A010874 -id:A010874 - cp's OEIS frontend

A010882 Period 3: repeat [1, 2, 3].

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

Offset: 0

N. J. A. Sloane

Partial sums are given by A130481(n)+n+1. - Hieronymus Fischer, Jun 08 2007
41/333 = 0.123123123... - Eric Desbiaux, Nov 03 2008
Terms of the simple continued fraction for 3/(sqrt(37)-4). - Paolo P. Lava, Feb 16 2009
This is the lexicographically earliest sequence with no substring of more than 1 term being a palindrome. - Franklin T. Adams-Watters, Nov 24 2013

Index entries for linear recurrences with constant coefficients, signature (0,0,1).
Index entries for sequences that are fixed points of mappings

Cf. A010872, A010873, A010874, A010875, A010876, A004526, A002264, A002265, A002266, A177036 (decimal expansion of (4+sqrt(37))/7), A214090.

Cf. A130481, A180593.

a010882 = (+ 1) . (`mod` 3)
a010882_list = cycle [1,2,3]
-- Reinhard Zumkeller, Mar 20 2013

&cat[[1..3]^^30]; // Vincenzo Librandi, Feb 04 2016

seq(op([1, 2, 3]), n=0..50); # Wesley Ivan Hurt, Jul 05 2016

Nest[ Flatten[ # /. {1 -> {1, 2}, 2 -> {3, 1}, 3 -> {2, 3}}] &, {1}, 7] (* Robert G. Wilson v, Mar 08 2005 *)
PadRight[{},120,{1,2,3}] (* Harvey P. Dale, Apr 09 2018 *)

a(n) = 1 + n%3; \\ Michel Marcus, Feb 04 2016

G.f.: (1+2x+3x^2)/(1-x^3). - Paul Barry, May 25 2003
a(n) = 1 + (n mod 3). - Paolo P. Lava, Nov 21 2006
a(n) = A010872(n) + 1. - Hieronymus Fischer, Jun 08 2007
a(n) = 6 - a(n-1) - a(n-2) for n > 1. - Reinhard Zumkeller, Apr 13 2008
a(n) = n+1-3*floor(n/3) = floor(41*10^(n+1)/333)-floor(41*10^n/333)*10; a(n)-a(n-3)=0 with n>2. - Bruno Berselli, Jun 28 2010
a(n) = A180593(n+1)/3. - Reinhard Zumkeller, Oct 25 2010
a(n) = floor((4*n+3)/3) mod 4. - Gary Detlefs, May 15 2011
a(n) = -cos(2/3*Pi*n)-1/3*3^(1/2)*sin(2/3*Pi*n)+2. - Leonid Bedratyuk, May 13 2012
E.g.f.: 2*(3*exp(3*x/2) - sqrt(3)*cos(Pi/6-sqrt(3)*x/2))*exp(-x/2)/3. - Ilya Gutkovskiy, Jul 05 2016

A130484 a(n) = Sum_{k=0..n} (k mod 6) (Partial sums of A010875).

Original entry on oeis.org

0, 1, 3, 6, 10, 15, 15, 16, 18, 21, 25, 30, 30, 31, 33, 36, 40, 45, 45, 46, 48, 51, 55, 60, 60, 61, 63, 66, 70, 75, 75, 76, 78, 81, 85, 90, 90, 91, 93, 96, 100, 105, 105, 106, 108, 111, 115, 120, 120, 121, 123, 126, 130, 135, 135, 136, 138, 141, 145, 150, 150, 151, 153

Offset: 0

Hieronymus Fischer, May 31 2007

Let A be the Hessenberg n X n matrix defined by A[1,j] = j mod 6, 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 (1,0,0,0,0,1,-1).

Cf. A010872, A010873, A010874, A010876, A010877, A130481, A130482, A130483, A130485.

a:=[0,1,3,6,10,15,15];; for n in [8..71] do a[n]:=a[n-1]+a[n-6]-a[n-7]; od; a; # G. C. Greubel, Aug 31 2019

I:=[0,1,3,6,10,15,15]; [n le 7 select I[n] else Self(n-1) + Self(n-6) - Self(n-7): n in [1..71]]; // G. C. Greubel, Aug 31 2019

seq(coeff(series(x*(1-6*x^5+5*x^6)/((1-x^6)*(1-x)^3), x, n+1), x, n), n = 0 .. 70); # G. C. Greubel, Aug 31 2019

Accumulate[Mod[Range[0,70],6]] (* or *) Accumulate[PadRight[ {},70, Range[0,5]]] (* Harvey P. Dale, Jul 12 2016 *)

a(n) = sum(k=0, n, k % 6); \\ Michel Marcus, Apr 28 2018

a(n)=n\6*15 + binomial(n%6+1,2) \\ Charles R Greathouse IV, Jan 24 2022

def A130484_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P(x*(1-6*x^5+5*x^6)/((1-x^6)*(1-x)^3)).list()
A130484_list(70) # G. C. Greubel, Aug 31 2019

a(n) = 15*floor(n/6) + A010875(n)*(A010875(n) + 1)/2.

G.f.: (Sum_{k=1..5} k*x^k)/((1-x^6)*(1-x)) = x*(1 - 6*x^5 + 5*x^6)/((1-x^6)*(1-x)^3).

A130485 a(n) = Sum_{k=0..n} (k mod 7) (Partial sums of A010876).

Original entry on oeis.org

0, 1, 3, 6, 10, 15, 21, 21, 22, 24, 27, 31, 36, 42, 42, 43, 45, 48, 52, 57, 63, 63, 64, 66, 69, 73, 78, 84, 84, 85, 87, 90, 94, 99, 105, 105, 106, 108, 111, 115, 120, 126, 126, 127, 129, 132, 136, 141, 147, 147, 148, 150, 153, 157, 162, 168, 168, 169, 171, 174, 178, 183

Offset: 0

Hieronymus Fischer, May 31 2007

Let A be the Hessenberg n X n matrix defined by A[1,j] = j mod 7, 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 (1,0,0,0,0,0,1,-1).

Cf. A010872, A010873, A010874, A010875, A010877, A130481, A130482, A130483, A130484.

a:=[0,1,3,6,10,15,21,21];; for n in [9..71] do a[n]:=a[n-1]+a[n-7]-a[n-8]; od; a; # G. C. Greubel, Aug 31 2019

I:=[0,1,3,6,10,15,21,21]; [n le 8 select I[n] else Self(n-1) + Self(n-7) - Self(n-8): n in [1..71]]; // G. C. Greubel, Aug 31 2019

a:=n->add(chrem( [n,j], [1,7] ),j=1..n):seq(a(n), n=1..70); # Zerinvary Lajos, Apr 07 2009

LinearRecurrence[{1,0,0,0,0,0,1,-1},{0,1,3,6,10,15,21,21},70] (* Harvey P. Dale, Jul 30 2017 *)

concat(0,Vec((1-7*x^6+6*x^7)/(1-x^7)/(1-x)^3+O(x^70))) \\ Charles R Greathouse IV, Dec 22 2011

def A130485_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P(x*(1-7*x^6+6*x^7)/((1-x^7)*(1-x)^3)).list()
A130485_list(70) # G. C. Greubel, Aug 31 2019

a(n) = 21*floor(n/7) + A010876(n)*(A010876(n) + 1)/2.
G.f.: (Sum_{k=1..6} k*x^k)/((1-x^7)*(1-x)).
G.f.: x*(1 - 7*x^6 + 6*x^7)/((1-x^7)*(1-x)^3).

A130518 a(n) = Sum_{k=0..n} floor(k/3). (Partial sums of A002264.)

Original entry on oeis.org

0, 0, 0, 1, 2, 3, 5, 7, 9, 12, 15, 18, 22, 26, 30, 35, 40, 45, 51, 57, 63, 70, 77, 84, 92, 100, 108, 117, 126, 135, 145, 155, 165, 176, 187, 198, 210, 222, 234, 247, 260, 273, 287, 301, 315, 330, 345, 360, 376, 392, 408, 425, 442, 459, 477, 495, 513, 532, 551, 570

Offset: 0

Hieronymus Fischer, Jun 01 2007

Complementary with A130481 regarding triangular numbers, in that A130481(n) + 3*a(n) = n(n+1)/2 = A000217(n).
Apart from offset, the same as A062781. - R. J. Mathar, Jun 13 2008
Apart from offset, the same as A001840. - Michael Somos, Sep 18 2010
The sum of any three consecutive terms is a triangular number. - J. M. Bergot, Nov 27 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).

Cf. A002265, A002266, A004526, A010872, A010873, A010874, A062781, A130482, A130483.

List([0..60], n-> Int(n*(n-1)/6)); # G. C. Greubel, Aug 31 2019

[Round(n*(n-1)/6): n in [0..60]]; // Vincenzo Librandi, Jun 25 2011

seq(floor(n*(n-1)/6),n=0..60); # Robert Israel, Nov 27 2014

Table[n, {n, 0, 19}, {3}] // Flatten // Accumulate (* Jean-François Alcover, Jun 05 2013 *)

a(n)=n*(n-1)\/6 \\ Charles R Greathouse IV, Jun 05 2013

[floor(binomial(n,2)/3) for n in range(0,60)] # Zerinvary Lajos, Dec 01 2009

G.f.: x^3 / ((1-x^3)*(1-x)^2).
a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + a(n-5).
a(n) = (1/2)*floor(n/3)*(2*n - 1 - 3*floor(n/3)) = A002264(n)*(2n - 1 - 3*A002264(n))/2.
a(n) = (1/2)*A002264(n)*(n - 1 + A010872(n)).
a(n) = round(n*(n-1)/6) = round((n^2-n-1)/6) = floor(n*(n-1)/6) = ceiling((n+1)*(n-2)/6). - Mircea Merca, Nov 28 2010
a(n) = a(n-3) + n - 2, n > 2. - Mircea Merca, Nov 28 2010
a(n) = A214734(n, 1, 3). - Renzo Benedetti, Aug 27 2012
a(3n) = A000326(n), a(3n+1) = A005449(n), a(3n+2) = 3*A000217(n) = A045943(n). - Philippe Deléham, Mar 26 2013
a(n) = (3*n*(n-1) - (-1)^n*((1+i*sqrt(3))^(n-2) + (1-i*sqrt(3))^(n-2))/2^(n-3) - 2)/18, where i=sqrt(-1). - Bruno Berselli, Nov 30 2014
Sum_{n>=3} 1/a(n) = 20/3 - 2*Pi/sqrt(3). - Amiram Eldar, Sep 17 2022

A130520 a(n) = Sum_{k=0..n} floor(k/5). (Partial sums of A002266.)

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 7, 9, 11, 13, 15, 18, 21, 24, 27, 30, 34, 38, 42, 46, 50, 55, 60, 65, 70, 75, 81, 87, 93, 99, 105, 112, 119, 126, 133, 140, 148, 156, 164, 172, 180, 189, 198, 207, 216, 225, 235, 245, 255, 265, 275, 286, 297, 308, 319, 330, 342, 354, 366

Offset: 0

Hieronymus Fischer, Jun 01 2007

Complementary with A130483 regarding triangular numbers, in that A130483(n) + 5*a(n) = n*(n+1)/2 = A000217(n).

Given a sequence b(n) defined by variables b(0) to b(5) and recursion b(n) = -(b(n-6) * a(n-2) * (b(n-4) * b(n-2)^3 - b(n-3)^3 * b(n-1)) - b(n-5) * b(n-3) * b(n-1) * (b(n-5) * b(n-2)^2 - b(n-4)^2 * b(n-1)))/(b(n-4) * (b(n-5) * b(n-3)^3 - b(n-4)^3 * b(n-2))). The denominator of b(n+1) has a factor of (b(1) * b(3)^3 - b(2)^3 * b(4))^a(n+1). For example, if b(0) = 2, b(1) = b(2) = b(3) = 1, b(4) = 1+x, b(5) = 4, then the denominator of b(n+1) is x^a(n+1). - Michael Somos, Nov 15 2023

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,1,-2,1).

Cf. A002264, A002265, A004526, A010872, A010873, A010874, A130481, A130482.

List([0..70], n-> Int((n-1)*(n-2)/10)); # G. C. Greubel, Aug 31 2019

[Round(n*(n-3)/10): n in [0..70]]; // Vincenzo Librandi, Jun 25 2011

seq(floor((n-1)*(n-2)/10), n=0..70); # G. C. Greubel, Aug 31 2019

Accumulate[Floor[Range[0,70]/5]] (* Harvey P. Dale, May 25 2016 *)

a(n) = sum(k=0, n, k\5); \\ Michel Marcus, May 13 2016

[floor((n-1)*(n-2)/10) for n in (0..70)] # G. C. Greubel, Aug 31 2019

a(n) = floor(n/5)*(2*n - 3 - 5*floor(n/5))/2.
a(n) = A002266(n)*(2*n - 3 - 5*A002266(n))/2.
a(n) = A002266(n)*(n -3 +A010874(n))/2.
G.f.: x^5/((1-x^5)*(1-x)^2) =  x^5/( (1+x+x^2+x^3+x^4)*(1-x)^3 ).
a(n) = floor((n-1)*(n-2)/10). - Mitch Harris, Sep 08 2008
a(n) = round(n*(n-3)/10) = ceiling((n+1)*(n-4)/10) = round((n^2 - 3*n - 1)/10). - Mircea Merca, Nov 28 2010
a(n) = A008732(n-5), n > 4. - R. J. Mathar, Nov 22 2008
a(n) = a(n-5) + n - 4, n > 4. - Mircea Merca, Nov 28 2010
a(5n) = A000566(n), a(5n+1) = A005476(n), a(5n+2) = A005475(n), a(5n+3) = A147875(n), a(5n+4) = A028895(n). - Philippe Deléham, Mar 26 2013
From Amiram Eldar, Sep 17 2022: (Start)
Sum_{n>=5} 1/a(n) = 518/45 - 2*sqrt(2*(sqrt(5)+5))*Pi/3.
Sum_{n>=5} (-1)^(n+1)/a(n) = 8*sqrt(5)*arccoth(3/sqrt(5))/3 + 92*log(2)/15 - 418/45. (End)

A039703 a(n) = n-th prime modulo 5.

Original entry on oeis.org

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

Offset: 1

Clark Kimberling

a(A049084(A045356(n-1))) = even; a(A049084(A045429(n-1))) = odd. - Reinhard Zumkeller, Feb 25 2008

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000

Cf. A000040, A010874, A039701, A039702, A039704-A039706, A038194, A007652, A039709-A039715.

[p mod 5: p in PrimesUpTo(500)]; // Vincenzo Librandi, May 06 2014

seq(ithprime(n) mod 5, n=1..105); # Nathaniel Johnston, Jun 29 2011

Table[Mod[Prime[n], 5], {n, 105}] (* Nathaniel Johnston, Jun 29 2011 *)
Mod[Prime[Range[100]], 5] (* Vincenzo Librandi, May 06 2014 *)

primes(105) % 5 \\ Zak Seidov, Apr 09 2013

Sum_k={1..n} a(k) ~ (5/2)*n. - Amiram Eldar, Dec 11 2024

A076314 a(n) = floor(n/10) + (n mod 10).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 8, 9, 10, 11, 12, 13, 14

Offset: 0

Reinhard Zumkeller, Oct 06 2002

For n<100 this is equal to the digital sum of n (see A007953). - Hieronymus Fischer, Jun 17 2007

			a(15) = floor(15 / 10) + (15 mod 10) = 1 + 5 = 6. - _Indranil Ghosh_, Feb 13 2017

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

Cf. A076313, A010879, A076309, A076310, A076311, A076312, A055017, A007953, A003132.

a076314 = uncurry (+) . flip divMod 10 -- Reinhard Zumkeller, Jun 01 2013

[n-9*Floor(n/10):n in [0..100]]; // Vincenzo Librandi, Dec 10 2016

Table[n - 9 Floor[n / 10], {n, 0, 100}] (* Vincenzo Librandi Dec 10 2016 *)

a(n)=n\10+n%10 \\ Charles R Greathouse IV, Sep 24 2015

def A076314(n): return (n/10)+(n%10) # Indranil Ghosh, Feb 13 2017

From Hieronymus Fischer, Jun 17 2007: (Start)
a(n) = n - 9*floor(n/10).
a(n) = (n + 9*(n mod 10))/10.
a(n) = n - 9*A002266(A004526(n)) = n - 9*A004526(A002266(n)).
a(n) = (n + 9*A010879(n))/10.
a(n) = (n + 9*A000035(n) + 18*A010874(A004526(n)))/10.
a(n) = (n + 9*A010874(n) + 45*A000035(A002266(n)))/10.
G.f.: x*(8*x^10 - 9*x^9 + 1)/((1 - x^10)*(1 - x)^2). (End)
a(n) = A033930(n) for 1 <= n < 100. - R. J. Mathar, Sep 21 2008
a(n) = +a(n-1) + a(n-10) - a(n-11). - R. J. Mathar, Feb 20 2011

A130486 a(n) = Sum_{k=0..n} (k mod 8) (Partial sums of A010877).

Original entry on oeis.org

0, 1, 3, 6, 10, 15, 21, 28, 28, 29, 31, 34, 38, 43, 49, 56, 56, 57, 59, 62, 66, 71, 77, 84, 84, 85, 87, 90, 94, 99, 105, 112, 112, 113, 115, 118, 122, 127, 133, 140, 140, 141, 143, 146, 150, 155, 161, 168, 168, 169, 171, 174, 178, 183, 189, 196, 196, 197, 199, 202, 206

Offset: 0

Hieronymus Fischer, May 31 2007

Let A be the Hessenberg n X n matrix defined by A[1,j] = j mod 8, 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 (1,0,0,0,0,0,0,1,-1).

Cf. A010872, A010873, A010874, A010875, A010876, A010878. A130481, A130482, A130483, A130484, A130485, A130487.

a:=[0,1,3,6,10,15,21,28,28];; for n in [10..71] do a[n]:=a[n-1]+a[n-8]-a[n-9]; od; a; # G. C. Greubel, Aug 31 2019

I:=[0,1,3,6,10,15,21,28,28]; [n le 9 select I[n] else Self(n-1) + Self(n-8) - Self(n-9): n in [1..71]]; // G. C. Greubel, Aug 31 2019

seq(coeff(series(x*(1-8*x^7+7*x^8)/((1-x^8)*(1-x)^3), x, n+1), x, n), n = 0 .. 40); # G. C. Greubel, Aug 31 2019

Array[28 Floor[#1/8] + #2 (#2 + 1)/2 & @@ {#, Mod[#, 8]} &, 61, 0] (* Michael De Vlieger, Apr 28 2018 *)
Accumulate[PadRight[{},100,Range[0,7]]] (* Harvey P. Dale, Dec 21 2018 *)

a(n) = sum(k=0, n, k % 8); \\ Michel Marcus, Apr 28 2018

def A130486_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P(x*(1-8*x^7+7*x^8)/((1-x^8)*(1-x)^3)).list()
A130486_list(70) # G. C. Greubel, Aug 31 2019

a(n) = 28*floor(n/8) + A010877(n)*(A010877(n) + 1)/2.
G.f.: (Sum_{k=1..7} k*x^k)/((1-x^8)*(1-x)).
G.f.: x*(1 - 8*x^7 + 7*x^8)/((1-x^8)*(1-x)^3).

A131296 a(n) = ds_5(a(n-1))+ds_5(a(n-2)), a(0)=0, a(1)=1; where ds_5=digital sum base 5.

Original entry on oeis.org

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

Offset: 0

Hieronymus Fischer, Jun 27 2007

The digital sum analog (in base 5) of the Fibonacci recurrence.
When starting from index n=3, periodic with Pisano period A001175(4)=6.
a(n) and Fib(n)=A000045(n) are congruent modulo 4 which implies that (a(n) mod 4) is equal to (Fib(n) mod 4)=A079343(n). Thus (a(n) mod 4) is periodic with the Pisano period A001175(4)=6 too.
For general bases p>2, the inequality 2<=a(n)<=2p-3 holds for n>2. Actually, a(n)<=5=A131319(5) for the base p=5.

			a(10)=3, since a(8)=5=10(base 5), ds_5(5)=1,
a(9)=2, ds_5(2)=2 and so a(10)=1+2.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for Colombian or self numbers and related sequences

Cf. A000045, A010073, A010074, A010075, A010076, A010077, A131294, A131295, A131297, A131318, A131319, A131320.

nxt[{a_,b_}]:={b,Total[IntegerDigits[a,5]]+Total[IntegerDigits[b,5]]}; NestList[nxt,{0,1},100][[;;,1]] (* Harvey P. Dale, Sep 01 2024 *)

a(n) = a(n-1)+a(n-2)-4*(floor(a(n-1)/5)+floor(a(n-2)/5)).
a(n) = floor(a(n-1)/5)+floor(a(n-2)/5)+(a(n-1)mod 5)+(a(n-2)mod 5).
a(n) = A002266(a(n-1))+A002266(a(n-2))+A010874(a(n-1))+A010874(a(n-2)).
a(n) = Fib(n)-4*sum{1A000045(n).

Incorrect comment removed by Michel Marcus, Apr 29 2018

A130487 a(n) = Sum_{k=0..n} (k mod 9) (Partial sums of A010878).

Original entry on oeis.org

0, 1, 3, 6, 10, 15, 21, 28, 36, 36, 37, 39, 42, 46, 51, 57, 64, 72, 72, 73, 75, 78, 82, 87, 93, 100, 108, 108, 109, 111, 114, 118, 123, 129, 136, 144, 144, 145, 147, 150, 154, 159, 165, 172, 180, 180, 181, 183, 186, 190, 195, 201, 208, 216, 216, 217, 219, 222, 226

Offset: 0

Hieronymus Fischer, May 31 2007

Let A be the Hessenberg n X n matrix defined by A[1,j]=j mod 9, 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 (1,0,0,0,0,0,0,0,1,-1).

Cf. A010872, A010873, A010874, A010875, A010876, A010877, A130481, A130482, A130483, A130484, A130485, A130486.

a:=[0,1,3,6,10,15,21,28,36,36];; for n in [11..71] do a[n]:=a[n-1]+a[n-9]-a[n-10]; od; a; # G. C. Greubel, Aug 31 2019

I:=[0,1,3,6,10,15,21,28,36,36]; [n le 10 select I[n] else Self(n-1) + Self(n-9) - Self(n-10): n in [1..71]]; // G. C. Greubel, Aug 31 2019

seq(coeff(series(x*(1-9*x^8+8*x^9)/((1-x^9)*(1-x)^3), x, n+1), x, n), n = 0 .. 70); # G. C. Greubel, Aug 31 2019

Accumulate[PadRight[{},120,Range[0,8]]] (* Harvey P. Dale, Dec 19 2018 *)
Accumulate[Mod[Range[0,100],9]] (* Harvey P. Dale, Oct 16 2021 *)

a(n) = sum(k=0, n, k % 9); \\ Michel Marcus, Apr 28 2018

def A130487_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P(x*(1-9*x^8+8*x^9)/((1-x^9)*(1-x)^3)).list()
A130487_list(70) # G. C. Greubel, Aug 31 2019

a(n) = 36*floor(n/9) + A010878(n)*(A010878(n) + 1)/2.
G.f.: (Sum_{k=1..8} k*x^k)/((1-x^9)*(1-x)).
G.f.: x*(1 - 9*x^8 + 8*x^9)/((1-x^9)*(1-x)^3).

cp's OEIS Frontend

A010882 Period 3: repeat [1, 2, 3].

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A130484 a(n) = Sum_{k=0..n} (k mod 6) (Partial sums of A010875).

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A130485 a(n) = Sum_{k=0..n} (k mod 7) (Partial sums of A010876).

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A130518 a(n) = Sum_{k=0..n} floor(k/3). (Partial sums of A002264.)

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A130520 a(n) = Sum_{k=0..n} floor(k/5). (Partial sums of A002266.)

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A039703 a(n) = n-th prime modulo 5.

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