A010873 - OEIS (original) (raw)

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

COMMENTS

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

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))).

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) = 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

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

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)

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 *)

PROG

(Haskell)

a010873 n = (`mod` 4)

EXTENSIONS

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