OFFSET
0,3
COMMENTS
LINKS
Muniru A Asiru, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,1).
FORMULA
Complex representation: a(n) = (1/5)*(1-r^n)*Sum{1<=k<5, k*Product{1<=m<5,m<>k, (1-r^(n-m))}} where r=exp(2*Pi/5*i) and i=sqrt(-1).
G.f.: g(x)=(4*x^4+3*x^3+2*x^2+x)/(1-x^5). - Hieronymus Fischer, May 29 2007
Trigonometric representation: a(n) = (16/5)^2*(sin(n*Pi/5))^2*Sum{1<=k<5, k*Product{1<=m<5,m<>k, (sin((n-m)*Pi/5))^2}}. Clearly, the squared terms may be replaced by their absolute values '|.|'. This formula can be easily adapted to represent any periodic sequence.
G.f.: also g(x) = x*(5*x^6 - 6*x^5 + 1)/((1-x^5)*(1-x)^2). - Hieronymus Fischer, Jun 01 2007
a(n) = -cos(4/5*Pi*n)-cos(2/5*Pi*n)+1/20*5^(1/2)*(10-2*5^(1/2))^(1/2)* sin(4/5*Pi*n)-1/4*(10-2*5^(1/2))^(1/2)*sin(4/5*Pi*n)-1/4*(10+2*5^(1/2))^(1/2)*sin(2/5*Pi*n)-1/20*5^(1/2)*(10+2*5^(1/2))^(1/2)*sin(2/5*Pi*n) + 2. - Leonid Bedratyuk, May 14 2012
a(n) = floor(1234/99999*10^(n+1)) mod 10. - Hieronymus Fischer, Jan 03 2013
a(n) = floor(97/1562*5^(n+1)) mod 5. - Hieronymus Fischer, Jan 04 2013
From Wesley Ivan Hurt, Jul 23 2016: (Start)
a(n) = a(n-5) for n>4.
a(n) = 4*(1 - floor(n/5)) + Sum_{k=1..4} floor((n-k)/5).
a(n) = 4 - 4*floor(n/5) + floor((n-1)/5) + floor((n-2)/5) + floor((n-3)/5) + floor((n-4)/5).
a(n) = n - 5*floor(n/5). (End)
a(n) = 2 + (2/5)*Sum_{k=1..4} k*((cos(2*(n-k)*Pi/5) + cos(4*(n-k)*Pi/5)). - Wesley Ivan Hurt, Sep 27 2018
MAPLE
seq(chrem( [n, n], [1, 5] ), n=0..80); # Zerinvary Lajos, Mar 25 2009
MATHEMATICA
Mod[Range[0, 100], 5] (* Wesley Ivan Hurt, Jul 23 2016 *)
PROG
(PARI) a(n)=n%5 \\ Charles R Greathouse IV, Sep 24 2015
(Magma) [n mod 5 : n in [0..100]]; // Wesley Ivan Hurt, Jul 23 2016
(GAP) List([0..100], n->n mod 5); # Muniru A Asiru, Sep 28 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved