A130482 -id:A130482 - 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).

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

A328943 a(n) = 2 + (n mod 4).

Original entry on oeis.org

2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3, 4, 5, 2, 3

Offset: 0

David Nacin, Oct 31 2019

nonn

Terms of the simple continued fraction of (36+sqrt(1806))/34.
2345/9999=0.234523452345...
Partial sums are given by A130482(n) + 2*n + 2.
Example of a sequence where the largest of any four consecutive terms equals the sum of the two smallest.

Michael De Vlieger, Table of n, a(n) for n = 0..10000
David Nacin, Van der Laan Sequences and a Conjecture on Padovan Numbers, J. Int. Seq., Vol. 26 (2023), Article 23.1.2.
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 1).

Cf. A010883, A010873, A130482

Mathematica
```
PadRight[{}, 120, {2, 3, 4, 5}]
```
Python
```
def a(n):
   return n%4+2
```

a(n) = 2 + (n mod 4).
G.f.: (5x^3 + 4x^2 + 3x + 2)/(1 - x^4).
a(n) = A010873(n) + 2 = A010883(n) + 1.
a(n) = 14 - a(n-1) - a(n-2) - a(n-3) for n > 2.

A244953 a(n) = Sum_{i=0..n} (-i mod 4).

Original entry on oeis.org

0, 3, 5, 6, 6, 9, 11, 12, 12, 15, 17, 18, 18, 21, 23, 24, 24, 27, 29, 30, 30, 33, 35, 36, 36, 39, 41, 42, 42, 45, 47, 48, 48, 51, 53, 54, 54, 57, 59, 60, 60, 63, 65, 66, 66, 69, 71, 72, 72, 75, 77, 78, 78, 81, 83, 84, 84, 87, 89, 90, 90, 93, 95, 96, 96, 99

Offset: 0

Wesley Ivan Hurt, Jul 08 2014

Partial sums of A158459.

Similar to A047271 with every third term repeated.

			To quickly generate terms of the sequence: start with zero for n=0, then add 3 more for n=1, then add 2 more for n=2, add 1 more..., then add 0..., and repeat.

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A158459. Same members as A047271. Similar to A130482.

[&+[-i mod 4: i in [0..n]]: n in [0..70]]; // Bruno Berselli, Jul 09 2014

A244953:=n->add(-i mod 4, i=0..n): seq(A244953(n), n=0..50);

Table[Sum[Mod[-i, 4], {i, 0, n}], {n, 0, 50}]
Table[1 + n + (2 (1 + n) - (1 + (-1)^n) (1 + 2 I^(n (n + 1))))/4, {n, 0, 70}] (* Bruno Berselli, Jul 09 2014 *)
LinearRecurrence[{1,0,0,1,-1},{0,3,5,6,6},70] (* Harvey P. Dale, Oct 29 2023 *)

a(n) = sum(i=0, n, -i % 4); \\ Michel Marcus, Jul 09 2014

a(n) = Sum_{i=0..n} A158459(i).
From Bruno Berselli, Jul 09 2014: (Start)
G.f.: (3 + 2*x + x^2)/((1 + x)*(1 - x)^2*(1 + x^2)).
a(n) = 1 + n + ( 2*(1 + n) - (1 + (-1)^n)*(1 + 2*i^(n*(n+1))) )/4, where i = sqrt(-1).
a(n) = 6 + Sum_{i=1..3}((4-i)*floor((n-i)/4)). (End)
a(n) = a(n-1) + a(n-4) - a(n-5). - Robert Israel, Jul 09 2014
a(n) = (3*n + 4 - (n mod 4 - 2)^2)/2. - Thomas Klemm, Aug 21 2022

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Python

Sage

Formula

A130909 Simple periodic sequence (n mod 16).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI

Python

Formula

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Sage

Formula

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Sage

Formula

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

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Formula

A328943 a(n) = 2 + (n mod 4).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Python

Formula

A244953 a(n) = Sum_{i=0..n} (-i mod 4).

Views