A130482 -id:A130482 - cp's OEIS frontend

A010879 Final digit of n.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0

Offset: 0

N. J. A. Sloane

Also decimal expansion of 137174210/1111111111 = 0.1234567890123456789012345678901234... - Jason Earls, Mar 19 2001
In general the base k expansion of A062808(k)/A048861(k) (k>=2) will produce the numbers 0,1,2,...,k-1 repeated with period k, equivalent to the sequence n mod k. The k-digit number in base k 123...(k-1)0 (base k) expressed in decimal is A062808(k), whereas A048861(k) = k^k-1. In particular, A062808(10)/A048861(10)=1234567890/9999999999=137174210/1111111111.
a(n) = n^5 mod 10. - Zerinvary Lajos, Nov 04 2009

Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,1).
Index entries for sequences related to final digits of numbers

Cf. A034948, A059988, A048861, A062808, A086457, A086458.
Cf. A008959, A008960, A070514. - Doug Bell, Jun 15 2015
Partial sums: A130488. Other related sequences A130481, A130482, A130483, A130484, A130485, A130486, A130487.

a010879 = (`mod` 10)
a010879_list = cycle [0..9]  -- Reinhard Zumkeller, Mar 26 2012

[n mod(10): n in [0..90]]; // Vincenzo Librandi, Jun 17 2015

A010879 := proc(n)
    n mod 10 ;
end proc: # R. J. Mathar, Jul 12 2013

Table[10*FractionalPart[n/10], {n, 1, 300}] (* Enrique Pérez Herrero, Jul 30 2009 *)
LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 0, 0, 1},{0, 1, 2, 3, 4, 5, 6, 7, 8, 9},81] (* Ray Chandler, Aug 26 2015 *)
PadRight[{},100,Range[0,9]] (* Harvey P. Dale, Oct 04 2021 *)

a(n)=n%10 \\ Charles R Greathouse IV, Jun 16 2011

def a(n): return n % 10 # Martin Gergov, Oct 17 2022

[power_mod(n,5,10)for n in range(0, 81)] # Zerinvary Lajos, Nov 04 2009

a(n) = n mod 10.
Periodic with period 10.
From Hieronymus Fischer, May 31 and Jun 11 2007: (Start)
Complex representation: a(n) = 1/10*(1-r^n)*sum{1<=k<10, k*product{1<=m<10,m<>k, (1-r^(n-m))}} where r=exp(Pi/5*i) and i=sqrt(-1).
Trigonometric representation: a(n) = (256/5)^2*(sin(n*Pi/10))^2 * sum{1<=k<10, k*product{1<=m<10,m<>k, (sin((n-m)*Pi/10))^2}}.
G.f.: g(x) = (Sum_{k=1..9} k*x^k)/(1-x^10) = -x*(1 +2*x +3*x^2 +4*x^3 +5*x^4 +6*x^5 +7*x^6 +8*x^7 +9*x^8) / ( (x-1) *(1+x) *(x^4+x^3+x^2+x+1) *(x^4-x^3+x^2-x+1) ).
Also: g(x) = x*(9*x^10-10*x^9+1)/((1-x^10)*(1-x)^2).
a(n) = n mod 2+2*(floor(n/2)mod 5) = A000035(n) + 2*A010874(A004526(n)).
Also: a(n) = n mod 5+5*(floor(n/5)mod 2) = A010874(n)+5*A000035(A002266(n)). (End)
a(n) = 10*{n/10}, where {x} means fractional part of x. - Enrique Pérez Herrero, Jul 30 2009
a(n) = n - 10*A059995(n). - Reinhard Zumkeller, Jul 26 2011
a(n) = n^k mod 10, for k > 0, where k mod 4 = 1. - Doug Bell, Jun 15 2015

Formula section edited for better readability by Hieronymus Fischer, Jun 13 2012

A010873 a(n) = n mod 4.

Original entry on oeis.org

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

N. J. A. Sloane

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

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

Partial sums: A130482. Other related sequences A130481, A130483, A130484, A130485.
Cf. A004526, A002264, A002265, A002266.
Cf. A000035, A010877.

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

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

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 *)

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

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

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)

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

A010872 a(n) = n mod 3.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

Fixed point of morphism 0 -> 01, 1 -> 20, 2 -> 12.
Complement of A002264, since 3*A002264(n) + a(n) = n. - Hieronymus Fischer, Jun 01 2007
Decimal expansion of 4/333. - Elmo R. Oliveira, Feb 19 2024
Period 3: repeat [0, 1, 2]. - Elmo R. Oliveira, Jun 20 2024

			G.f. = x + 2*x^2 + x^4 + 2*x^5 + x^7 + 2*x^8 + x^10 + 2*x^11 + x^13 + ...

Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
Ralph E. Griswold, Shaft Sequences
Ralph E. Griswold, Shaft Sequences [From the Wayback machine]
Index entries for linear recurrences with constant coefficients, signature (0,0,1).
Index entries for sequences that are fixed points of mappings

Cf. A000035, A002264, A002265, A002266, A004526, A010873, A080425, A102283.
Cf. A010882, A130481 (partial sums), A131555.
Other related sequences are A130482, A130483, A130484, A130485.

a010872 = (`mod` 3)
a010872_list = cycle [0,1,2]  -- Reinhard Zumkeller, May 26 2012

[n mod 3 : n in [0..100]]; // Wesley Ivan Hurt, May 27 2015

A010872:=n->(n mod 3): seq(A010872(n), n=0..100); # Wesley Ivan Hurt, May 27 2015

Nest[ Function[ l, {Flatten[(l /. {0 -> {0, 1}, 1 -> {2, 0}, 2 -> {1, 2}})]}], {0}, 7] (* Robert G. Wilson v, Feb 28 2005 *)
PadRight[{},120,{0,1,2}] (* or *) Mod[Range[0,120],3] (* Harvey P. Dale, Jul 20 2025 *)

my(x='x+O('x^200)); concat(0, Vec((2*x^2+x)/(1-x^3))) \\ Altug Alkan, Mar 23 2016

a(n) = n - 3*floor(n/3) = a(n-3).
G.f.: (2*x^2+x)/(1-x^3). - Mario Catalani (mario.catalani(AT)unito.it), Jan 08 2003
From Hieronymus Fischer, May 29 2007: (Start)
a(n) = 1 + (1-2*cos(2*Pi*(n-1)/3)) * sin(2*Pi*(n-1)/3) / sqrt(3).
a(n) = (1-r^n)*(1+r^n/(1-r)) where r=exp(2*Pi*i/3)=(-1+sqrt(3)*i)/2 and i=sqrt(-1). [corrected by Guenther Schrack, Sep 23 2019] (End)
From Hieronymus Fischer, Jun 01 2007: (Start)
a(n) = (16/9)*((sin(Pi*(n-2)/3))^2+2*(sin(Pi*(n-1)/3))^2)*(sin(Pi*n/3))^2.
a(n) = (4/3)*(|sin(Pi*(n-2)/3)|+2*|sin(Pi*(n-1)/3)|)*|sin(Pi*n/3)|.
a(n) = (4/9)*((1-cos(2*Pi*(n-2)/3))+2*(1-cos(2*Pi*(n-1)/3)))*(1-cos(2*Pi*n/3)). (End)
a(n) = 3 - a(n-1) - a(n-2) for n > 1. - Reinhard Zumkeller, Apr 13 2008
a(n) = 1-2*sin(4*Pi*(n+2)/3)/sqrt(3). - Jaume Oliver Lafont, Dec 05 2008
From Wesley Ivan Hurt, May 27 2015, Mar 22 2016: (Start)
a(n) = 1 - 0^((-1)^(n/3)-(-1)^n) + 0^((-1)^((n+1)/3)+(-1)^n).
a(n) = 1 + (-1)^((2*n+4)/3)/3 + (-1)^((-2*n-4)/3)/3 + 2*(-1)^((2*n+2)/3)/3 + 2*(-1)^((-2*n-2)/3)/3.
a(n) = 1 + 2*cos(Pi*(2*n+4)/3)/3 + 4*cos(Pi*(2*n+2)/3)/3. (End)
a(n) = (r^n*(r-1) - r^(2*n)*(r + 2) + 3)/3 where r = (-1 + sqrt(-3))/2. - Guenther Schrack, Sep 23 2019
E.g.f.: exp(x) - exp(-x/2)*(cos(sqrt(3)*x/2) + sin(sqrt(3)*x/2)/sqrt(3)). - Stefano Spezia, Mar 01 2020
a(n) = A010882(n) - 1 = A131555(2*n) = A131555(2*n+1). - Elmo R. Oliveira, Jun 25 2024
From Nicolas Bělohoubek, May 26 2025: (Start)
a(n) = (3*a(n-1)+1)*(2-a(n-1))/2 for n > 0.
a(n) = (2*a(n-1)-4)/(3*a(n-1)-4) for n > 0. (End)

Edited by Joerg Arndt, Apr 21 2014

A010874 a(n) = n mod 5.

Original entry on oeis.org

0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0

Offset: 0

N. J. A. Sloane

Complement of A002266, since 5*A002266(n) + a(n) = n. - Hieronymus Fischer, Jun 01 2007

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).

Partial sums: A130483.

Cf. A130481, A130482, A130484, A130485, A004526, A002264, A002265, A002266.

List([0..100],n->n mod 5); # Muniru A Asiru, Sep 28 2018

[n mod 5 : n in [0..100]]; // Wesley Ivan Hurt, Jul 23 2016

seq(chrem( [n,n], [1,5] ), n=0..80); # Zerinvary Lajos, Mar 25 2009

Mod[Range[0, 100], 5] (* Wesley Ivan Hurt, Jul 23 2016 *)

a(n)=n%5 \\ Charles R Greathouse IV, Sep 24 2015

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

A010875 a(n) = n mod 6.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0

Offset: 0

N. J. A. Sloane

Period 6: repeat [0, 1, 2, 3, 4, 5].
The rightmost digit in the base-6 representation of n. - Hieronymus Fischer, Jun 11 2007
[a(n) * a(m)] mod 6 == a(n*m mod 6) == a(n*m). - Jon Perry, Nov 11 2014
If n > 3 and (a(n) is in {0,2,3,4}), then n is not prime. - Jean-Marc Rebert, Jul 22 2015, corrected by M. F. Hasler, Jul 24 2015

Antti Karttunen, Table of n, a(n) for n = 0..65538
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,1).

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

Cf. also A079979, A097325, A122841.

[n mod 6: n in [0..100]]; // Wesley Ivan Hurt, Jul 06 2014

A010875:=n->n mod 6; seq(A010875(n), n=0..100); # Wesley Ivan Hurt, Jul 06 2014

Mod[Range[0, 100], 6] (* Wesley Ivan Hurt, Jul 06 2014 *)

a(n)=n%6 \\ Charles R Greathouse IV, Dec 05 2011

[power_mod(n,3,6 )for n in range(0, 81)] # Zerinvary Lajos, Oct 29 2009

(define (A010875 n) (modulo n 6)) ;; Antti Karttunen, Dec 22 2017

Complex representation: a(n) = (1/6) * (1 - r^n) * Sum_{k = 1..6} k * Product_{1 <= m < 6, m <> k} (1-r^(n-m)), where r = exp((Pi/3)*i) = (1 + sqrt(3)*i)/2 and i = sqrt(-1).
Trigonometric representation: a(n) = (16/3)^2 * (sin(n*Pi/6))^2 * Sum_{k = 1..6} k * Product_{1 <= m < 6, m<>k} (sin((n-m)*Pi/6))^2.
G.f.: g(x) = (Sum_{k = 1..6} k*x^k)/(1-x^6).
Also: g(x) = x*(5*x^6 - 6*x^5 + 1)/((1 - x^6)*(1 - x)^2). - Hieronymus Fischer, May 31 2007
a(n) = (n mod 2) + 2(floor(n/2) mod 3) = A000035(n) +2*A010872(A004526(n));
a(n) = (n mod 3) + 3(floor(n/3) mod 2) = A010872(n) +3*A000035(A002264(n)). - Hieronymus Fischer, Jun 11 2007
a(n) = 2.5 - 0.5*(-1)^n - cos(Pi*n/3) - 3^0.5*sin(Pi*n/3) -cos(2*Pi*n/3) - 3^0.5/3*sin(2*Pi*n/3). - Richard Choulet, Dec 11 2008
a(n) = n^3 mod 6. - Zerinvary Lajos, Oct 29 2009
a(n) = floor(12345/999999*10^(n+1)) mod 10. - Hieronymus Fischer, Jan 03 2013
a(n) = floor(373/9331*6^(n+1)) mod 6. - Hieronymus Fischer, Jan 04 2013
a(n) = 5/2 - (-1)^n/2 - 2*0^((-1)^(n/6 - 1/12 + (-1)^n/12) - (-1)^(n/2 - 1/4 +(-1)^n/4)) + 2*0^((-1)^(n/6 + 1/4 + (-1)^n/12) + (-1)^(n/2 - 1/4 + (-1)^n/4)). - Wesley Ivan Hurt, Jun 23 2015
E.g.f.: -sqrt(3)*exp(x/2)*sin(sqrt(3)*x/2) - 2*cosh(x/2)*cos(sqrt(3)*x/2). - Robert Israel, Jul 22 2015

Formulas 1 to 6 re-edited for better readability by Hieronymus Fischer, Dec 05 2011

More terms from Antti Karttunen, Dec 22 2017

A010877 a(n) = n mod 8.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3, 4, 5, 6, 7, 0

Offset: 0

N. J. A. Sloane

The rightmost digit in the base-8 representation of n. Also, the equivalent value of the three rightmost digits in the base-2 representation of n. - Hieronymus Fischer, Jun 12 2007

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

Partial sums: A130486. Other related sequences A130481, A130482, A130483, A130484, A130485.

Cf. A000035, A010887, A010873, A130909, A168181, A244413, A253513.

Table[Mod[n,8],{n,0,120}]   (* Harvey P. Dale, Apr 21 2011 *)

vector(100,i,i)%8 \\ Charles R Greathouse IV, Jul 16 2011

def A010877(n): return n&7 # Chai Wah Wu, Jul 09 2022

Complex representation: a(n) = (1/8)*(1-r^n)*Sum_{k=1..7} k*Product_{m=1..7, m<>k} (1 - r^(n-m)) where r = exp(Pi/4*i) = (1+i)*sqrt(2)/2 and i=sqrt(-1).
Trigonometric representation: a(n) = 256*(sin(n*Pi/8))^2*Sum_{k=1..7} k*Product_{m=1..7, m<>k} (sin((n-m)*Pi/8))^2.
G.f.: g(x) = (Sum_{k=1..7}, k*x^k)/(1-x^8).
Also: g(x) = x(7x^8-8x^7+1)/((1-x^8)(1-x)^2). - Hieronymus Fischer, May 31 2007
a(n) = n mod 2 + 2*(floor(n/2) mod 4) = A000035(n) + 2*A010873(A004526(n)).
a(n) = n mod 4 + 4*(floor(n/4) mod 2) = A010873(n) + 4*A000035(A002265(n)).
a(n) = n mod 2 + 2*(floor(n/2) mod 2) + 4*(floor(n/4) mod 2) = A000035(n) + 2*A000035(A004526(n)) + 4*A000035(A002265(n)). - Hieronymus Fischer, Jun 12 2007
a(n) = (1/2)*(7 - (-1)^n - 2*(-1)^(b/4) - 4*(-1)^((b - 2 + 2*(-1)^(b/4))/8)) where b = 2n - 1 + (-1)^n. - Hieronymus Fischer, Jun 12 2007
General formula for period 2^k: a(n) = (1/2)*(2^k - 1 - Sum_{j=0..k-1} 2^j*(-1)^p(j,n)) where p(j,n) is defined recursively by p(0,n)=n, p(j,n) = (1/4)*(2*p(j-1,n) - 1 + (-1)^p(j-1,n)). - Hieronymus Fischer, Jun 14 2007
a(n) = floor(1234567/99999999*10^(n+1)) mod 10. - Hieronymus Fischer, Jan 03 2013
a(n) = floor(48913/2396745*8^(n+1)) mod 8. - Hieronymus Fischer, Jan 04 2013

Formula section re-edited for better readability by Hieronymus Fischer

A010876 a(n) = n mod 7.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 0, 1, 2, 3, 4, 5, 6, 0, 1, 2, 3, 4, 5, 6, 0, 1, 2, 3, 4, 5, 6, 0, 1, 2, 3, 4, 5, 6, 0, 1, 2, 3, 4, 5, 6, 0, 1, 2, 3, 4, 5, 6, 0, 1, 2, 3, 4, 5, 6, 0, 1, 2, 3, 4, 5, 6, 0, 1, 2, 3, 4, 5, 6, 0, 1, 2, 3, 4, 5, 6, 0, 1, 2, 3, 4, 5, 6, 0, 1, 2, 3

Offset: 0

N. J. A. Sloane

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

Partial sums: A130485.

Other related sequences: A130481, A130482, A130483, A130484.

&cat [[0..6]^^20]; // Bruno Berselli, Jun 09 2016

a(n)=n%7 \\ Charles R Greathouse IV, Dec 05 2011

[power_mod(n,7,7) for n in range(0, 81)] # Zerinvary Lajos, Nov 07 2009

Complex representation: a(n) = (1/7)*(1-r^n) * Sum_{1<=k<7} k * Product_{1<=m<7, m<>k} (1-r^(n-m)) where r=exp(2*pi/7*i) and i=sqrt(-1).
Trigonometric representation: a(n) = (64/7)^2*(sin(n*pi/7))^2*Sum_{1<=k<7} k*Product_{1<=m<7,m<>k} sin((n-m)*pi/7)^2.
G.f.: ( Sum_{1<=k<7} k*x^k ) / (1 - x^7).
G.f.: x*(6*x^7-7*x^6+1)/((1-x^7)*(1-x)^2). - Hieronymus Fischer, May 31 2007
a(n) = floor(41152/3333333*10^(n+1)) mod 10. - Hieronymus Fischer, Jan 03 2013
a(n) = floor(7625/274514*7^(n+1)) mod 7. - Hieronymus Fischer, Jan 04 2013

Formula section re-edited for better readability by Hieronymus Fischer, Dec 05 2011

A010878 a(n) = n mod 9.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 2, 3, 4, 5

Offset: 0

N. J. A. Sloane

Periodic with period of length 9. The digital root of n (A010888) is a very similar sequence.

The rightmost digit in the base-9 representation of n. Also, the equivalent value of the two rightmost digits in the base-3 representation of n. - Hieronymus Fischer, Jun 11 2007

Ely Golden, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,1).

Partial sums: A130487. Other related sequences A130481, A130482, A130483, A130484, A130485, A130486.

a010878 = (`mod` 9)
a010878_list = cycle [0..8]  -- Reinhard Zumkeller, Jan 09 2013

A010878 := proc(n)
    modp(n,9) ;
end proc:
seq(A010878(n),n=0..100) ; # R. J. Mathar, Sep 09 2015

Array[Mod[#, 9]&, 105, 0] (* Jean-François Alcover, Jan 30 2018 *)
PadRight[{},120,Range[0,8]] (* Harvey P. Dale, Dec 19 2018 *)

a(n)=n%9 \\ Charles R Greathouse IV, Sep 24 2015

Complex representation: a(n)=(1/9)*(1-r^n)*sum{1<=k<9, k*product{1<=m<9,m<>k, (1-r^(n-m))}} where r=exp(2*pi/9*i) and i=sqrt(-1). Trigonometric representation: a(n)=(256/9)^2*(sin(n*pi/9))^2*sum{1<=k<9, k*product{1<=m<9,m<>k, (sin((n-m)*pi/9))^2}}. G.f.: g(x)=(sum{1<=k<9, k*x^k})/(1-x^9). Also: g(x)=x(8x^9-9x^8+1)/((1-x^9)(1-x)^2). - Hieronymus Fischer, May 31 2007
a(n) = n mod 3 + 3*(floor(n/3)mod 3) = A010872(n) + 3*A010872(A002264(n)). - Hieronymus Fischer, Jun 11 2007
a(n) = floor(12345678/999999999*10^(n+1)) mod 10. - Hieronymus Fischer, Jan 03 2013
a(n) = floor(1513361/96855122*9^(n+1)) mod 9. - Hieronymus Fischer, Jan 04 2013

A130481 a(n) = Sum_{k=0..n} (k mod 3) (i.e., partial sums of A010872).

Original entry on oeis.org

0, 1, 3, 3, 4, 6, 6, 7, 9, 9, 10, 12, 12, 13, 15, 15, 16, 18, 18, 19, 21, 21, 22, 24, 24, 25, 27, 27, 28, 30, 30, 31, 33, 33, 34, 36, 36, 37, 39, 39, 40, 42, 42, 43, 45, 45, 46, 48, 48, 49, 51, 51, 52, 54, 54, 55, 57, 57, 58, 60, 60, 61, 63, 63, 64, 66, 66, 67, 69, 69, 70, 72, 72

Offset: 0

Hieronymus Fischer, May 29 2007

Essentially the same as A092200. - R. J. Mathar, Jun 13 2008
Let A be the Hessenberg n X n matrix defined by: A[1,j]=j mod 3, A[i,i]:=1, A[i,i-1]=-1. Then, for n>=1, a(n)=det(A). - Milan Janjic, Jan 24 2010
2-adic valuation of A104537(n+1). - Gerry Martens, Jul 14 2015
Conjecture: a(n) is the exponent of the largest power of 2 that divides all the entries of the matrix {{3,1},{1,-1}}^n. - Greg Dresden, Sep 09 2018

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Cf. A010872, A010873, A010874, A010875, A010876, A010877, A130482, A130483, A130484.

Cf. A092200, A104537.

List([0..80], n-> Int((n+1)/3) + Int(2*(n+1)/3)); # G. C. Greubel, Aug 31 2019

[Floor((n+1)/3) + Floor(2*(n+1)/3): n in [0..80]]; // G. C. Greubel, Aug 31 2019

seq(coeff(series(x*(1+2*x)/((1-x^3)*(1-x)), x, n+1), x, n), n = 0..80); # G. C. Greubel, Aug 31 2019

a[n_]:= Floor[(n+1)/3] + Floor[2(n+1)/3]; Table[a[n], {n, 0, 80}] (* Clark Kimberling, May 28 2012 *)
a[n_]:= IntegerExponent[A104537[n + 1], 2];
Table[a[n], {n, 0, 80}]  (* Gerry Martens, Jul 14 2015 *)
CoefficientList[Series[x(1+2x)/((1-x^3)(1-x)), {x, 0, 80}], x] (* Stefano Spezia, Sep 09 2018 *)
LinearRecurrence[{1,0,1,-1},{0,1,3,3},100] (* Harvey P. Dale, Jun 14 2021 *)

main(size)=my(n,k);vector(size,n,sum(k=0,n,k%3)) \\ Anders Hellström, Jul 14 2015

first(n)=my(s); concat(0, vector(n,k,s+=k%3)) \\ Charles R Greathouse IV, Jul 14 2015

a(n)=n\3*3+[0,1,3][n%3+1] \\ Charles R Greathouse IV, Jul 14 2015

def A130481_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P(x*(1+2*x)/((1-x^3)*(1-x))).list()
A130481_list(80) # G. C. Greubel, Aug 31 2019

a(n) = 3*floor(n/3) + A010872(n)*(A010872(n) + 1)/2.
G.f.: x*(1 + 2*x)/((1-x^3)*(1-x)).
a(n) = n + 1 - (Fibonacci(n+1) mod 2). - Gary Detlefs, Mar 13 2011
a(n) = floor((n+1)/3) + floor(2*(n+1)/3). - Clark Kimberling, May 28 2010
a(n) = n when n+1 is not a multiple of 3, and a(n) = n+1 when n+1 is a multiple of 3. - Dennis P. Walsh, Aug 06 2012
a(n) = n + 1 - sign((n+1) mod 3). - Wesley Ivan Hurt, Sep 25 2017
a(n) = n + (1-cos(2*(n+2)*Pi/3))/3 + sin(2*(n+2)*Pi/3)/sqrt(3). - Wesley Ivan Hurt, Sep 27 2017
a(n) = n + 1 - (n+1)^2 mod 3. - Ammar Khatab, Aug 14 2020
E.g.f.: ((1 + 3*x)*cosh(x) - (cos(sqrt(3)*x/2) + sqrt(3)*sin(sqrt(3)*x/2))*(cosh(x/2) - sinh(x/2)) + (1 + 3*x)*sinh(x))/3. - Stefano Spezia, May 28 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/(3*sqrt(3)) + log(2)/3. - Amiram Eldar, Sep 17 2022

A130483 a(n) = Sum_{k=0..n} (k mod 5) (Partial sums of A010874).

Original entry on oeis.org

0, 1, 3, 6, 10, 10, 11, 13, 16, 20, 20, 21, 23, 26, 30, 30, 31, 33, 36, 40, 40, 41, 43, 46, 50, 50, 51, 53, 56, 60, 60, 61, 63, 66, 70, 70, 71, 73, 76, 80, 80, 81, 83, 86, 90, 90, 91, 93, 96, 100, 100, 101, 103, 106, 110, 110, 111, 113, 116, 120, 120, 121, 123, 126, 130, 130

Offset: 0

Hieronymus Fischer, May 29 2007

Let A be the Hessenberg n X n matrix defined by: A[1,j]=j mod 5, A[i,i]=1, A[i,i-1]=-1. Then, for n>=1, a(n)=det(A). - Milan Janjic, Jan 24 2010

Shawn A. Broyles, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,1,-1).

Cf. A010872, A010873, A010875, A010876, A010877, A130481, A130482, A130484, A130485.

a:=[0,1,3,6,10,10];; for n in [7..71] do a[n]:=a[n-1]+a[n-5]-a[n-6]; od; a; # G. C. Greubel, Aug 31 2019

I:=[0,1,3,6,10,10]; [n le 6 select I[n] else Self(n-1) + Self(n-5) - Self(n-6): n in [1..71]]; // G. C. Greubel, Aug 31 2019

seq(coeff(series(x*(1+2*x+3*x^2+4*x^3)/((1-x^5)*(1-x)), x, n+1), x, n), n = 0..70); # G. C. Greubel, Aug 31 2019

Accumulate[Mod[Range[0,70],5]] (* or *) Accumulate[PadRight[{},70,{0,1,2,3,4}]] (* Harvey P. Dale, Nov 11 2016 *)

a(n) = sum(k=0, n, k % 5); \\ Michel Marcus, Apr 28 2018

def A130483_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P(x*(1+2*x+3*x^2+4*x^3)/((1-x^5)*(1-x))).list()
A130483_list(70) # G. C. Greubel, Aug 31 2019

a(n) = 10*floor(n/5) + A010874(n)*(A010874(n)+1)/2.
G.f.: x*(1 + 2*x + 3*x^2 + 4*x^3)/((1-x^5)*(1-x)).
From Wesley Ivan Hurt, Jul 23 2016: (Start)
a(n) = a(n-5) - a(n-6) for n>5; a(n) = a(n-5) + 10 for n>4.
a(n) = 10 + Sum_{k=1..4} k*floor((n-k)/5). (End)
a(n) = ((n mod 5)^2 - 3*(n mod 5) + 4*n)/2. - Ammar Khatab, Aug 13 2020

cp's OEIS Frontend

A010879 Final digit of n.

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A010873 a(n) = n mod 4.

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A010872 a(n) = n mod 3.

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A010874 a(n) = n mod 5.

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A010875 a(n) = n mod 6.

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A010877 a(n) = n mod 8.