A010873 a(n) = n mod 4.

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

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

Complement of A002265, since 4*A002265(n)+a(n) = n. - Hieronymus Fischer, Jun 01 2007

The rightmost digit in the base-4 representation of n. Also, the equivalent value of the two rightmost digits in the base-2 representation of n. - Hieronymus Fischer, Jun 11 2007

Periodic sequences of this type can be also calculated by a(n) = floor(q/(p^m-1)*p^n) mod p, where q is the number representing the periodic digit pattern and m is the period length. p and q can be calculated as follows: Let D be the array representing the number pattern to be repeated, m = size of D, max = maximum value of elements in D. Than p := max + 1 and q := p^m*sum_{i=1..m} D(i)/p^i. Example: D = (0, 1, 2, 3), p = 4 and q = 57 for this sequence. - Hieronymus Fischer, Jan 04 2013

Links

• Antti Karttunen, Table of n, a(n) for n = 0..65536
• Index entries for linear recurrences with constant coefficients, signature (0,0,0,1).

Crossrefs

Partial sums: A130482. Other related sequences A130481, A130483, A130484, A130485.

Cf. A004526, A002264, A002265, A002266.

Cf. A000035, A010877.

Programs

Haskell

a010873 n = (`mod` 4)
a010873_list = cycle [0..3]  -- Reinhard Zumkeller, Jun 05 2012

Maple

seq(chrem( [n,n], [1,4] ), n=0..80); # Zerinvary Lajos, Mar 25 2009

Mathematica

nn=40; CoefficientList[Series[(x+2x^2+3x^3)/(1-x^4), {x,0,nn}], x] (* Geoffrey Critzer, Jul 26 2013 *)
Table[Mod[n,4], {n, 0, 100}] (* T. D. Noe, Jul 26 2013 *)
PadRight[{},120,{0,1,2,3}] (* Harvey P. Dale, Mar 29 2018 *)

PARI

a(n)=n%4 \\ Charles R Greathouse IV, Dec 05 2011

Scheme

(define (A010873 n) (modulo n 4)) ;; Antti Karttunen, Nov 07 2017

Formula

a(n) = (1/2)*(3-(-1)^n-2*(-1)^floor(n/2));

also a(n) = (1/2)*(3-(-1)^n-2*(-1)^((2*n-1+(-1)^n)/4));

also a(n) = (1/2)*(3-(-1)^n-2*sin(Pi/4*(2n+1+(-1)^n))).

G.f.: (3x^3+2x^2+x)/(1-x^4). - Hieronymus Fischer, May 29 2007

From Hieronymus Fischer, Jun 11 2007: (Start)

Trigonometric representation: a(n)=2^2*(sin(n*Pi/4))^2*sum{1<=k<4, k*product{1<=m<4,m<>k, (sin((n-m)*Pi/4))^2}}. Clearly, the squared terms may be replaced by their absolute values '|.|'.

Complex representation: a(n)=1/4*(1-r^n)*sum{1<=k<4, k*product{1<=m<4,m<>k, (1-r^(n-m))}} where r=exp(Pi/2*i)=i=sqrt(-1). All these formulas can be easily adapted to represent any periodic sequence.

a(n) = n mod 2+2*(floor(n/2)mod 2) = A000035(n)+2*A000035(A004526(n)). (End)

a(n) = 6 - a(n-1) - a(n-2) - a(n-3) for n > 2. - Reinhard Zumkeller, Apr 13 2008

a(n) = 3/2 + cos((n+1)*Pi)/2 + sqrt(2)*cos((2*n+3)*Pi/4). - Jaume Oliver Lafont, Dec 05 2008

From Hieronymus Fischer, Jan 04 2013: (Start)

a(n) = floor(41/3333*10^(n+1)) mod 10.

a(n) = floor(19/85*4^(n+1)) mod 4. (End)

E.g.f.: 2*sinh(x) - sin(x) + cosh(x) - cos(x). - Stefano Spezia, Apr 20 2021

From Nicolas Bělohoubek, May 30 2024: (Start)

a(n) = (2*a(n-1)-1)*(2-a(n-2)) for n > 1.

a(n) = (2*a(n-1)^2+1)*(3-a(n-1))/3 for n > 0. (End)

Extensions

First to third formulas re-edited for better readability by Hieronymus Fischer, Dec 05 2011

Incorrect g.f. removed by Georg Fischer, May 18 2019