A003842 -id:A003842 - cp's OEIS frontend

A182028 Take first n bits of the infinite Fibonacci word A003849, regard them as a binary number, then convert it to base 10.

0, 1, 2, 4, 9, 18, 37, 74, 148, 297, 594, 1188, 2377, 4754, 9509, 19018, 38036, 76073, 152146, 304293, 608586, 1217172, 2434345, 4868690, 9737380, 19474761, 38949522, 77899045, 155798090, 311596180, 623192361, 1246384722, 2492769444, 4985538889, 9971077778

Offset: 0

Reinhard Zumkeller, Apr 07 2012

a(n) mod 2 = A003849(n);

a(n) = A000225(n+1) - A044432(n).

			0 ->                            0 -> a(0) = 0,
0,1 ->                         01 -> a(1) = 1,
0,1,0 ->                      010 -> a(2) = 2,
0,1,0,0 ->                   0100 -> a(3) = 4,
0,1,0,0,1 ->                01001 -> a(4) = 9,
0,1,0,0,1,0 ->             010010 -> a(5) = 18,
0,1,0,0,1,0,1 ->          0100101 -> a(6) = 37
0,1,0,0,1,0,1,0 ->       01001010 -> a(7) = 74
0,1,0,0,1,0,1,0,0 ->    010010100 -> a(8) = 148,
0,1,0,0,1,0,1,0,0,1 -> 0100101001 -> a(9) = 297.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000

Cf. A003842, A003849, A000225, A044432, A214318.

a182028 n = a182028_list !! n
a182028_list = scanl1 (\v b -> 2 * v + b) a003849_list

nesting = 7; A003849 = Flatten[Nest[{#, #[[1]]}&, {0, 1}, nesting]]; a[n_] := FromDigits[Take[A003849, n+1], 2]; Table[a[n], {n, 0, Length[A003849] - 1}] (* Jean-François Alcover, Feb 13 2016 *)

a(n) = 2*a(n-1) + A003849(n) for n > 0, a(0) = 0.

A103269 Apply the tribonacci morphism 1 -> {1, 2}, 2 -> {1, 3}, 3 -> {1} n times to 1, and concatenate the resulting string.

Original entry on oeis.org

1, 12, 1213, 1213121, 1213121121312, 121312112131212131211213, 12131211213121213121121312131211213121213121, 121312112131212131211213121312112131212131211213121121312121312112131213121121312

Offset: 0

Jonathan Vos Post and Robert G. Wilson v, May 16 2005

nonn

The number of letters in the n-th iteration is tribonacci(n+3) (that is, A000073(n+3)).

N. J. A. Sloane, Table of n, a(n) for n = 0..10 (shortened by _N. J. A. Sloane_, Jan 18 2019)
Elena Barcucci, Luc Belanger and Srecko Brlek, On tribonacci sequences, Fib. Q., 42 (2004), 314-320. See T on page 315.
Julien Cassaigne, Sebastien Ferenczi, and Luca Q. Zamboni, Imbalances in Arnoux-Rauzy sequences, Annales de l'institut Fourier, 50 (2000), 1265-1276.
David Damanik and Luca Q. Zamboni, Arnoux-Rauzy Subshifts: Linear Recurrence, Powers and Palindromes, arXiv:math/0208137 [math.CO], 2002.
C. Holton and L. Q. Zamboni, Directed Graphs and Substitutions, Theory Comput. Systems, 34 (2001) 545-564
S. Ito and M. Kimura, On Rauzy fractal, Japan J. Indust. Appl. Math. 8 (1991) 461-486.
A. Messaoudi, Propriétes arithmétiques et dynamiques du fractal de Rauzy, J. Th. Nombres Bordeaux, 10 (1998) 195-224.
Gérard Rauzy, Nombres algébriques et substitutions, Bull. Soc. Math. France, 110 (1982), 147-178.
Bo Tan and Zhi-Ying Wen, Some properties of the Tribonacci sequence, European Journal of Combinatorics, 28 (2007) 1703-1719. See A_n.
Index entries for sequences that are fixed points of mappings

A092782 is limit of these strings.

Cf. A000073, A106748, A106749, A003842, A106750, A092782, A317196.

FromDigits /@ NestList[ Flatten[ # /. {1 -> {1, 2}, 2 -> {1, 3}, 3 -> 1}] &, {1}, 7]
(* Second program: *)
FromDigits /@ SubstitutionSystem[{1 -> {1, 2}, 2 -> {1, 3}, 3 -> {1}}, {1}, 7] (* Jean-François Alcover, Nov 12 2018 *)

Definition edited by N. J. A. Sloane, Aug 06 2018

A106750 Define the "Fibonacci" morphism f: 1->12, 2->1 and let a(0) = 2; then a(n+1) = f(a(n)).

Original entry on oeis.org

2, 1, 12, 121, 12112, 12112121, 1211212112112, 121121211211212112121, 1211212112112121121211211212112112, 1211212112112121121211211212112112121121211211212112121

Offset: 0

N. J. A. Sloane, May 16 2005. Initial term 2 added by N. J. A. Sloane, Jul 05 2012

nonn

a(n) converges to the Fibonacci word A003842.

a(n) has length Fibonacci(n+1) (cf. A000045).

Berstel, Jean. "Fibonacci words—a survey." In The book of L, pp. 13-27. Springer Berlin Heidelberg, 1986.
E. Bombieri and J. Taylor, Which distribution of matter diffracts? An initial investigation, in International Workshop on Aperiodic Crystals (Les Houches, 1986), J. de Physique, Colloq. C3, 47 (1986), C3-19 to C3-28.

N. J. A. Sloane, Table of n, a(n) for n = 0..15

Cf. A106748, A106749, A003842, A000045, A213975, A213976.

FromDigits /@ NestList[ Flatten[ # /. {1 -> {1, 2}, 2 -> 1}] &, {2}, 8] (* Robert G. Wilson v, May 17 2005 *)

More terms from Robert G. Wilson v, May 17 2005

A112658 Dean's Word: Omega 2,1: the trajectory of 0 -> 01, 1 -> 21, 2 -> 03, 3 -> 23.

Original entry on oeis.org

0, 1, 2, 1, 0, 3, 2, 1, 0, 1, 2, 3, 0, 3, 2, 1, 0, 1, 2, 1, 0, 3, 2, 3, 0, 1, 2, 3, 0, 3, 2, 1, 0, 1, 2, 1, 0, 3, 2, 1, 0, 1, 2, 3, 0, 3, 2, 3, 0, 1, 2, 1, 0, 3, 2, 3, 0, 1, 2, 3, 0, 3, 2, 1, 0, 1, 2, 1, 0, 3, 2, 1, 0, 1, 2, 3, 0, 3, 2, 1, 0, 1, 2, 1, 0, 3, 2, 3, 0, 1, 2, 3, 0, 3, 2, 3, 0, 1, 2, 1, 0, 3, 2, 1, 0

Offset: 1

Jeremy Gardiner, Dec 27 2005

nonn

Even-indexed terms of this sequence are the sequence A099545. - Alexandre Wajnberg, Jan 02 2006

Fractal sequence: odd terms are 0, 2, 0, 2,...; the subsets formed with the terms of index (2^i)n, with i>0, are identical: a(2n)=a(4n)=a(8n)=a(16n)=... - Alexandre Wajnberg, Jan 02 2006

			The first few iterations of the morphism, starting with 0:
Start: 0
Rules:
  0 --> 01
  1 --> 21
  2 --> 03
  3 --> 23
-------------
0:   (#=1)
  0
1:   (#=2)
  01
2:   (#=4)
  0121
3:   (#=8)
  01210321
4:   (#=16)
  0121032101230321
5:   (#=32)
  01210321012303210121032301230321
6:   (#=64)
  0121032101230321012103230123032101210321012303230121032301230321
/* _Joerg Arndt_, Jul 18 2012 */

J.-P. Allouche and M. Mendes France, Automata and Automatic Sequences, in: Axel F. and Gratias D. (eds), Beyond Quasicrystals. Centre de Physique des Houches, vol 3. Springer, Berlin, Heidelberg, pp. 293-367, 1995; DOI https://doi.org/10.1007/978-3-662-03130-8_11. See page 6.
J.-P. Allouche and M. Mendes France, Automata and Automatic Sequences, in: Axel F. and Gratias D. (eds), Beyond Quasicrystals. Centre de Physique des Houches, vol 3. Springer, Berlin, Heidelberg, pp. 293-367, 1995; DOI https://doi.org/10.1007/978-3-662-03130-8_11. See page 6. [Local copy]
Kirby A. Baker, George F. McNulty, Walter Taylor, Growth Problems For Avoidable Words, Theoretical Computer Science, volume 69, number 3, 1989, pages 319-345. (See morphism start of section 3, page 325.)
Richard A. Dean, A sequence without repeats on x, ..., Amer. Math. Monthly 72, 1965. pp. 383-385. MR 31 #350.
George F. McNulty, Avoidable Words, conference slides, 2003, slides 38-39. (Also conference abstract.)
Index entries for sequences that are fixed points of mappings
Index entries for sequences related to squarefree words

Essentially the same: A343180, also A122002 (map 0123 -> 1537), A125047 (map 0123 -> 2134).

Cf. A003324.

Nest[ Flatten[ # /. {0 -> {0, 1}, 1 -> {2, 1}, 2 -> {0, 3}, 3 -> {2, 3}}] &, {0}, 7] (* Robert G. Wilson v, Dec 27 2005 *)

a(n) = 2*bittest(n,valuation(n,2)+1) + !(n%2); \\ Kevin Ryde, Sep 09 2020

It should be easy to prove that a(4n) = 0, a(4n+2) = 2, a(8n+1) = 1, a(8n+5) = 3, a(4n+3) = a(2n+1). This would imply that a(2n) = 2(n mod 2), a(2n+1) = 1 + 2*A014707(n), with A014707(n) the classical paperfolding curve. - Ralf Stephan, Dec 28 2005

A214318 Replace the word A214317(n) with its position in A007931.

Original entry on oeis.org

1, 4, 9, 19, 40, 81, 164, 329, 659, 1320, 2641, 5283, 10568, 21137, 42276, 84553, 169107, 338216, 676433, 1352868, 2705737, 5411475, 10822952, 21645905, 43291811, 86583624, 173167249, 346334500, 692669001, 1385338003, 2770676008, 5541352017, 11082704035

Offset: 1

N. J. A. Sloane, Jul 12 2012

nonn

			A214317(5) = 12112 is the 40th term of A007931, so a(5)=40.

Alois P. Heinz, Table of n, a(n) for n = 1..1000

Cf. A003842, A214317, A007931.

S:= proc(n) option remember;
      `if`(n<2, [2-n], [S(n-1)[], S(n-2)[]])
    end:
a:= proc(n) option remember; local k;
      for k while nops(S(k))Alois P. Heinz, Jul 19 2012

nesting = 6; A003849 = Flatten[Nest[{#, #[[1]]}&, {0, 1}, nesting]]; A182028[n_] := FromDigits[Take[A003849, n+1], 2]; a[n_] := A182028[n-1] + 2^n - 1; Table[a[n], {n, 1, Length[A003849]}] (* Jean-François Alcover, Feb 13 2016 *)

a(n) = A182028(n-1)+2^n-1.

a(n) = 2*a(n-1) + A003842(n-1) for n>1, a(1) = 1. - Alois P. Heinz, Jul 19 2012

More terms from Alois P. Heinz, Jul 19 2012

A361756 Irregular triangle T(n, k), n >= 0, k = 1..A361757(n), read by rows; the n-th row lists the numbers k such that the Fibonacci numbers that appear in the dual Zeckendorf representation of k also appear in that of n.

Original entry on oeis.org

0, 0, 1, 0, 2, 0, 1, 2, 3, 0, 1, 4, 0, 2, 5, 0, 1, 2, 3, 4, 5, 6, 0, 2, 7, 0, 1, 2, 3, 7, 8, 0, 1, 4, 9, 0, 2, 5, 7, 10, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 4, 12, 0, 2, 5, 13, 0, 1, 2, 3, 4, 5, 6, 12, 13, 14, 0, 2, 7, 15, 0, 1, 2, 3, 7, 8, 15, 16

Offset: 0

Rémy Sigrist, Mar 23 2023

In other words, the n-th row lists the numbers k such that A003754(1+n) AND A003754(1+k) = A003754(1+k) (where AND denotes the bitwise AND operator).

The dual Zeckendorf representation is also known as the lazy Fibonacci representation (see A356771 for further details).

			Triangle T(n, k) begins:
  n   n-th row
  --  -------------------------------------
   0  0
   1  0, 1
   2  0, 2
   3  0, 1, 2, 3
   4  0, 1, 4
   5  0, 2, 5
   6  0, 1, 2, 3, 4, 5, 6
   7  0, 2, 7
   8  0, 1, 2, 3, 7, 8
   9  0, 1, 4, 9
  10  0, 2, 5, 7, 10
  11  0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11
  12  0, 1, 4, 12

Rémy Sigrist, Table of n, a(n) for n = 0..9956 (rows for n = 0..377 flattened)
Rémy Sigrist, PARI program
Index entries for sequences related to Zeckendorf expansion of n

See A361755 for a similar sequence.

Cf. A003754, A003842, A356771, A361757.

PARI
```
See Links section.
```

T(n, 1) = 0.
T(n, 2) = A003842(n - 1) for any n > 0.
T(n, A361757(n)) = n.

A380560 Rectangular array R, read by descending antidiagonals: (row 1) = (R(1,k)) = (A006337(k)), k >= 1; (row n+1) = inverse runlength sequence of row n; and R(n,1) = 1 for n >=1, See Comments.

Original entry on oeis.org

1, 2, 1, 1, 2, 1, 2, 2, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 1, 1, 2, 2, 1

Offset: 1

Clark Kimberling, Jan 27 2025

For present purposes, all sequences to be considered consist entirely of 1s and 2s. If u and v are such sequences (infinite or finite), we call v an inverse runlength sequence of u if u is the runlength sequence of v. Each u has two inverse runlength sequences, one with first term 1 and the other with first term 2. Consequently, an inverse runlength array (in which each row after the first is an inverse runlength sequence of the preceding row) is determined by its first column. In this array, the first column consists solely of 1s. No two rows are identical.
Row 1: A006337; limiting row: A000002 (Kolakoski sequence).
Guide to related sequences (arrays):
A378282, limiting sequences of periodic inverse runlength arrays
A380560, (row 1)=A006337, periodic (column 1)=(1,...)
A380561, (row 1)=A006337, periodic (column 1)=(1,1,2,...)
A380562, (row 1)=A006337, periodic (column 1)=(1,2,2,...)
A380563, (row 1)=1+A010060, periodic (column 1)=(1,...)
A378303, (ro1 1)=A006337, periodic (column 1)=(2,2,1,...)
A378396, self-inverse runlength array, u = v = 1+A010060
A378397, self-inverse runlength array, u = v = A003842
A378398, self-inverse runlength array, u = v = A014675
A378399, self-inverse runlength array, u = v = A006337

			Corner:
  1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 2 1 1 2 1
  1 2 2 1 2 2 1 2 1 1 2 1 1 2 1 2 2 1 2 2
  1 2 2 1 1 2 1 1 2 2 1 2 2 1 2 1 1 2 1 2
  1 2 2 1 1 2 1 2 2 1 2 1 1 2 2 1 2 2 1 1
  1 2 2 1 1 2 1 2 2 1 2 2 1 1 2 1 1 2 1 2
  1 2 2 1 1 2 1 2 2 1 2 2 1 1 2 1 1 2 2 1
  1 2 2 1 1 2 1 2 2 1 2 2 1 1 2 1 1 2 2 1

Cf. A000002, A000012 (column 1), A006337.

invR[seq_] := Flatten[Map[ConstantArray[#[[2]], #[[1]]] &,
    Partition[Riffle[seq, {1, 2}, {2, -1, 2}], 2]]];
s = Differences[Table[Floor[n*Sqrt[2]], {n, 1, 21}]]; (* A006337 *)
t = NestList[invR, s, 12];
u[n_] := Take[t[[n]], 20];
Table[u[n], {n, 1, 12}]  (* array *)
v[n_, k_] := t[[n]][[k]];
Table[v[n - k + 1, k], {n, 12}, {k, n, 1, -1}] // Flatten (* sequence *)
(* Peter J. C. Moses, Nov 13 2024 *)

A110006 a(n) = n-F(F(n)) where F(x)=A120613(x)=floor(phi*floor(x/phi)) and phi=(1+sqrt(5))/2.

Original entry on oeis.org

1, 2, 3, 3, 2, 3, 3, 4, 3, 2, 3, 3, 2, 3, 3, 4, 3, 2, 3, 3, 4, 3, 2, 3, 3, 2, 3, 3, 4, 3, 2, 3, 3, 2, 3, 3, 4, 3, 2, 3, 3, 4, 3, 2, 3, 3, 2, 3, 3, 4, 3, 2, 3, 3, 4, 3, 2, 3, 3, 2, 3, 3, 4, 3, 2, 3, 3, 2, 3, 3, 4, 3, 2, 3, 3, 4, 3, 2, 3, 3, 2, 3, 3, 4, 3, 2, 3, 3, 2, 3, 3, 4, 3, 2, 3, 3, 4, 3, 2, 3, 3, 2, 3, 3, 4

Offset: 1

Benoit Cloitre, Sep 02 2005

To built the sequence start from the infinite Fibonacci word : b(n)=floor(n/phi)-floor((n-1)/phi) for n>=1 giving 0,1,0,1,1,0,1,0,1,1,0,1,1,0,1,0,1,1,..... Then replace each 0 by the block {2,3,3} and each 1 by the block {2,3,3,4,3}. Append an initial 1.

Benoit Cloitre, On properties of irrational numbers related to the floor function, in preparation, 2005.

Cf. A005614 (infinite Fibonacci binary word), A120613.

Cf. sequences for a(n) = n-F^k(n): A003842 (k=1), A110007 (k=3), A110010 (k=4), A110011 (k=5).

a(n)=n-floor((1+sqrt(5))/2*floor((-1+sqrt(5))/2*floor((1+sqrt(5))/2*floor((-1+sqrt(5))/2*n))))

A110007 a(n) = n-F(F(F(n))) where F(x)=A120613(x)=floor(phi*floor(x/phi)) and phi=(1+sqrt(5))/2.

Original entry on oeis.org

1, 2, 3, 4, 4, 5, 4, 5, 5, 4, 5, 4, 4, 5, 4, 5, 5, 4, 5, 4, 5, 5, 4, 5, 4, 4, 5, 4, 5, 5, 4, 5, 4, 4, 5, 4, 5, 5, 4, 5, 4, 5, 5, 4, 5, 4, 4, 5, 4, 5, 5, 4, 5, 4, 5, 5, 4, 5, 4, 4, 5, 4, 5, 5, 4, 5, 4, 4, 5, 4, 5, 5, 4, 5, 4, 5, 5, 4, 5, 4, 4, 5, 4, 5, 5, 4, 5, 4, 4, 5, 4, 5, 5, 4, 5, 4, 5, 5, 4, 5, 4, 4, 5, 4, 5

Offset: 1

Benoit Cloitre, Sep 02 2005

To build the sequence start from the infinite Fibonacci word: b(k)=floor(k/phi)-floor((k-1)/phi) for k>=1 giving 0,1,0,1,1,0,1,0,1,1,0,1,1,0,1,0,1,1,..... Then replace each 0 by the block {4,5,4} and each 1 by the block {5,5,4,5,4}. Append the initial string {1,2,3,4}.

Benoit Cloitre, On properties of irrational numbers related to the floor function, in preparation, 2005.

Cf. A005614 (infinite Fibonacci binary word), A120613.

Cf. sequences for a(n) = n-F^k(n): A003842 (k=1), A110006 (k=2), A110010 (k=4), A110011 (k=5).

Join[{1,2,3,4},Flatten[Table[Floor[k/GoldenRatio]-Floor[(k-1)/ GoldenRatio],{k,30}]/.{0->{4,5,4},1->{5,5,4,5,4}}]] (* Harvey P. Dale, Dec 12 2017 *)

F(x)=floor((1+sqrt(5))/2*floor((-1+sqrt(5))/2*x))
a(n)=n-F(F(F(n)))

A110010 a(n) = n-F(F(F(F(n)))) where F(x)=A120613(x)=floor(phi*floor(x/phi)) and phi=(1+sqrt(5))/2.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 6, 7, 6, 6, 7, 6, 5, 6, 6, 7, 6, 6, 7, 6, 7, 6, 6, 7, 6, 5, 6, 6, 7, 6, 6, 7, 6, 5, 6, 6, 7, 6, 6, 7, 6, 7, 6, 6, 7, 6, 5, 6, 6, 7, 6, 6, 7, 6, 7, 6, 6, 7, 6, 5, 6, 6, 7, 6, 6, 7, 6, 5, 6, 6, 7, 6, 6, 7, 6, 7, 6, 6, 7, 6, 5, 6, 6, 7, 6, 6, 7, 6, 5, 6, 6, 7, 6, 6, 7, 6, 7, 6, 6, 7, 6, 5, 6, 6, 7

Offset: 1

Benoit Cloitre, Sep 02 2005

To built the sequence start from the infinite Fibonacci word b(k)=floor(k/phi)-floor((k-1)/phi) for k>=1 giving 0,1,0,1,1,0,1,0,1,1,0,1,1,0,1,0,1,1,..... Then replace each 0 by the block {5,6,6} and each 1 by the block {7, 6, 6, 7, 6}. Append the initial string {1,2,3,4}.

Benoit Cloitre, On properties of irrational numbers related to the floor function, in preparation, 2005.

Cf. A005614 (infinite Fibonacci binary word), A120613.

Cf. sequences for a(n) = n-F^k(n): A003842 (k=1), A110006 (k=2), A110007 (k=3), A110011 (k=5).

F(x)=floor((1+sqrt(5))/2*floor((-1+sqrt(5))/2*x)); a(n)=n-F(F(F(F(n))))

cp's OEIS Frontend

A182028 Take first n bits of the infinite Fibonacci word A003849, regard them as a binary number, then convert it to base 10.

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A103269 Apply the tribonacci morphism 1 -> {1, 2}, 2 -> {1, 3}, 3 -> {1} n times to 1, and concatenate the resulting string.

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A106750 Define the "Fibonacci" morphism f: 1->12, 2->1 and let a(0) = 2; then a(n+1) = f(a(n)).

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A112658 Dean's Word: Omega 2,1: the trajectory of 0 -> 01, 1 -> 21, 2 -> 03, 3 -> 23.

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A214318 Replace the word A214317(n) with its position in A007931.

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A361756 Irregular triangle T(n, k), n >= 0, k = 1..A361757(n), read by rows; the n-th row lists the numbers k such that the Fibonacci numbers that appear in the dual Zeckendorf representation of k also appear in that of n.

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Comments

Examples

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Programs

PARI

Formula

A380560 Rectangular array R, read by descending antidiagonals: (row 1) = (R(1,k)) = (A006337(k)), k >= 1; (row n+1) = inverse runlength sequence of row n; and R(n,1) = 1 for n >=1, See Comments.

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A110006 a(n) = n-F(F(n)) where F(x)=A120613(x)=floor(phi*floor(x/phi)) and phi=(1+sqrt(5))/2.

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