A131743 -id:A131743 - cp's OEIS frontend

A006368 The "amusical permutation" of the nonnegative numbers: a(2n)=3n, a(4n+1)=3n+1, a(4n-1)=3n-1.

0, 1, 3, 2, 6, 4, 9, 5, 12, 7, 15, 8, 18, 10, 21, 11, 24, 13, 27, 14, 30, 16, 33, 17, 36, 19, 39, 20, 42, 22, 45, 23, 48, 25, 51, 26, 54, 28, 57, 29, 60, 31, 63, 32, 66, 34, 69, 35, 72, 37, 75, 38, 78, 40, 81, 41, 84, 43, 87, 44, 90, 46, 93, 47, 96, 49, 99, 50, 102, 52, 105, 53

Offset: 0

N. J. A. Sloane and J. H. Conway

A permutation of the nonnegative integers.
There is a famous open question concerning the closed trajectories under this map - see A217218, A028393, A028394, and Conway (2013).
This is lodumo_3 of A131743. - Philippe Deléham, Oct 24 2011
Multiples of 3 interspersed with numbers other than multiples of 3. - Harvey P. Dale, Dec 16 2011
For n>0: a(2n+1) is the smallest number missing from {a(0),...,a(2n-1)} and a(2n) = a(2n-1) + a(2n+1). - Bob Selcoe, May 24 2017
From Wolfdieter Lang, Sep 21 2021: (Start)
The permutation P of positive natural numbers with P(n) = a(n-1) + 1, for n >= 1, is the inverse of the permutation given in A265667, and it maps the index n of A178414 to the index of A047529: A178414(n) = A047529(P(n)).
Thus each number {1, 3, 7} (mod 8) appears in the first column A178414 of the array A178415 just once. For the formulas see below. (End)
Starting at n = 1, the sequence equals the smallest unused positive number such that a(n)-a(n-1) does not appear as a term in the current sequence. - Scott R. Shannon, Dec 20 2023

			9 is odd so a(9) = round(3*9/4) = round(7-1/4) = 7.

J. H. Conway, Unpredictable iterations, in Proc. Number Theory Conf., Boulder, CO, 1972, pp. 49-52.
R. K. Guy, Unsolved Problems in Number Theory, E17.
J. C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, Amer. Math. Soc., 2010; see page 5.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

R. Zumkeller, Table of n, a(n) for n = 0..10000
David L. Applegate, Hans Havermann, Bob Selcoe, Vladimir Shevelev, N. J. A. Sloane, and Reinhard Zumkeller, The Yellowstone Permutation, arXiv preprint arXiv:1501.01669 [math.NT], 2015 and J. Int. Seq. 18 (2015) 15.6.7..
J. H. Conway, On unsettleable arithmetical problems, Amer. Math. Monthly, 120 (2013), 192-198. [Introduces the name "amusical permutation".]
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
S. Schreiber & N. J. A. Sloane, Correspondence, 1980
Index entries for sequences related to 3x+1 (or Collatz) problem
Index entries for two-way infinite sequences
Index entries for sequences that are permutations of the natural numbers
Index entries for linear recurrences with constant coefficients, signature (0,1,0,1,0,-1).

Inverse mapping to A006369.
Cf. A047529, A178414, A178415, A265667.
Cf. A028393, A028294, A028397, A180853, A180864, A182205, A217218.
Trajectories under A006368 and A006369: A180853, A217218, A185590, A180864, A028393, A028394, A094328, A094329, A028396, A028395, A217729, A182205, A223083-A223088, A185589, A185590.

a006368 n | u' == 0   = 3 * u
          | otherwise = 3 * v + (v' + 1) `div` 2
          where (u,u') = divMod n 2; (v,v') = divMod n 4
-- Reinhard Zumkeller, Apr 18 2012

[n mod 2 eq 1 select Round(3*n/4) else 3*n/2: n in [0..80]]; // G. C. Greubel, Jan 03 2024

f:=n-> if n mod 2 = 0 then 3*n/2 elif n mod 4 = 1 then (3*n+1)/4 else (3*n-1)/4; fi; # N. J. A. Sloane, Jan 21 2011
A006368:=(1+3*z+z**2+3*z**3+z**4)/(1+z**2)/(z-1)**2/(1+z)**2; # [Conjectured (correctly, except for the offset) by Simon Plouffe in his 1992 dissertation.]

Table[If[EvenQ[n],(3n)/2,Floor[(3n+2)/4]],{n,0,80}] (* or *) LinearRecurrence[ {0,1,0,1,0,-1},{0,1,3,2,6,4},80] (* Harvey P. Dale, Dec 16 2011 *)

PARI
```
a(n)=(3*n+n%2)\(2+n%2*2)
```
PARI
```
a(n)=if(n%2,round(3*n/4),3*n/2)
```

def a(n): return 0 if n == 0 else 3*n//2 if n%2 == 0 else (3*n+1)//4
print([a(n) for n in range(72)]) # Michael S. Branicky, Aug 12 2021

If n even, then a(n) = 3*n/2, otherwise, a(n) = round(3*n/4).
G.f.: x*(1+3*x+x^2+3*x^3+x^4)/((1-x^2)*(1-x^4)). - Michael Somos, Jul 23 2002
a(n) = -a(-n).
From Reinhard Zumkeller, Nov 20 2009: (Start)
a(n) = A006369(n) - A168223(n).
A168221(n) = a(a(n)).
A168222(a(n)) = A006369(n). (End)
a(n) = a(n-2) + a(n-4) - a(n-6); a(0)=0, a(1)=1, a(2)=3, a(3)=2, a(4)=6, a(5)=4. - Harvey P. Dale, Dec 16 2011
From Wolfdieter Lang, Sep 21 2021: (Start)
Formulas for the permutation P(n) = a(n-1) + 1 mentioned above:
P(n) = n + floor(n/2) if n is odd, and n - floor(n/4) if n is even.
P(n) = (3*n-1)/2 if n is odd; P(n) = (3*n+2)/4 if n == 2 (mod 4); and P(n) = 3*n/4 if n == 0 (mod 4). (End)

Edited by Michael Somos, Jul 23 2002

I replaced the definition with the original definition of Conway and Guy. - N. J. A. Sloane, Oct 03 2012

A109718 Periodic sequence with period {0,1,0,3}, or n^3 mod 4.

Original entry on oeis.org

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

Offset: 0

Bruce Corrigan (scentman(AT)myfamily.com), Aug 09 2005

Since n^(2k+1) mod 4 = n^3 mod 4 for k>1 this sequence also represents n^5 mod 4; and n^7 mod 4; etc.

Index entries for linear recurrences with constant coefficients, signature (0,0,0,1).

n mod 4 = A010873; n^2 mod 4 = A000035.

Cf. A110270; A131743. - Bruno Berselli, Mar 14 2011

&cat[[0,1,0,3]: k in [0..26]]; // Bruno Berselli, Mar 14 2011

a(n)=n^3%4 \\ Charles R Greathouse IV, Apr 06 2016

[power_mod(n,3,4 )for n in range(0, 105)] # Zerinvary Lajos, Oct 29 2009

a(n) = n^3 mod 4.
G.f. = (x+3*x^3)/(1-x^4).
a(n) = (n mod 2)*(n mod 4) = (1+(-1)^(n+1))*(2+i^(n+1))/2 with i=sqrt(-1). - Bruno Berselli, Mar 14 2011

A189998 Numerator of h(n+5) - h(n) where h(n) = Sum_{k=1..n} (1/k) are the Harmonic numbers.

Original entry on oeis.org

137, 29, 153, 743, 1879, 1627, 15797, 2021, 11899, 25381, 7793, 2627, 124877, 26987, 68879, 65003, 107699, 66167, 482897, 16167, 77293, 412561, 323959, 94781, 1323137, 255127, 587299, 504563, 255733, 145209, 2956637, 277681, 1247459, 2094661, 1558379, 433501

Offset: 0

Gary Detlefs, May 03 2011

nonn

a(n) = Numerator of (5*n^4 + 60*n^3 + 255*n^2 + 450*n + 274)/((n+1)*(n+2)*(n+3)*(n+4)*(n+5)).
(5*n^4 + 60*n^3 + 255*n^2 + 450*n + 274)/a(n) can be factored into 2^p(n)* 3^q(n) where p(n) is a sequence of period 4 repeating [1,2,1,3] and q(n) is of period 9,repeating [0,2,2,0,1,1,0,1,1].
p(n) = A131743(n) + 1.
q(n) = A011655(n) + [0,2,2,0,0,0,0,0,0]

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A011655, A131743.

[137] cat [Numerator(HarmonicNumber(n+5)-HarmonicNumber(n)): n in [0..30]]; // G. C. Greubel, Jan 11 2018

h:= n->sum(1/k,k=1..n):seq(numer(h(n+5)-h(n)), n=0..28);
q:=n-> (1-(-1)^n)*(3+I^(n+1))/4+1:
P:=(k,n)-> floor(1/2*cos(2*n*Pi/k)+1/2):
seq( (5*n^4+60*n^3+255*n^2+450*n+274)/(2^q(n)*3^(P(9,n-1)+P(9,n-2)+1-P(3,n))),n=0..28)

Numerator[Table[HarmonicNumber[n+5]-HarmonicNumber[n],{n,0,30}]] (* Harvey P. Dale, Sep 15 2016 *)

from sympy import harmonic,numer
print([numer(harmonic(n+5) - harmonic(n)) for n in range(0, 30)])
# Javier Rivera Romeu, May 22 2023

a(n) = (5*n^4 + 60*n^3 + 255*n^2 + 450*n + 274)/(2^q(n)*3^(P(9,n-1) + P(9,n-2) + 1 - P(3,n))), where q(n) = (1-(-1)^n)*(3+i^(n+1))/4 + 1 and P(k,n) = floor(1/2*cos(2*n*Pi/k)+1/2).

A364447 Repeat [1,2,1,3].

Original entry on oeis.org

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

Offset: 1

Rok Cestnik, Jul 25 2023

Continued fraction of sqrt(5) - 3/2 = 0.7360679... (without integer part); and (4*sqrt(5) + 6)/11 = 1.3585701... (with integer part).

Lexicographically earliest sequence in which n is banned for n terms after each appearance (see A364448 for n^2 and A364449 for n^3).

Index entries for linear recurrences with constant coefficients, signature (0,0,0,1).

Cf. A131743 (repeat [0,1,0,2]).
Cf. A000034, A010882, A131534.
Cf. A364448, A364449.

PadRight[{}, 100, {1, 2, 1, 3}] (* Paolo Xausa, Jan 23 2025 *)

def A364447(n): return (3,1,2,1)[n&3] # Chai Wah Wu, Jul 29 2023

a(n) = A131743(n-1) + 1.

A180713 If n is even then a(n) = 3n, if n == 1 mod 4 then a(n) = 3n+1, if n == 3 mod 4 then a(n) = 3n+2.

Original entry on oeis.org

0, 4, 6, 11, 12, 16, 18, 23, 24, 28, 30, 35, 36, 40, 42, 47, 48, 52, 54, 59, 60, 64, 66, 71, 72, 76, 78, 83, 84, 88, 90, 95, 96, 100, 102, 107, 108, 112, 114, 119, 120, 124, 126, 131, 132, 136, 138, 143, 144, 148, 150, 155, 156, 160, 162, 167, 168, 172, 174, 179, 180, 184, 186, 191, 192, 196, 198, 203, 204

Offset: 0

N. J. A. Sloane, Jan 21 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for sequences related to 3x+1 (or Collatz) problem
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

A variant of A006368.

U:=n->if n mod 2 = 0 then 3*n elif n mod 4 = 1 then 3*n+1 else 3*n+2; fi;

fn[n_]:=Which[EvenQ[n],3n,Mod[n,4]==1,3n+1,Mod[n,4]==3,3n+2]; Array[fn,70,0] (* Harvey P. Dale, May 03 2013 *)
CoefficientList[Series[x (4 + 2 x + 5 x^2 + x^3) / ((1 - x) (1 - x^4)), {x, 0, 40}], x] (* Vincenzo Librandi, Jun 19 2013 *)

From Bruno Berselli, Jan 23 2011: (Start)
G.f.:  x*(4+2*x+5*x^2+x^3)/((1-x)*(1-x^4)).
a(n) = (12*n+i*(i^n-(-i)^n)-3*(-1)^n+3)/4, where i is the imaginary unit.
a(n) = A131743(n) + 3*n. (End)
a(n) = +1*a(n-1)+1*a(n-4)-1*a(n-5) for n>=5. [Joerg Arndt, Jan 25 2011]

Definition corrected by N. J. A. Sloane, Jan 23 2011

A281098 a(n) is the GCD of the sequence d(n) = A261327(k+n) - A261327(k) for all k.

Original entry on oeis.org

0, 1, 1, 3, 4, 1, 3, 1, 8, 3, 5, 1, 12, 1, 7, 3, 16, 1, 9, 1, 20, 3, 11, 1, 24, 1, 13, 3, 28, 1, 15, 1, 32, 3, 17, 1, 36, 1, 19, 3, 40, 1, 21, 1, 44, 3, 23, 1, 48, 1, 25, 3, 52, 1, 27, 1, 56, 3, 29, 1, 60, 1, 31, 3, 64, 1, 33, 1, 68, 3, 35, 1, 72, 1, 37, 3, 76, 1, 39, 1

Offset: 0

Paul Curtz, Jan 14 2017

Successive sequences:
   0,   0,  0,    0, ...    = 0 * ( )
   4,  -3,  11,  -8, ...    = 1 * ( )
   1,   8,   3,  16, ...    = 1 * ( )                 A195161
  12,   0,  27,  -3, ...    = 3 * (4, 0, 9, -1, ...)
   4,  24,   8,  40, ...    = 4 * (1, 6, 2, 10, ...)  A064680
5;   28,   5,  51,   4, ...    = 1 * ( )
   9,  48,  15,  72, ...    = 3 * (3, 16, 5, 24, ...) A195161
  52,  12,  83,  13, ...    = 1 * ( )
  16,  80,  24, 112, ...    = 8 * (2, 10, 3, 14, ...) A064080
  84   21, 123,  24, ...    = 3 * (28, 7, 41, 8, ...)
 25, 120,  35, 160, ...    = 5 * (5, 24, 7, 32, ...) A195161

Antti Karttunen, Table of n, a(n) for n = 0..16384
Index entries for linear recurrences with constant coefficients, signature (0,-1,0,1,0,2,0,1,0,-1,0,-1).

Cf. A064680, A144433 or A195161.

Cf. A000012, A005408, A008586, A010701, A109007 (bisection), A016825, A165988 (via A022998), A166138, A166304, A280579.

CoefficientList[Series[(-x (-1 - x - 4 x^2 - 5 x^3 - 3 x^4 - 6 x^5 + 3 x^6 - 5 x^7 + 4 x^8 - x^9 + x^10))/((x^2 - x + 1) (1 + x + x^2) (x - 1)^2*(1 + x)^2*(1 + x^2)^2), {x, 0, 79}], x] (* Michael De Vlieger, Feb 02 2017 *)

f(n) = numerator((4 + n^2)/4);
a(n) = gcd(vector(1000, k, f(k+n) - f(k))); \\ Michel Marcus, Jan 15 2017

A281098(n) = if(n%2, gcd((n\2)-1,3), n>>(bitand(n,2)/2)); \\ Antti Karttunen, Feb 15 2023

G.f.: -x*( -1 - x - 4*x^2 - 5*x^3 - 3*x^4 - 6*x^5 + 3*x^6 - 5*x^7 + 4*x^8 - x^9 + x^10 )/( (x^2 - x + 1)*(1 + x + x^2)*(x - 1)^2*(1 + x)^2*(1 + x^2)^2 ). - R. J. Mathar, Jan 31 2017
a(2*k)   = A022998(k).
a(2*k+1) = A109007(k-1).
a(3*k)   = interleave 3*k*(3 +(-1)^k)/2, 3.
a(3*k+1) = interleave 1, A166304(k).
a(3*k+2) = interleave A166138(k), 1.
a(4*k)   = 4*k.
a(4*k+1) = period 3: repeat [1, 1, 3].
a(4*k+2) = 1 + 2*k.
a(4*k+3) = period 3: repeat [3, 1, 1].
a(n+12) - a(n) = 6*A131743(n+3).
a(n) = (18*n + 40 - 16*cos(n*Pi/3) + 9*n*cos(n*Pi/2) + 32*cos(2*n*Pi/3) + (18*n - 40)*cos(n*Pi) + 3*n*cos(3*n*Pi/2) - 16*cos(5*n*Pi/3))/48. - Wesley Ivan Hurt, Oct 04 2018

Corrected and extended by Michel Marcus, Jan 15 2017

cp's OEIS Frontend

A006368 The "amusical permutation" of the nonnegative numbers: a(2n)=3n, a(4n+1)=3n+1, a(4n-1)=3n-1.

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A109718 Periodic sequence with period {0,1,0,3}, or n^3 mod 4.

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A189998 Numerator of h(n+5) - h(n) where h(n) = Sum_{k=1..n} (1/k) are the Harmonic numbers.

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A364447 Repeat [1,2,1,3].

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A180713 If n is even then a(n) = 3n, if n == 1 mod 4 then a(n) = 3n+1, if n == 3 mod 4 then a(n) = 3n+2.

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A281098 a(n) is the GCD of the sequence d(n) = A261327(k+n) - A261327(k) for all k.

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