 if n mod 3 = 1; a(n) = n/2 if n mod 6 = 0 or n mod 6 = 2; a(n) = (3n+1)/2 if n mod 6 = 3 or n mod 6 = 5.

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A375910 Row sums of A350279.

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A350548 Irregular triangle T(n,k) read by rows in which row n lists the iterates of the A350515 map from n to 0.

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