A010872 -id:A010872 - cp's OEIS frontend

A010875 a(n) = n mod 6.

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

A244477 a(1)=3, a(2)=2, a(3)=1; thereafter a(n) = a(n-a(n-1)) + a(n-a(n-2)).

Original entry on oeis.org

3, 2, 1, 3, 5, 4, 3, 8, 7, 3, 11, 10, 3, 14, 13, 3, 17, 16, 3, 20, 19, 3, 23, 22, 3, 26, 25, 3, 29, 28, 3, 32, 31, 3, 35, 34, 3, 38, 37, 3, 41, 40, 3, 44, 43, 3, 47, 46, 3, 50, 49, 3, 53, 52, 3, 56, 55, 3, 59, 58, 3, 62, 61, 3, 65, 64, 3, 68, 67, 3, 71, 70, 3, 74, 73, 3, 77, 76, 3, 80

Offset: 1

N. J. A. Sloane, Jul 02 2014

Similar to Hofstadter's Q-sequence A005185 but with different starting values.

Golomb describes this as "quasi-periodic sequence with a quasi-period of 3".

Higham, J.; Tanny, S. More well-behaved meta-Fibonacci sequences. Proceedings of the Twenty-fourth Southeastern International Conference on Combinatorics, Graph Theory, and Computing (Boca Raton, FL, 1993). Congr. Numer. 98(1993), 3-17.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Altug Alkan, Nathan Fox, and Orhan Ozgur Aybar, On Hofstadter Heart Sequences, Complexity, Volume 2017, Article ID 2614163, 8 pages.
Nathan Fox, Linear-Recurrent Solutions to Meta-Fibonacci Recurrences, Part 1 (video), Rutgers Experimental Math Seminar, Oct 01 2015. Part 2 is vimeo.com/141111991.
S. W. Golomb, Discrete chaos: sequences satisfying "strange" recursions, unpublished manuscript, circa 1990 [cached copy, with permission (annotated)]
Index entries for Hofstadter-type sequences
Index entries for linear recurrences with constant coefficients, signature (0, 0, 2, 0, 0, -1).

Cf. A005185.

Cf. A010872.

a244477 n = a244477_list !! (n-1)
a244477_list = 3 : 2 : 1 : zipWith (+)
   (map a244477 $ zipWith (-) [4..] $ tail a244477_list)
   (map a244477 $ zipWith (-) [4..] $ drop 2 a244477_list)
-- Reinhard Zumkeller, Jul 05 2014

[n le 3 select 4-n else Self(n-Self(n-1)) + Self(n-Self(n-2)): n in [1..80]]; // Vincenzo Librandi, Nov 24 2015

f := proc(n) option remember;
    if n<=3 then
        4-n
    elif n > procname(n-1) and n > procname(n-2) then
        RETURN(procname(n-procname(n-1))+procname(n-procname(n-2)));
    else
        ERROR(" died at n= ", n);
    fi;
end proc;
[seq(f(n),n=0..200)];

a[1] = 3; a[2] = 2; a[3] = 1; a[n_] := a[n] = a[n - a[n - 1]] + a[n - a[n - 2]]; Array[a, 75] (* or *)
Flatten@ Table[{Mod[3n, 3] +3, 3n -1, 3n -2}, {n, 25}] (* Robert G. Wilson v, Nov 23 2015 *)

From Colin Barker, Nov 23 2015: (Start)
a(n) = 2*a(n-3) - a(n-6) for n>6.
G.f.: x*(2*x^5 + x^4 - 3*x^3 + x^2 + 2*x + 3)/((x - 1)^2*(x^2 + x + 1)^2). (End)
a(3*k) = 3*k-2, a(3*k+1) = 3, a(3*k+2) = 3*k+2. - Nathan Fox, Apr 02 2017
a(n) = 3*(m-1)^2*floor(n/3) - (3*m^2-8*m+2), where m = n mod 3. - Luce ETIENNE, Oct 17 2018

A052223 Numbers whose sum of digits is 9.

Original entry on oeis.org

9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 108, 117, 126, 135, 144, 153, 162, 171, 180, 207, 216, 225, 234, 243, 252, 261, 270, 306, 315, 324, 333, 342, 351, 360, 405, 414, 423, 432, 441, 450, 504, 513, 522, 531, 540, 603, 612, 621, 630, 702, 711, 720, 801, 810

Offset: 1

Henry Bottomley, Feb 01 2000

Any term of this sequence with an 11 appended cannot have 11 as prime factor. See A075154. [Lekraj Beedassy, Sep 27 2009]
A007953(a(n)) = 9; number of repdigits = #{9,333,1^9} = A242627(9) = 3. - Reinhard Zumkeller, Jul 17 2014
A010872(a(n)) = A010878(a(n)) = 0. - Ilya Gutkovskiy, Jun 04 2016

Alois P. Heinz, Table of n, a(n) for n = 1..10000 (first 1000 terms from Vincenzo Librandi)

Cf. A007953.
Row n=9 of A245062.
Cf. A011557 (1), A052216 (2), A052217 (3), A052218 (4), A052219 (5), A052220 (6), A052221 (7), A052222 (8), A052224 (10), A166311 (11), A235151 (12), A143164 (13), A235225(14), A235226 (15), A235227 (16), A166370 (17), A235228 (18), A166459 (19), A235229 (20).
Cf. A242614, A242627.

a052223 n = a052223_list !! (n-1)
a052223_list = filter ((== 9) . a007953) [0..]
-- Reinhard Zumkeller, Jul 17 2014

[n: n in [1..1500] | &+Intseq(n) eq 9 ]; // Vincenzo Librandi, Mar 08 2013

Select[Range[1500], Total[IntegerDigits[#]] == 9 &] (* Vincenzo Librandi, Mar 08 2013 *)

More terms from Larry Reeves (Larryr(AT)acm.org), Sep 05 2000

Offset changed by Bruno Berselli, Mar 07 2013

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

A039701 a(n) = n-th prime modulo 3.

Original entry on oeis.org

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

Offset: 1

Clark Kimberling

If n > 2 and prime(n) is a Mersenne prime then a(n) = 1. Proof: prime(n) = 2^p - 1 for some odd prime p, so prime(n) = 2*4^((p-1)/2) - 1 == 2 - 1 = 1 (mod 3). - Santi Spadaro, May 03 2002; corrected and simplified by Dean Hickerson, Apr 20 2003
Except for n = 2, a(n) is the smallest number k > 0 such that 3 divides prime(n)^k - 1. - T. D. Noe, Apr 17 2003
a(n) <> 0 for n <> 2; a(A049084(A003627(n))) = 2; a(A049084(A002476(n))) = 1; A134323(n) = (1 - 0^a(n)) * (-1)^(a(n)+1). - Reinhard Zumkeller, Oct 21 2007
Probability of finding 1 (or 2) in this sequence is 1/2. This follows from the Prime Number Theorem in arithmetic progressions. Examples: There are 4995 1's in terms (10^9 +1) to (10^9+10^4); there are 10^9/2-1926 1's in the first 10^9 terms. - Jerzy R Borysowicz, Mar 06 2022

Nathaniel Johnston, Table of n, a(n) for n = 1..10000

Cf. A091178 (indices of 1's), A091177 (indices of 2's).
Cf. A120326 (partial sums).
Cf. A185934, A217659.
Cf. A010872.
Other moduli: A039702-A039706, A038194, A007652, A039709-A039715.

a039701 = (`mod` 3) . a000040
a039701_list = map (`mod` 3) a000040_list
-- Reinhard Zumkeller, Nov 16 2012

[p mod(3): p in PrimesUpTo(500)]; // Vincenzo Librandi, May 06 2014

seq(ithprime(n) mod 3, n=1..105); # Nathaniel Johnston, Jun 29 2011

Mathematica
```
Table[Mod[Prime[n], 3], {n, 100}]
```

primes(100)%3 \\ Charles R Greathouse IV, May 06 2014

Sum_k={1..n} a(k) ~ (3/2)*n. - Amiram Eldar, Dec 11 2024

A004773 Numbers congruent to {0, 1, 2} mod 4: a(n) = floor(4*n/3).

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 8, 9, 10, 12, 13, 14, 16, 17, 18, 20, 21, 22, 24, 25, 26, 28, 29, 30, 32, 33, 34, 36, 37, 38, 40, 41, 42, 44, 45, 46, 48, 49, 50, 52, 53, 54, 56, 57, 58, 60, 61, 62, 64, 65, 66, 68, 69, 70, 72, 73, 74, 76, 77, 78, 80, 81, 82, 84, 85, 86, 88, 89, 90

Offset: 0

N. J. A. Sloane

The sequence b(n) = floor((4/3)*(n+2)) appears as an upper bound in Fijavz and Wood.
Binary expansion does not end in 11.
From Guenther Schrack, May 04 2023: (Start)
The sequence is the interleaving of the sequences A008586, A016813, A016825, in that order.
Let S(n) = a(n) + a(n+1) + a(n+2). Then floor(S(n)/3) = A042968(n+1), round(S(n)/3) = a(n+1), ceiling(S(n)/3) = A042965(n+2). (End)

Daniel Starodubtsev, Table of n, a(n) for n = 0..10000
Gasper Fijavz and David R. Wood, Graph Minors and Minimum Degree, arXiv:0812.1064 [math.CO], 2008.
Niall Graham and Frank Harary, Edge Sums of Hypercubes, Bull. Irish Math. Soc., Vol. 21 (1988), pp. 8-12.
Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Cf. A177702 (first differences), A000969 (partial sums).
Cf. A032766, this sequence, A001068, A047226, A047368, A004777.
Cf. similar sequences with formula n+i*floor(n/3) listed in A281899.
Cf. A002264, A004396, A004523, A004767, A004772, A008586, A010872, A016813, A016825, A042965, A042968.

[n: n in [0..100] | n mod 4 in [0..2]]; // Vincenzo Librandi, Dec 23 2010

seq(floor(n/3)+n,n=0..68); # Gary Detlefs, Mar 20 2010

f[n_] := Floor[4 n/3]; Array[f, 69, 0] (* Robert G. Wilson v, Dec 24 2010 *)
fQ[n_] := Mod[n, 4] != 3; Select[ Range[0, 90], fQ] (* Robert G. Wilson v, Dec 24 2010 *)
a[0] = 0; a[n_] := a[n] = a[n - 1] + 2 - If[ Mod[a[n - 1], 4] < 2, 1, 0]; Array[a, 69, 0] (* Robert G. Wilson v, Dec 24 2010 *)
CoefficientList[ Series[x (1 + x + 2 x^2)/((1 - x) (1 - x^3)), {x, 0, 68}], x] (* Robert G. Wilson v, Dec 24 2010 *)

a(n)=4*n\3 \\ Charles R Greathouse IV, Sep 27 2012

G.f.: x*(1+x+2*x^2)/((1-x)*(1-x^3)).
a(0) = 0, a(n+1) = a(n) + a(n) mod 4 + 0^(a(n) mod 4). - Reinhard Zumkeller, Mar 23 2003
a(n) = A004396(n) + A004523(n); complement of A004767. - Reinhard Zumkeller, Aug 29 2005
a(n) = floor(n/3) + n. - Gary Detlefs, Mar 20 2010
a(n) = (12*n-3+3*cos(2*n*Pi/3)+sqrt(3)*sin(2*n*Pi/3))/9. - Wesley Ivan Hurt, Sep 30 2017
E.g.f.: (3*exp(x)*(4*x - 1) + exp(-x/2)*(3*cos((sqrt(3)*x)/2) + sqrt(3)*sin((sqrt(3)*x)/2)))/9. - Stefano Spezia, Jun 09 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = (sqrt(2)-1)*Pi/8 + sqrt(2)*log(sqrt(2)+2)/4 + (2-sqrt(2))*log(2)/8. - Amiram Eldar, Dec 05 2021
From Guenther Schrack, May 04 2023: (Start)
a(n) = (12*n - 3 +  w^(2*n)*(w + 2) - w^n*(w - 1))/9 where w = (-1 + sqrt(-3))/2.
a(n) = 2*floor(n/3) + floor((n+1)/3) + floor((n+2)/3).
a(n) = (4*n - n mod 3)/3.
a(n) = a(n-3) + 4.
a(n) = a(n-1) + a(n-3) - a(n-4).
a(n) = 4*A002264(n) + A010872(n).
a(n) = A042968(n+1) - 1.
(End)

A130482 a(n) = Sum_{k=0..n} (k mod 4) (Partial sums of A010873).

Original entry on oeis.org

0, 1, 3, 6, 6, 7, 9, 12, 12, 13, 15, 18, 18, 19, 21, 24, 24, 25, 27, 30, 30, 31, 33, 36, 36, 37, 39, 42, 42, 43, 45, 48, 48, 49, 51, 54, 54, 55, 57, 60, 60, 61, 63, 66, 66, 67, 69, 72, 72, 73, 75, 78, 78, 79, 81, 84, 84, 85, 87, 90, 90, 91, 93, 96, 96, 97, 99, 102, 102, 103, 105

Offset: 0

Hieronymus Fischer, May 29 2007

Let A be the Hessenberg n X n matrix defined by: A[1,j]=j mod 4, A[i,i]:=1, A[i,i-1]=-1. Then, for n>=1, a(n)=det(A). - Milan Janjic, Jan 24 2010

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

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

a:=[0,1,3,6,6];; for n in [6..71] do a[n]:=a[n-1]+a[n-4]-a[n-5]; od; a; # G. C. Greubel, Aug 31 2019

I:=[0,1,3,6,6]; [n le 5 select I[n] else Self(n-1) + Self(n-4) - Self(n-5): n in [1..71]]; // G. C. Greubel, Aug 31 2019

a:=n->add(chrem( [n,j], [1,4] ),j=1..n):seq(a(n), n=0..70); # Zerinvary Lajos, Apr 07 2009

Table[(6*n +(1-(-1)^n)*(1+2*I^(n+1)))/4, {n,0,70}] (* G. C. Greubel, Aug 31 2019 *)
LinearRecurrence[{1,0,0,1,-1},{0,1,3,6,6},80] (* Harvey P. Dale, Feb 16 2024 *)

a(n) = (1 - (-1)^n - (2*I)*(-I)^n + (2*I)*I^n + 6*n) / 4 \\ Colin Barker, Oct 15 2015

def A130482_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P(x*(1+2*x+3*x^2)/((1-x^4)*(1-x))).list()
A130482_list(70) # G. C. Greubel, Aug 31 2019

a(n) = 6*floor(n/4) + A010873(n)*(A010873(n)+1)/2.
G.f.: x*(1 + 2*x + 3*x^2)/((1-x^4)*(1-x)).
a(n) = (1 - (-1)^n - (2*i)*(-i)^n + (2*i)*i^n + 6*n) / 4 where i = sqrt(-1). - Colin Barker, Oct 15 2015
a(n) = 3*n/2 + (n mod 2)* ( (n-1) mod 4 ) - (n mod 2)/2. - Ammar Khatab, Aug 27 2020
E.g.f.: (3*x*exp(x) - 2*sin(x) + sinh(x))/2. - Stefano Spezia, Apr 22 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/(4*sqrt(3)) + log(3)/4. - 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

A130519 a(n) = Sum_{k=0..n} floor(k/4). (Partial sums of A002265.)

Original entry on oeis.org

0, 0, 0, 0, 1, 2, 3, 4, 6, 8, 10, 12, 15, 18, 21, 24, 28, 32, 36, 40, 45, 50, 55, 60, 66, 72, 78, 84, 91, 98, 105, 112, 120, 128, 136, 144, 153, 162, 171, 180, 190, 200, 210, 220, 231, 242, 253, 264, 276, 288, 300, 312, 325, 338, 351, 364, 378, 392, 406, 420, 435, 450

Offset: 0

Hieronymus Fischer, Jun 01 2007

Complementary to A130482 with respect to triangular numbers, in that A130482(n) + 4*a(n) = n(n+1)/2 = A000217(n).
Disregarding the first three 0's the resulting sequence a'(n) is the sum of the positive integers <= n that have the same residue modulo 4 as n. This is the additive counterpart of the quadruple factorial numbers. - Peter Luschny, Jul 06 2011
From Heinrich Ludwig, Dec 23 2017: (Start)
Column sums of (shift of rows = 4):
  1 2 3 4 5 6  7  8  9 10 11 12 13 14 ...
          1 2  3  4  5  6  7  8  9 10 ...
                     1  2  3  4  5  6 ...
                                 1  2 ...
  .......................................
  ---------------------------------------
  1 2 3 4 6 8 10 12 15 18 21 24 28 32 ...
  shift of rows = 1 see A000217
  shift of rows = 2 see A002620
  shift of rows = 3 see A001840
  shift of rows = 5 see A130520
(End)
Conjecture: a(n+2) is the maximum effective weight of a numerical semigroup S of genus n (see Nathan Pflueger). - Stefano Spezia, Jan 04 2019

			G.f. = x^4 + 2*x^5 + 3*x^6 + 4*x^7 + 6*x^8 + 8*x^9 + 10*x^10 + 12*x^11 + ...
[ n] a(n)
---------
[ 4] 1
[ 5] 2
[ 6] 3
[ 7] 4
[ 8] 1 + 5
[ 9] 2 + 6
[10] 3 + 7
[11] 4 + 8

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Bakir Farhi, On the Representation of the Natural Numbers as the Sum of Three Terms of the Sequence floor(n^2/a), Journal of Integer Sequences, Vol. 16 (2013), Article 13.6.4.
Darren Glass and Joshua Wagner, Arithmetical Structures on Paths With a Doubled Edge, arXiv:1903.01398 [math.CO], 2019.
Nathan Pflueger, On non-primitive Weierstrass points, Alg. Number Th. 12 (2018) 1923-1947 and arXiv:1608.0566 [math.AG], 2016.
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,1,-2,1).

Cf. A000217, A001840, A002264, A002265, A002266, A002620, A004526, A010872, A010873, A010874, A130481, A130483, A130520.

Cf. A000290, A007590, A000212, A118015, A056827, A056834, A056838, A056865.

a:=List([0..65],n->Sum([0..n],k->Int(k/4)));; Print(a); # Muniru A Asiru, Jan 04 2019

[Round(n*(n-2)/8): n in [0..70]]; // Vincenzo Librandi, Jun 25 2011

quadsum := n -> add(k, k = select(k -> k mod 4 = n mod 4, [$1 .. n])):
A130519 := n ->`if`(n<3,0,quadsum(n-3)); seq(A130519(n),n=0..58); # Peter Luschny, Jul 06 2011

a[ n_] := Quotient[ (n - 1)^2, 8]; (* Michael Somos, Oct 14 2011 *)

makelist(floor((n-1)^2/8), n, 0, 70); /* Stefano Spezia, Jan 04 2019 */

{a(n) = (n - 1)^2 \ 8}; /* Michael Somos, Oct 14 2011 */

def A130519(n): return (n-1)**2>>3  # Chai Wah Wu, Jul 30 2022

G.f.: x^4/((1-x^4)*(1-x)^2) = x^4/((1+x)*(1+x^2)*(1-x)^3).
a(n) = +2*a(n-1) -1*a(n-2) +1*a(n-4) -2*a(n-5) +1*a(n-6).
a(n) = floor(n/4)*(n - 1 - 2*floor(n/4)) = A002265(n)*(n - 1 - 2*A002265(n)).
a(n) = (1/2)*A002265(n)*(n - 2 + A010873(n)).
a(n) = floor((n-1)^2/8). - Mitch Harris, Sep 08 2008
a(n) = round(n*(n-2)/8) = round((n^2-2*n-1)/8) = ceiling((n+1)*(n-3)/8). - Mircea Merca, Nov 28 2010
a(n) = A001972(n-4), n>3. - Franklin T. Adams-Watters, Jul 10 2009
a(n) = a(n-4)+n-3, n>3. - Mircea Merca, Nov 28 2010
Euler transform of length 4 sequence [ 2, 0, 0, 1]. - Michael Somos, Oct 14 2011
a(n) = a(2-n) for all n in Z. - Michael Somos, Oct 14 2011
a(n) = A214734(n, 1, 4). - Renzo Benedetti, Aug 27 2012
a(4n) = A000384(n), a(4n+1) = A001105(n), a(4n+2) = A014105(n), a(4n+3) = A046092(n). - Philippe Deléham, Mar 26 2013
a(n) = Sum_{i=1..ceiling(n/2)-1} (i mod 2) * (n - 2*i - 1). - Wesley Ivan Hurt, Jan 23 2014
a(n) = ( 2*n^2-4*n-1+(-1)^n+2*((-1)^((2*n-1+(-1)^n)/4)-(-1)^((6*n-1+(-1)^n)/4)) )/16 = ( 2*n*(n-2) - (1-(-1)^n)*(1-2*i^(n*(n-1))) )/16, where i=sqrt(-1). - Luce ETIENNE, Aug 29 2014
E.g.f.: (1/8)*((- 1 + x)*x*cosh(x) + 2*sin(x) + (- 1 - x + x^2)*sinh(x)). - Stefano Spezia, Jan 15 2019
a(n) = (A002620(n-1) - A011765(n+1)) / 2, for n > 0. - Yuchun Ji, Feb 05 2021
Sum_{n>=4} 1/a(n) = Pi^2/12 + 5/2. - Amiram Eldar, Aug 13 2022

Partially edited by R. J. Mathar, Jul 11 2009

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

Original entry on oeis.org

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

cp's OEIS Frontend

A010875 a(n) = n mod 6.

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A244477 a(1)=3, a(2)=2, a(3)=1; thereafter a(n) = a(n-a(n-1)) + a(n-a(n-2)).

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A052223 Numbers whose sum of digits is 9.

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A010878 a(n) = n mod 9.

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A039701 a(n) = n-th prime modulo 3.

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A004773 Numbers congruent to {0, 1, 2} mod 4: a(n) = floor(4*n/3).

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