A130482 -id:A130482 - cp's OEIS frontend

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

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

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)

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).

A010880 a(n) = n mod 11.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

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

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

PadRight[{},80,Range[0,10]] (* Harvey P. Dale, Nov 13 2012 *)

a(n)=n%11 \\ Charles R Greathouse IV, Oct 07 2015

[power_mod(n,11,11) for n in range(0, 78)] # Zerinvary Lajos, Nov 07 2009

From Hieronymus Fischer, Sep 30 2007: (Start)
a(n) = (1/11)*(1-r^n)*sum{1<=k<11, k*product{1<=m<11,m<>k, (1-r^(n-m))}} where r=exp(2*Pi/11*i) and i=sqrt(-1).
a(n) = (1024/11)^2*(sin(n*Pi/11))^2*sum{1<=k<11, k*product{1<=m<11,m<>k, (sin((n-m)*Pi/11))^2}}.
G.f.: (sum{1<=k<11, k*x^k})/(1-x^11).
G.f.: x*(10*x^11-11*x^10+1)/((1-x^11)*(1-x)^2). (End)

More terms from Correction. Typo at the sum formula for the g.f.: the summation index should not read "1<=k<10" but "1<=k<11" (see corrected formula).

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

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).

A010883 Simple periodic sequence: repeat 1,2,3,4.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

Partial sums are given by A130482(n) + n + 1. - Hieronymus Fischer, Jun 08 2007

1234/9999 = 0.123412341234... - Eric Desbiaux, Nov 03 2008

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

Cf. A010872, A010873, A010874, A010875, A010876, A004526, A002264, A002265, A002266.

Cf. A177037 (decimal expansion of (9+2*sqrt(39))/15). - Klaus Brockhaus, May 01 2010

PadRight[{},120,{1,2,3,4}] (* Harvey P. Dale, Aug 02 2016 *)

a(n)=(n-1)%4+1 \\ Charles R Greathouse IV, Jun 11 2015

def A010883(n): return 1 + (n & 3) # Chai Wah Wu, May 25 2022

a(n) = 1 + (n mod 4). - Paolo P. Lava, Nov 21 2006
From Hieronymus Fischer, Jun 08 2007: (Start)
a(n) = A010873(n) + 1.
Also a(n) = (1/2)*(5 - (-1)^n - 2*(-1)^((2*n - 1 + (-1)^n)/4)).
G.f.: g(x) = (4*x^3 + 3*x^2 + 2*x + 1)/(1 - x^4) = (4*x^5 - 5*x^4 + 1)/((1 - x^4)*(1-x)^2). (End)
a(n) = 5/2 - cos(Pi*n/2) - sin(Pi*n/2) - (-1)^n/2. - R. J. Mathar, Oct 08 2011

cp's OEIS Frontend

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

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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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