A010872 -id:A010872 - cp's OEIS frontend

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

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).

A080425 Period 3: repeat [0, 2, 1].

Original entry on oeis.org

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

Offset: 0

Paul Barry, Feb 20 2003

Previous name was: Jacobsthal selector sequence.
The Jacobsthal sequence A001045 can be defined by A001045(n) = Sum_{k=0..floor(n,3)} C(n, a(n-1)+3*k).
The floor of the area under the polygon connecting the lattice points: (n, a(n)) from 0..n is A001477(n), the nonnegative integers. - Wesley Ivan Hurt, Jun 16 2014

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

Cf. A001045, A001477, A007318, A010872, A130196, A131534.

a080425 = (`mod` 3) . (3 -) . (`mod` 3)
a080425_list = cycle [0, 2, 1]  -- Reinhard Zumkeller, Feb 22 2013

[-n mod 3 : n in [0..100]]; // Wesley Ivan Hurt, Jun 16 2014

seq(modp(2*n,3), n=0..90); # Zerinvary Lajos, Dec 01 2006

Table[Mod[-n, 3], {n, 0, 100}] (* Wesley Ivan Hurt, Jun 16 2014 *)

a(n)=2*n%3 \\ Charles R Greathouse IV, Sep 24 2015

a(n) = ceiling(((n mod 3) + 1)/2) + (-1)^((n mod 3) + 1).
G.f.: x*(x+2)/(1-x^3). - Paul Barry, May 25 2003
a(n) = (3 - (n mod 3)) mod 3. - Reinhard Zumkeller, Jul 30 2005
a(n) = 2 * A001045(L(n/3)), where L(j/p) is the Legendre symbol of j and p.
a(n) = (-n) mod 3; also a(n) = 3*ceiling(n/3)-n. - Hieronymus Fischer, May 29 2007
a(n) = A130196(n) + A131534(n) - 2. - Reinhard Zumkeller, Nov 12 2009
a(n) = (2n) mod 3. - Wesley Ivan Hurt, Jun 23 2013
From Wesley Ivan Hurt, Jul 02 2016: (Start)
a(n) = a(n-3) for n>2.
a(n) = 2*sin(n*Pi/3)*(3*sin(n*Pi/3) + sqrt(3)*cos(n*Pi/3))/3. (End)

More terms from Reinhard Zumkeller, Jul 30 2005

New name from Joerg Arndt, Apr 21 2014

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)

A053838 a(n) = (sum of digits of n written in base 3) modulo 3.

Original entry on oeis.org

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

Offset: 0

Henry Bottomley, Mar 28 2000

Start with 0, repeatedly apply the morphism 0->012, 1->120, 2->201. This is a ternary version of the Thue-Morse sequence A010060. See Brlek (1989). - N. J. A. Sloane, Jul 10 2012

A090193 is generated by the same mapping starting with 1. A090239 is generated by the same mapping starting with 2. - Andrey Zabolotskiy, May 04 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
S. Brlek, Enumeration of factors in the Thue-Morse word, Discrete Applied Math. 24 (1989), 83-96.
Arthur Dolgopolov, Equitable Sequencing and Allocation Under Uncertainty, Preprint, 2016.
Glen Joyce C. Dulatre, Jamilah V. Alarcon, Vhenedict M. Florida, and Daisy Ann A. Disu, On Fractal Sequences, DMMMSU-CAS Science Monitor (2016-2017) Vol. 15 No. 2, 109-113.
Michael Gilleland, Some Self-Similar Integer Sequences
Michel Rigo, Relations on words, arXiv preprint arXiv:1602.03364 [cs.FL], 2016. See Example 17.
Robert Walker, Self Similar Sloth Canon Number Sequences
Index entries for sequences that are fixed points of mappings

Cf. A004128, A010060, A053837, A053839-A053844.

Equals A026600(n+1) - 1.

A053838 := proc(n)
    add(d,d=convert(n,base,3)) ;
    modp(%,3) ;
end proc:
seq(A053838(n),n=0..100) ; # R. J. Mathar, Nov 04 2017

Nest[ Flatten[ # /. {0 -> {0, 1, 2}, 1 -> {1, 2, 0}, 2 -> {2, 0, 1}}] &, {0}, 7] (* Robert G. Wilson v, Mar 08 2005 *)

a(n) = vecsum(digits(n, 3)) % 3; \\ Michel Marcus, May 04 2016

from sympy.ntheory import digits
def A053838(n): return sum(digits(n,3)[1:])%3 # Chai Wah Wu, Feb 28 2025

a(n) = A010872(A053735(n)) =(n+a(floor[n/3])) mod 3. So one can construct sequence by starting with 0 and mapping 0->012, 1->120 and 2->201 (e.g. 0, 012, 012120201, 012120201120201012201012120, ...) and looking at n-th digit of a term with sufficient digits.

a(n) = A004128(n) mod 3. [Gary W. Adamson, Aug 24 2008]

A126043 Exponents p of the Mersenne primes 2^p - 1 (see A000043) read mod 3.

Original entry on oeis.org

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

Offset: 1

Artur Jasinski, Dec 17 2006

Cf. A000043, A010872 (n mod 3), A126044-A126059.

Array[Mod[MersennePrimeExponent@ #, 3] &, 45] (* Michael De Vlieger, Apr 07 2018 *)

forprime(p=1, 1e3, if(isprime(2^p-1), print1(p%3, ", "))) \\ Felix Fröhlich, Aug 12 2014

a(n) = A010872(A000043(n)). - Michel Marcus, Aug 12 2014

a(45)-a(47) from Ivan Panchenko, Apr 08 2018

a(48) from Amiram Eldar, Oct 14 2024

A074941 a(n) = sigma(n) mod 3.

Original entry on oeis.org

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

Offset: 1

Benoit Cloitre, Oct 04 2002

Antti Karttunen, Table of n, a(n) for n = 1..16384

Cf. A000203, A010872.

Differs from A002324 for the first time at n=49, where a(49) = 0, while A002324(49) = 3.

A074941:= n-> (numtheory[sigma](n) mod 3):
seq (A074941(n), n=1..105); # Jani Melik, Jan 26 2011

a[n_] := Mod[DivisorSigma[1, n], 3]; Array[a, 100] (* Amiram Eldar, Jun 06 2022 *)

PARI
```
a(n)=sigma(n)%3
```

a(n) = A010872(A000203(n)). - Antti Karttunen, Nov 05 2017

A109594 n followed by n^3 followed by n^2.

Original entry on oeis.org

1, 1, 1, 2, 8, 4, 3, 27, 9, 4, 64, 16, 5, 125, 25, 6, 216, 36, 7, 343, 49, 8, 512, 64, 9, 729, 81, 10, 1000, 100, 11, 1331, 121, 12, 1728, 144, 13, 2197, 169, 14, 2744, 196, 15, 3375, 225, 16, 4096, 256, 17, 4913, 289, 18, 5832, 324, 19, 6859, 361, 20, 8000, 400

Offset: 1

Mohammad K. Azarian, Aug 30 2005

Iain Fox, Table of n, a(n) for n = 1..10000 (first 1000 terms from Colin Barker)
Index entries for linear recurrences with constant coefficients, signature (0,0,4,0,0,-6,0,0,4,0,0,-1).

Cf. A000463, A010872.

map(t -> (t,t^3,t^2), [$1..100]); # Robert Israel, Dec 17 2017

Flatten[Table[{n,n^3,n^2},{n,20}]] (* Harvey P. Dale, Jul 22 2012 *)

Vec(x*(1+x+x^2-2*x^3+4*x^4+x^6+x^7-x^8)/((1-x)^4*(1+x+x^2)^4) + O(x^100)) \\ Colin Barker, Dec 02 2015

From Colin Barker, Dec 02 2015: (Start)
a(n) = 4*a(n-3) - 6*a(n-6) + 4*a(n-9) - a(n-12) for n > 12.
G.f.: x*(1+x+x^2-2*x^3+4*x^4+x^6+x^7-x^8) / ((1-x)^4*(1+x+x^2)^4).
(End)
a(n) = floor((n+2)/3)^((3*(n mod 3)^2-5*(n mod 3)+4)/2). - Luce ETIENNE, Dec 17 2017

A131294 a(n)=ds_3(a(n-1))+ds_3(a(n-2)), a(0)=0, a(1)=1; where ds_3=digital sum base 3.

Original entry on oeis.org

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

Offset: 0

Hieronymus Fischer, Jun 27 2007

The digital sum analog (in base 3) of the Fibonacci recurrence.
When starting from index n=3, periodic with Pisano period A001175(2)=3.
a(n) and Fib(n)=A000045(n) are congruent modulo 2 which implies that (a(n) mod 2) is equal to (Fib(n) mod 2)=A011655(n). Thus (a(n) mod 2) is periodic with the Pisano period A001175(2)=3 too.
For general bases p>2, we have the inequality 2<=a(n)<=2p-3 (for n>2). Actually, a(n)<=3=A131319(3) for the base p=3.

			a(5)=3, since a(3)=2, ds_3(2)=2, a(4)=3=10(base 3),
ds_3(3)=1 and so a(5)=2+1.

Index entries for Colombian or self numbers and related sequences
Index entries for linear recurrences with constant coefficients, signature (0, 0, 1).

Cf. A000045, A010073, A010074, A010075, A010076, A010077, A131295, A131296, A131297, A131318, A131319, A131320.

nxt[{a_,b_}]:={b,Total[IntegerDigits[a,3]]+Total[IntegerDigits[b,3]]}; Transpose[NestList[nxt,{0,1},100]][[1]] (* Harvey P. Dale, Aug 02 2016 *)

a(n) = a(n-1)+a(n-2)-2*(floor(a(n-1)/3)+floor(a(n-2)/3)).
a(n) = floor(a(n-1)/3)+floor(a(n-2)/3)+(a(n-1)mod 3)+(a(n-2)mod 3).
a(n) = A002264(a(n-1))+A002264(a(n-2))+A010872(a(n-1))+A010872(a(n-2)).
a(n) = Fib(n)-2*sum{1A000045(n).

Incorrect comment removed by Michel Marcus, Apr 29 2018

cp's OEIS Frontend

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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A080425 Period 3: repeat [0, 2, 1].

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

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Sage

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

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A053838 a(n) = (sum of digits of n written in base 3) modulo 3.

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