A130481 -id:A130481 - cp's OEIS frontend

A130488 a(n) = Sum_{k=0..n} (k mod 10) (Partial sums of A010879).

0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 45, 46, 48, 51, 55, 60, 66, 73, 81, 90, 90, 91, 93, 96, 100, 105, 111, 118, 126, 135, 135, 136, 138, 141, 145, 150, 156, 163, 171, 180, 180, 181, 183, 186, 190, 195, 201, 208, 216, 225, 225, 226, 228, 231, 235, 240, 246, 253

Offset: 0

Hieronymus Fischer, May 31 2007

nonn

Let A be the Hessenberg n X n matrix defined by A[1,j] = j mod 10, 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 (1,0,0,0,0,0,0,0,0,1,-1).

Cf. A010872, A010873, A010874, A010875, A010876, A010877, A010878, A130481, A130482, A130483, A130484, A130485, A130486, A130487.

a:=[0,1,3,6,10,15,21,28,36,45,45];; for n in [12..61] do a[n]:=a[n-1]+a[n-10]-a[n-11]; od; a; # G. C. Greubel, Aug 31 2019

I:=[0,1,3,6,10,15,21,28,36,45,45]; [n le 11 select I[n] else Self(n-1) + Self(n-10) - Self(n-11): n in [1..61]]; // G. C. Greubel, Aug 31 2019

seq(coeff(series(x*(1-10*x^9+9*x^10)/((1-x^10)*(1-x)^3), x, n+1), x, n), n = 0..60); # G. C. Greubel, Aug 31 2019

LinearRecurrence[{1,0,0,0,0,0,0,0,0,1,-1}, {0,1,3,6,10,15,21,28,36,45, 45}, 60] (* G. C. Greubel, Aug 31 2019 *)

a(n) = sum(k=0, n, k % 10); \\ Michel Marcus, Apr 28 2018

def A130488(n):
    a, b = divmod(n,10)
    return 45*a+(b*(b+1)>>1) # Chai Wah Wu, Jul 27 2022

def A130488_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P(x*(1-10*x^9+9*x^10)/((1-x^10)*(1-x)^3)).list()
A130488_list(60) # G. C. Greubel, Aug 31 2019

a(n) = 45*floor(n/10) + A010879(n)*(A010879(n) + 1)/2.
G.f.: (Sum_{k=1..9} k*x^k)/((1-x^10)*(1-x)).
G.f.: x*(1 - 10*x^9 + 9*x^10)/((1-x^10)*(1-x)^3).

A130909 Simple periodic sequence (n mod 16).

Original entry on oeis.org

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

Offset: 0

Hieronymus Fischer, Jun 11 2007

The value of the rightmost digit in the base-16 representation of n. Also, the equivalent value of the two rightmost digits in the base-4 representation of n. Also, the equivalent value of the four rightmost digits in the base-2 representation of n.

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

Cf. partial sums A130910. Other related sequences A010872, A010873, A130481, A130482, A130483, A130486.

See A010877 for a general formula in terms of powers of -1 (for period 2^k).

a(n)=n%16 \\ Charles R Greathouse IV, Jul 13 2016

def A130909(n): return n&15 # Chai Wah Wu, Jan 18 2023

a(n) = n mod 16 = n-16*floor(n/16).
G.f.: g(x) = (Sum_{k=1..15} k*x^k)/(1-x^16).
G.f.: g(x) = x(15x^16-16x^15+1)/((1-x^16)(1-x)^2).
a(n) = A000035(n) + 2*A010877(A004526(n)).
a(n) = A010873(n) + 4*A010873(A002265(n)).
a(n) = A010877(n) + 8*A000035(floor(n/8)).
a(n) = (1/2)*(15 - ( - 1)^n - 2*( - 1)^(b/4) - 4*( - 1)^((b - 2 + 2*( - 1)^(b/4))/8) - 8*( - 1)^((b - 6 + ( - 1)^n + 2*( - 1)^(b/4) + 4*( - 1)^((b - 2 + 2*( - 1)^(b/4))/8))/16)) where b = 2n - 1 + ( - 1)^n.
a(n) = n mod 2+2*(floor(n/2)mod 2)+4*(floor(n/4)mod 2)+8*(floor(n/8)mod 2).
a(n) = (1/2)*(15-(-1)^n-2*(-1)^floor(n/2)-4*(-1)^floor(n/4)-8*(-1)^floor(n/= 8)).
Complex representation: a(n) = (1/16)*(1-r^n)*sum{1<=k<16, k*product{1<=m<16,m<>k, (1-r^(n-m))}} where r=exp(Pi/8*i)=(sqrt(2+sqrt(2))+i*sqrt(2-sqrt(2)))/2 and i=sqrt(-1).
Trigonometric representation: a(n) = 2^22*(sin(n*Pi/16))^2*sum{1<=k<16, k*product{1<=m<16,m<>k, (sin((n-m)*Pi/16))^2}}.
a(n) = (1/2)*(15-(-1)^p(0,n)-2*(-1)^p(1,n)-4*(-1)^p(2,n)-8*(-1)^p(3,n)) where p(k,n) is defined recursively by p(0,n)=n, p(k,n)=1/4*(2*p(k-1,n)-1+(-1)^p(k-1,n)).

A130489 a(n) = Sum_{k=0..n} (k mod 11) (Partial sums of A010880).

Original entry on oeis.org

0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 55, 56, 58, 61, 65, 70, 76, 83, 91, 100, 110, 110, 111, 113, 116, 120, 125, 131, 138, 146, 155, 165, 165, 166, 168, 171, 175, 180, 186, 193, 201, 210, 220, 220, 221, 223, 226, 230, 235, 241, 248, 256, 265, 275, 275, 276

Offset: 0

Hieronymus Fischer, May 31 2007

nonn

Let A be the Hessenberg n X n matrix defined by A[1,j] = j mod 11, 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 (1,0,0,0,0,0,0,0,0,0,1,-1).

Cf. A010872, A010873, A010874, A010875, A010876, A010877, A010878, A010879, A130481, A130482, A130483, A130484, A130485, A130486, A130487, A130488.

a:=[0,1,3,6,10,15,21,28,36,45, 55,55];; for n in [13..61] do a[n]:=a[n-1]+a[n-11]-a[n-12]; od; a; # G. C. Greubel, Aug 31 2019

I:=[0,1,3,6,10,15,21,28,36,45,55,55]; [n le 12 select I[n] else Self(n-1) + Self(n-11) - Self(n-12): n in [1..61]]; // G. C. Greubel, Aug 31 2019

seq(coeff(series(x*(1-11*x^10+10*x^11)/((1-x^11)*(1-x)^3), x, n+1), x, n), n = 0 .. 60); # G. C. Greubel, Aug 31 2019

LinearRecurrence[{1,0,0,0,0,0,0,0,0,0,1,-1}, {0,1,3,6,10,15,21,28,36,45, 55,55}, 60] (* G. C. Greubel, Aug 31 2019 *)
Accumulate[PadRight[{},80,Range[0,10]]] (* Harvey P. Dale, Jul 21 2021 *)

a(n) = sum(k=0, n, k % 11); \\ Michel Marcus, Apr 28 2018

def A130489_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P(x*(1-11*x^10+10*x^11)/((1-x^11)*(1-x)^3)).list()
A130489_list(60) # G. C. Greubel, Aug 31 2019

a(n) = 55*floor(n/11) + A010880(n)*(A010880(n) + 1)/2.
G.f.: (Sum_{k=1..10} k*x^k)/((1-x^11)*(1-x)).
G.f.: x*(1 - 11*x^10 + 10*x^11)/((1-x^11)*(1-x)^3).

A130490 a(n) = Sum_{k=0..n} (k mod 12) (Partial sums of A010881).

Original entry on oeis.org

0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 66, 67, 69, 72, 76, 81, 87, 94, 102, 111, 121, 132, 132, 133, 135, 138, 142, 147, 153, 160, 168, 177, 187, 198, 198, 199, 201, 204, 208, 213, 219, 226, 234, 243, 253, 264, 264, 265, 267, 270, 274, 279, 285, 292, 300

Offset: 0

Hieronymus Fischer, May 31 2007

nonn

Let A be the Hessenberg n X n matrix defined by: A[1,j] = j mod 12, 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 (1,0,0,0,0,0,0,0,0,0,0,1,-1).

Cf. A010872, A010873, A010874, A010875, A010876, A010877, A010878, A010879, A010880, A130481, A130482, A130483, A130484, A130485, A130486, A130487, A130488, A130489.

List([0..60], n-> Sum([0..n], k-> k mod 12 )); # G. C. Greubel, Sep 01 2019

[&+[(k mod 12): k in [0..n]]: n in [0..60]]; // G. C. Greubel, Sep 01 2019

seq(coeff(series(x*(1-12*x^11+11*x^12)/((1-x^12)*(1-x)^3), x, n+1), x, n), n = 0..60); # G. C. Greubel, Sep 01 2019

Sum[Mod[k, 12], {k, 0, Range[0, 60]}] (* G. C. Greubel, Sep 01 2019 *)
LinearRecurrence[{1,0,0,0,0,0,0,0,0,0,0,1,-1},{0,1,3,6,10,15,21,28,36,45,55,66,66},60] (* Harvey P. Dale, Jan 16 2024 *)

a(n) = sum(k=0, n, k % 12); \\ Michel Marcus, Apr 29 2018

[sum(k%12 for k in (0..n)) for n in (0..60)] # G. C. Greubel, Sep 01 2019

a(n) = 66*floor(n/12) + A010881(n)*(A010881(n) + 1)/2.
G.f.: (Sum_{k=1..11} k*x^k)/((1-x^12)*(1-x)).
G.f.: x*(1 - 12*x^11 + 11*x^12)/((1-x^12)*(1-x)^3).

A171452 a(n) = C(n,2) + floor(n/3).

Original entry on oeis.org

0, 0, 1, 4, 7, 11, 17, 23, 30, 39, 48, 58, 70, 82, 95, 110, 125, 141, 159, 177, 196, 217, 238, 260, 284, 308, 333, 360, 387, 415, 445, 475, 506, 539, 572, 606, 642, 678, 715, 754, 793, 833, 875, 917, 960, 1005, 1050, 1096, 1144, 1192, 1241, 1292, 1343, 1395

Offset: 0

Paul Barry, Dec 09 2009

Exponents in Hankel transform A171451.

For n>=2 a(n) is the smallest addend in the sums of n terms where all the natural numbers are used once 1+2=3, 4+5+6=15, 7+8+9+10=34, 11+12+13+14+16=66, 17+18+19+20+21+22=117 23+24+25+26+27+28+29=182, 30+31+32+33+35+36+37+38=272. - Anton Zakharov, Aug 28 2016

Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).

Cf. A130481 (first differences)

a(n)=n*(3*n-1)\6 \\ Charles R Greathouse IV, Aug 28 2016

G.f.: (1+2x)/((1-x)^3*(1+x+x^2));

a(n) = (3n^2-n-2)/6 + sqrt(3)*cos(2*Pi*n/3+Pi/6)/9 + sin(2*Pi*n/3+Pi/6)/3.

A092200 Expansion of (1+2x)/((1-x)(1-x^3)).

Original entry on oeis.org

1, 3, 3, 4, 6, 6, 7, 9, 9, 10, 12, 12, 13, 15, 15, 16, 18, 18, 19, 21, 21, 22, 24, 24, 25, 27, 27, 28, 30, 30, 31, 33, 33, 34, 36, 36, 37, 39, 39, 40, 42, 42, 43, 45, 45, 46, 48, 48, 49, 51, 51, 52, 54, 54, 55, 57, 57, 58, 60, 60, 61, 63, 63, 64, 66, 66, 67, 69, 69, 70, 72, 72

Offset: 0

Paul Barry, Feb 24 2004

Partial sums of A010872(n+1).

Essentially the same as A130481. - R. J. Mathar, Jun 13 2008

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

Cf. A010872, A130481.

a:=n->add(chrem( [n,j], [1,3] ),j=1..n):seq(a(n), n=1..72);# Zerinvary Lajos, Apr 08 2009

f[n_]:=Mod[n,3];s=0;lst={};Do[AppendTo[lst,s+=f[n]],{n,5!}];lst (* Vladimir Joseph Stephan Orlovsky, Feb 07 2010 *)
CoefficientList[Series[(1+2x)/((1-x)(1-x^3)),{x,0,80}],x] (* or *) LinearRecurrence[{1,0,1,-1},{1,3,3,4},81] (* Harvey P. Dale, Sep 15 2011 *)

G.f.: (1+2x)/(1-x-x^3+x^4);
a(n) = 4/3 + n + 2*cos(Pi*2(n-1)/3)/3;
a(n) = Sum_{k=0..n} (k+1) mod 3;
a(n) = (n+1)*(n+2)/2 - 3*Sum_{k=0..n} floor((k+1)/3);
a(n) = 1 + n + Sum_{k=0..n} Jacobi(k, 3).
a(n) = a(n-1) + a(n-3) - a(n-4); a(0)=1, a(1)=3, a(2)=3, a(3)=4. - Harvey P. Dale, Sep 15 2011
a(n) = n + 1 when n + 2 is not a multiple of 3, and a(n) = n + 2 when n + 2 is a multiple of 3. - Dennis P. Walsh, Aug 06 2012
Sum_{n>=0} (-1)^n/a(n) = Pi/(3*sqrt(3)) + log(2)/3. - Amiram Eldar, Feb 14 2023

A130910 Sum {0<=k<=n, k mod 16} (Partial sums of A130909).

Original entry on oeis.org

0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 120, 121, 123, 126, 130, 135, 141, 148, 156, 165, 175, 186, 198, 211, 225, 240, 240, 241, 243, 246, 250, 255, 261, 268, 276, 285, 295, 306, 318, 331, 345, 360, 360, 361, 363, 366, 370, 375, 381, 388

Offset: 0

Hieronymus Fischer, Jun 11 2007

nonn

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Cf. A130486, A010872, A010873, A010874, A010875, A010876, A010878, A130481, A130482, A130483, A130484, A130485, A130487.

Accumulate[Mod[Range[0,60],16]] (* Harvey P. Dale, May 30 2020 *)

a(n)=120*floor(n/16)+A130909(n)*(A130909(n)+1)/2. - G.f.: g(x)=(sum{1<=k<16, k*x^k})/((1-x^16)(1-x)). Also: g(x)=x(15x^16-16x^15+1)/((1-x^16)(1-x)^3).

a(n) = +a(n-1) +a(n-16) -a(n-17). G.f. ( x*(1 +2*x +3*x^2 +4*x^3 +5*x^4 +6*x^5 +7*x^6 +8*x^7 +9*x^8 +10*x^9 +11*x^10 +12*x^11 +13*x^12 +14*x^13 +15*x^14) ) / ( (1+x) *(1+x^2) *(1+x^4) *(1+x^8) *(x-1)^2 ). - R. J. Mathar, Nov 05 2011

A319053 a(n) is the exponent of the largest power of 2 that appears in the factorization of the entries in the matrix {{3,1},{1,-1}}^n.

Original entry on oeis.org

0, 1, 5, 3, 4, 8, 6, 7, 12, 9, 10, 15, 12, 13, 18, 15, 16, 20, 18, 19, 25, 21, 22, 28, 24, 25, 31, 27, 28, 32, 30, 31, 36, 33, 34, 39, 36, 37, 42, 39, 40, 44, 42, 43, 50, 45, 46, 53, 48, 49, 56, 51, 52, 56, 54, 55, 60, 57, 58, 63, 60, 61, 66, 63, 64, 68, 66, 67, 73, 69

Offset: 1

Greg Dresden, Sep 09 2018

nonn

a(n) appears to equal n-1 for n not a multiple of 3.
The matrix entries of M^n, with n >= 0, are M^n(1, 1) = 2^(n-1)*F(n+3) = A063782(n), M^n(2, 2) = 2^(n-1)*F(n-3) = A319196(n), M^n(1, 2) = M^n(2, 1) = 2^(n-1)*F(n) = A085449(n), where i = sqrt(-1), F = A000045, and F(-1) = 1, F(-2) = -1, F(-3) = 2. Proof by Cayley-Hamilton, with S(n, -i) = (-i)^n*F(n+1), where S(n, x) is given in A049310. - Wolfdieter Lang, Oct 08 2018
The above conjecture is true. From the preceding formulas for the elements of M^n this claims that the Fibonacci numbers F(n-3), F(n) and F(n+3) are always odd for n == 1 or 2 (mod 3). This is true because F(n) is even iff n == 0 (mod 3) (see e.g. Vajda, p.73), and each of the three indices is == 1 or 2 (mod 3) for n == 1 or 2 (mod 3), respectively. - Wolfdieter Lang, Oct 09 2018

			For n = 3, the matrix {{3,1},{1,-1}}^3 = {{32,8},{8,0}} and the largest power of 2 appearing in the factorization of any entry is 2^5 = 32. Hence, a(3) = 5.

S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989, p. 73.

Cf. A000045, A049310, A063782, A085449, A130481, A319196.

Join[{0, 1, 5}, Table[Max[ IntegerExponent[Flatten[MatrixPower[{{3, 1}, {1, -1}}, n]], 2]], {n, 4, 40}]]

a(n) = vecmax(apply(x->if (x, valuation(x, 2), 0), [3,1;1,-1]^n)); \\ Michel Marcus, Sep 09 2018

A268291 a(n) = Sum_{k = 0..n} (k mod 13).

Original entry on oeis.org

0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 78, 79, 81, 84, 88, 93, 99, 106, 114, 123, 133, 144, 156, 156, 157, 159, 162, 166, 171, 177, 184, 192, 201, 211, 222, 234, 234, 235, 237, 240, 244, 249, 255, 262, 270, 279, 289, 300, 312, 312, 313, 315, 318, 322, 327, 333, 340, 348

Offset: 0

Ilya Gutkovskiy, Jan 31 2016

More generally, the ordinary generating function for the Sum_{k = 0..n} (k mod m) is (Sum_{k = 1..(m - 1)} k*x^k)/((1 - x^m)*(1 - x)).

Sum_{k = 0..n} (k mod m) = m*(m - 1)/2 + Sum_{k = 1..(m - 1)} k*floor((n - k)/m), m>0.

			(see Extended example in Links section)
a(0)  = 0;
a(1)  = 0+1 = 1;
a(2)  = 0+1+2 = 3;
a(3)  = 0+1+2+3 = 6;
a(4)  = 0+1+2+3+4 = 10;
a(5)  = 0+1+2+3+4+5 = 15;
...
a(11) = 0+1+2+3+4+5+6+7+8+9+10+11 = 66;
a(12) = 0+1+2+3+4+5+6+7+8+9+10+11+12 = 78;
a(13) = 0+1+2+3+4+5+6+7+8+9+10+11+12+0 = 78;
a(14) = 0+1+2+3+4+5+6+7+8+9+10+11+12+0+1 = 79;
a(15) = 0+1+2+3+4+5+6+7+8+9+10+11+12+0+1+2 = 81, etc.

Shawn A. Broyles, Table of n, a(n) for n = 0..1000
Ilya Gutkovskiy, Extended example
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,0,0,0,1,-1).

Cf. A004526, A130481-A130490.

Table[Sum[Mod[k, 13], {k, 0, n}], {n, 0, 60}]
Table[Sum[k - 13 Floor[k/13], {k, 0, n}], {n, 0, 60}]
LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1}, {0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 78}, 61]
CoefficientList[Series[(x + 2 x^2 + 3 x^3 + 4 x^4 + 5 x^5 + 6 x^6 + 7 x^7 + 8 x^8 + 9 x^9 + 10 x^10 + 11 x^11 + 12 x^12) / ((1 - x^13) (1 - x)), {x, 0, 70}], x] (* Vincenzo Librandi, Jan 31 2016 *)
Accumulate[Mod[Range[0,60],13]] (* Harvey P. Dale, May 10 2021 *)

a(n) = sum(k = 0, n, k % 13); \\ Michel Marcus, Jan 31 2016

G.f.: (x + 2*x^2 + 3*x^3 + 4*x^4 + 5*x^5 + 6*x^6 + 7*x^7 + 8*x^8 + 9*x^9 + 10*x^10 + 11*x^11 + 12*x^12)/((1 - x^13)*(1 - x)).
a(n) = 12*floor((n - 12)/13) + 11*floor((n - 11)/13) + 10*floor((n - 10)/13) + 9*floor((n - 9)/13) + 8*floor((n - 8)/13) + 7*floor((n - 7)/13) + 6*floor((n - 6)/13) + 5*floor((n - 5)/13) + 4*floor((n - 4)/13) + 3*floor((n - 3)/13) + 2*floor((n - 2)/13) + floor((n - 1)/13) + 78.
a(n) = 6*n + r*(r-11)/2 where r = (n mod 13). - Hoang Xuan Thanh, Jun 02 2025

A244590 a(n) = sum( floor(k*n/8), k=1..7 ).

Original entry on oeis.org

0, 0, 4, 7, 12, 14, 18, 21, 28, 28, 32, 35, 40, 42, 46, 49, 56, 56, 60, 63, 68, 70, 74, 77, 84, 84, 88, 91, 96, 98, 102, 105, 112, 112, 116, 119, 124, 126, 130, 133, 140, 140, 144, 147, 152, 154, 158, 161, 168, 168

Offset: 0

Gary Detlefs, Jun 30 2014

This sequence is G(n,8) where G(n,m) = sum(floor(k*n/m), k=1..m-1). This function is referenced in A109004 and is used in the following formula for gcd(n,m): gcd(n,m) = n+m-n*m+2*G(n,m).
Listed sequences of this form are:
G(n,2) ... A004526;
G(3,n) ... A130481;
G(n,4) ... A187326;
G(n,5) ... A187333;
G(n,6) ... A187336;
G(n,7) ... A187337;
G(n,k*n)/k = n*(n-1)/2  = G(n,n+k)-G(n,k).
It is of interest to note that this alternate form of gcd(n,m) will be undefined if m is a function having a zero in it. For example, gcd(n, n mod 4) would be undefined but gcd(n mod 4, n) would be defined.

Cf. A109004.

[&+[Floor(k*n/8): k in [1..7]]: n in [0..50]]; // Bruno Berselli, Jul 01 2014

G:=(n,m)-> sum(floor(k*n/m), k=1..m-1): seq(G(n,8), n = 0..60);

Table[Sum[Floor[k n/8], {k, 1, 7}], {n, 0, 50}] (* Bruno Berselli, Jul 01 2014 *)

[sum(floor(k*n/8) for k in (1..7)) for n in (0..50)] # Bruno Berselli, Jul 01 2014

a(n) = sum( floor(k*n/8), k=1..7 ).
a(n) = ( gcd(n,8) - (n+8) + n*8 )/2.
G.f.: x^2*(4 + 3*x + 5*x^2 + 2*x^3 + 4*x^4 + 3*x^5 + 7*x^6)/((1 + x)*(1 - x)^2*(1 + x^2)*(1 + x^4)). [Bruno Berselli, Jul 01 2014]

Some terms corrected by Bruno Berselli, Jul 01 2014

cp's OEIS Frontend

A130488 a(n) = Sum_{k=0..n} (k mod 10) (Partial sums of A010879).

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A130909 Simple periodic sequence (n mod 16).

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A130489 a(n) = Sum_{k=0..n} (k mod 11) (Partial sums of A010880).

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A130490 a(n) = Sum_{k=0..n} (k mod 12) (Partial sums of A010881).

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A171452 a(n) = C(n,2) + floor(n/3).

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A092200 Expansion of (1+2x)/((1-x)(1-x^3)).

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A130910 Sum {0<=k<=n, k mod 16} (Partial sums of A130909).