A010873 -id:A010873 - 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).

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)

A099545 Odd part of n, modulo 4.

Original entry on oeis.org

1, 1, 3, 1, 1, 3, 3, 1, 1, 1, 3, 3, 1, 3, 3, 1, 1, 1, 3, 1, 1, 3, 3, 3, 1, 1, 3, 3, 1, 3, 3, 1, 1, 1, 3, 1, 1, 3, 3, 1, 1, 1, 3, 3, 1, 3, 3, 3, 1, 1, 3, 1, 1, 3, 3, 3, 1, 1, 3, 3, 1, 3, 3, 1, 1, 1, 3, 1, 1, 3, 3, 1, 1, 1, 3, 3, 1, 3, 3, 1, 1, 1, 3, 1, 1, 3, 3, 3, 1, 1, 3, 3, 1, 3, 3, 3, 1, 1, 3, 1, 1, 3

Offset: 1

Ralf Stephan, Oct 23 2004

The terms of this sequence are the even-indexed terms of A112658. - Alexandre Wajnberg, Jan 02 2006
Fractal sequence: odd terms are 1, 3, 1, 3,...; the even terms are the sequence itself: a(n)=a(2n)=a(4n)=a(8n)=a(16n)=... - Alexandre Wajnberg, Jan 02 2006
From Micah D. Tillman, Jan 29 2021: (Start)
Has the same structure as the regular paper-folding (dragon curve) sequence (A014577, A014709). We can interpret a(n) as the number of 90-degree rotations to make in a single direction at the n-th "turn" in the dragon curve. After all, making three 90-degree rotations to the left (turning a total of 270 degrees) is equivalent to making one 90-degree rotation to the right, and vice versa.
We can likewise produce the dragon curve by interpreting A000265(n), the whole odd part of n, as the number of 90-degree rotations to make in a single direction at the n-th "turn" in the curve. (End)

			a(100) = 1: the odd part of 100 is 100/4 = 25, and 25 mod 4 = 1.

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Cf. A010873, A000265, A014707, A038189.

Cf. A099544, A099546, A099547, A099548, A099549, A099550, A099551.

Array[Mod[#/(2^IntegerExponent[#, 2]), 4] &, 105] (* Michael De Vlieger, Feb 24 2021 *)

a(n)=bitand(n/(2^valuation(n,2)), 3); /* Joerg Arndt, Jul 18 2012 */

def A099545(n): return n>>(~n&n-1).bit_length()&3 # Chai Wah Wu, Feb 26 2025

a(n) = 2 * A038189(n) + 1.
(a(n)-1)/2 = A014707(n). - Alexandre Wajnberg, Jan 02 2006
a(n) = A010873(A000265(n)).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 2. - Amiram Eldar, Aug 29 2024

A131295 a(n)=ds_4(a(n-1))+ds_4(a(n-2)), a(0)=0, a(1)=1; where ds_4=digital sum base 4.

Original entry on oeis.org

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

Offset: 0

Hieronymus Fischer, Jun 27 2007

The digital sum analog (in base 4) of the Fibonacci recurrence.
When starting from index n=3, periodic with Pisano period A001175(3)=8.
Also a(n)==A004090(n) modulo 3 (A004090(n)=digital sum of Fib(n)).
For general bases p>2, the inequality 2<=a(n)<=2p-3 holds for n>2. Actually, a(n)<=5=A131319(4) for the base p=4.
a(n) and Fib(n)=A000045(n) are congruent modulo 3 which implies that (a(n) mod 3) is equal to (Fib(n) mod 3)=A082115(n-1) (for n>0). Thus (a(n) mod 3) is periodic with the Pisano period = A001175(3)=8 too. - Hieronymus Fischer

			a(8)=3, since a(6)=5=11(base 4), ds_4(5)=2,
a(7)=4=10(base 4), ds_4(4)=1 and so a(8)=2+1.

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, A131296, A131297, A131318, A131319, A131320.

nxt[{a_,b_}]:={b,Total[IntegerDigits[a,4]]+Total[IntegerDigits[b,4]]}; NestList[ nxt,{0,1},110][[All,1]] (* Harvey P. Dale, Jul 30 2018 *)

a(n)=a(n-1)+a(n-2)-3*(floor(a(n-1)/4)+floor(a(n-2)/4)).
a(n)=floor(a(n-1)/4)+floor(a(n-2)/4)+(a(n-1)mod 4)+(a(n-2)mod 4).
a(n)=A002265(a(n-1))+A002265(a(n-2))+A010873(a(n-1))+A010873(a(n-2)).
a(n)=Fib(n)-3*sum{1A000045(n).

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).

A010881 Simple periodic sequence: n mod 12.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11

Offset: 0

N. J. A. Sloane

The value of the rightmost digit in the base-12 representation of n. - Hieronymus Fischer, Jun 11 2007

			a(27) = 3 since 27 = 12*2+3.

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

Partial sums: A130490. Other related sequences A130481, A130482, A130483, A130484, A130485, A130486, A130487, A130488, A130489.

Mod[Range[0, 100], 12] (* Paolo Xausa, Feb 02 2024 *)

A010881(n)=n%12 \\ M. F. Hasler, Sep 25 2014

From Hieronymus Fischer, May 31 2007: (Start)
a(n) = n mod 12.
Complex representation: a(n) = (1/12)*(1-r^n)*Sum_{k=1..11} k*Product_{m=1..11, m<>k} (1-r^(n-m)) where r = exp(Pi/6*i) = (sqrt(3)+i)/2 and i = sqrt(-1).
Trigonometric representation: a(n) = (512/3)^2*(sin(n*Pi/12))^2*Sum_{k=1..11} k*Product_{m=1..11, m<>k} (sin((n-m)*Pi/12))^2.
G.f.: (Sum_{k=1..11} k*x^k)/(1-x^12).
G.f.: x*(11*x^12-12*x^11+1)/((1-x^12)*(1-x)^2). (End)
From Hieronymus Fischer, Jun 11 2007: (Start)
a(n) = (n mod 2)+2*(floor(n/2) mod 6) = A000035(n)+2*A010875(A004526(n)).
a(n) = (n mod 3)+3*(floor(n/3) mod 4) = A010872(n)+3*A010873(A002264(n)).
a(n) = (n mod 4)+4*(floor(n/4) mod 3) = A010873(n)+4*A010872(A002265(n)).
a(n) = (n mod 6)+6*(floor(n/6) mod 2) = A010875(n)+6*A000035(A152467(n)).
a(n) = (n mod 2)+2*(floor(n/2) mod 2)+4*(floor(n/4) mod 3) = A000035(n)+2*A000035(A004526(n))+4*A010872(A002265(n)). (End)
a(A001248(k) + 17) = 6 for k>2. - Reinhard Zumkeller, May 12 2010
a(n) = A034326(n+1)-1. - M. F. Hasler, Sep 25 2014

A353486 Primorial base exp-function reduced modulo 4.

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Apr 24 2022

There are no zeros because A276086 is a permutation of A048103.

Antti Karttunen, Table of n, a(n) for n = 0..65537
Index entries for sequences related to primorial base

Cf. A010873, A048103, A276086, A353488, A353489, A353516, A353526.

Cf. A007395, A353487 (bisections).

A276086(n) = { my(m=1, p=2); while(n, m *= (p^(n%p)); n = n\p; p = nextprime(1+p)); (m); };
A353486(n) = (A276086(n)%4);

a(n) = A010873(A276086(n)).
For n >= 1, a(n) = A010873(A353516(n) * A353526(n)).
For all n >= 0, a(2n) = A353487(n), a(2n+1) = 2.

A105824 a(n) = sigma(n) mod 4.

Original entry on oeis.org

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

Offset: 1

Shyam Sunder Gupta, May 05 2005

Antti Karttunen, Table of n, a(n) for n = 1..16384
Index entries for sequences related to sums of divisors

Cf. A000203, A010873, A248150, A072461, A072462, A191217.

Sequences sigma(n) mod k: A053866 (k=2), A074941 (k=3), A105824 (k=4), A105825 (k=5), A084301 (k=6), A105826 (k=7), A105827 (k=8).

A105824:= n-> (numtheory[sigma](n) mod 4):
seq (A105824(n), n=1..105); # Jani Melik, Jan 26 2011

Table[Mod[DivisorSigma[1, n], 4], {n, 100}] (* Wesley Ivan Hurt, Nov 07 2017 *)

PARI
```
a(n)=sigma(n)%4
```

a(n) = A010873(A000203(n)). - Antti Karttunen, Nov 07 2017

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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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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A099545 Odd part of n, modulo 4.

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A131295 a(n)=ds_4(a(n-1))+ds_4(a(n-2)), a(0)=0, a(1)=1; where ds_4=digital sum base 4.

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