A118015 -id:A118015 - cp's OEIS frontend

A147874 a(n) = (5n-7)(n-1).

0, 3, 16, 39, 72, 115, 168, 231, 304, 387, 480, 583, 696, 819, 952, 1095, 1248, 1411, 1584, 1767, 1960, 2163, 2376, 2599, 2832, 3075, 3328, 3591, 3864, 4147, 4440, 4743, 5056, 5379, 5712, 6055, 6408, 6771, 7144, 7527, 7920, 8323, 8736, 9159, 9592, 10035

Offset: 1

Vladimir Joseph Stephan Orlovsky, Nov 16 2008

Zero followed by partial sums of A017305.
Appears to be related to various other sequences: a(n) = A036666(2*n-2) for n>1; a(n) = A115006(2*n-3) for n>1; a(n) = A118015(5*n-6) for n>1; a(n) = A008738(5*n-7) for n>1.
Even dodecagonal numbers divided by 4. - Omar E. Pol, Aug 19 2011

Vincenzo Librandi, Table of n, a(n) for n = 1..2000
Leo Tavares, Illustration: Triangular Badges
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A017305 (10n+3), A036666, A115006, A118015 (floor(n^2/5)), A008738 (floor((n^2+1)/5)), A294830.
Cf. A051624, A193872. - Omar E. Pol, Aug 19 2011
Cf. A014105, A002378.

List([1..50], n-> (5*n-7)*(n-1)); # G. C. Greubel, Jul 30 2019

[ 0 ] cat [ &+[ 10*k+3: k in [0..n-1] ]: n in [1..50] ]; // Klaus Brockhaus, Nov 17 2008

Magma
```
[ 5*n^2-2*n: n in [0..50] ];
```

s=0;lst={s};Do[s+=n++ +3;AppendTo[lst,s],{n,0,6!,10}];lst
Table[5n^2-12n+7,{n,50}] (* or *) LinearRecurrence[{3,-3,1},{0,3,16},50] (* or *) PolygonalNumber[12,Range[0,100,2]]/4 (* Harvey P. Dale, Aug 08 2021 *)

{m=50; a=7; for(n=0, m, print1(a=a+10*(n-1)+3, ","))} \\ Klaus Brockhaus, Nov 17 2008

[(5*n-7)*(n-1) for n in (1..50)] # G. C. Greubel, Jul 30 2019

a(n) = Sum_{k=0..n-2} 10*k+3 = Sum_{k=0..n-2} A017305(k).
G.f.: x*(3 + 7*x)/(1-x)^3.
a(n) = 10*(n-2) + 3 + a(n-1).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
a(n) = A193872(n-1)/4. - Omar E. Pol, Aug 19 2011
a(n+1) = A131242(10n+2). - Philippe Deléham, Mar 27 2013
E.g.f.: -7 + (7 - 7*x + 5*x^2)*exp(x). - G. C. Greubel, Jul 30 2019
Sum_{n>=2} 1/a(n) = A294830. - Amiram Eldar, Nov 15 2020
a(n) = A014105(n-1) + 3*A002378(n-2). - Leo Tavares, Mar 31 2025

Edited by R. J. Mathar and Klaus Brockhaus, Nov 17 2008, Nov 20 2008

A056838 a(n) = floor(n^2/9).

Original entry on oeis.org

0, 0, 0, 1, 1, 2, 4, 5, 7, 9, 11, 13, 16, 18, 21, 25, 28, 32, 36, 40, 44, 49, 53, 58, 64, 69, 75, 81, 87, 93, 100, 106, 113, 121, 128, 136, 144, 152, 160, 169, 177, 186, 196, 205, 215, 225, 235, 245, 256, 266, 277, 289, 300, 312, 324, 336, 348

Offset: 0

N. J. A. Sloane, Sep 02 2000

Paolo Xausa, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,0,0,0,0,1,-2,1).

Cf. A008740.

Cf. A000290, A007590, A000212, A002620, A118015, A056827, A056834, A130519, A056865.

Floor[Range[0, 100]^2/9] (* Paolo Xausa, Aug 21 2024 *)

G.f.: x^3*(1+x)*(x^2-x+1)^2/((1-x)^3*(1+x+x^2)(x^6+x^3+1)). [R. J. Mathar, Jan 05 2009]

A118013 Triangle read by rows: T(n,k) = floor(n^2/k), 1<=k<=n.

Original entry on oeis.org

1, 4, 2, 9, 4, 3, 16, 8, 5, 4, 25, 12, 8, 6, 5, 36, 18, 12, 9, 7, 6, 49, 24, 16, 12, 9, 8, 7, 64, 32, 21, 16, 12, 10, 9, 8, 81, 40, 27, 20, 16, 13, 11, 10, 9, 100, 50, 33, 25, 20, 16, 14, 12, 11, 10, 121, 60, 40, 30, 24, 20, 17, 15, 13, 12, 11, 144, 72, 48, 36, 28, 24, 20, 18, 16, 14

Offset: 1

Reinhard Zumkeller, Apr 10 2006

T(n,1) = A000290(n); T(n,n) = n;
T(n,2) = A007590(n) for n>1;
T(n,3) = A000212(n) for n>2;
T(n,4) = A002620(n) for n>3;
T(n,5) = A118015(n) for n>4;
T(n,6) = A056827(n) for n>5;
central terms give A008574: T(2*k-1,k) = 4*(k-1)+0^(k-1);
row sums give A118014.

			Triangle begins:
1,
4, 2,
9, 4, 3,
16, 8, 5, 4,

Reinhard Zumkeller, Rows n=1..100 of triangle, flattened

Cf. A010766.

a118013 n k = a118013_tabl !! (n-1) !! (k-1)
a118013_row n = map (div (n^2)) [1..n]
a118013_tabl = map a118013_row [1..]
-- Reinhard Zumkeller, Jan 22 2012

T(n,k)=n^2\k \\ Charles R Greathouse IV, Jan 15 2012

A277646 Triangle T(n,k) = floor(n^2/k) for 1 <= k <= n^2, read by rows.

Original entry on oeis.org

1, 4, 2, 1, 1, 9, 4, 3, 2, 1, 1, 1, 1, 1, 16, 8, 5, 4, 3, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 25, 12, 8, 6, 5, 4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 36, 18, 12, 9, 7, 6, 5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 49, 24, 16, 12, 9, 8, 7, 6

Offset: 1

Jason Kimberley, Nov 09 2016

			The first five rows of the triangle are:
1;
4, 2, 1, 1;
9, 4, 3, 2, 1, 1, 1, 1, 1;
16, 8, 5, 4, 3, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1;
25, 12, 8, 6, 5, 4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;

Jason Kimberley, Table of n, a(n) for n = 1..10416 (the first 31 rows of the triangle)

Cf. Related triangles: A010766, A277647, A277648.
Rows of this triangle (with infinite trailing zeros):
T(1,k) = A000007(k-1),
T(2,k) = A033324(k),
T(3,k) = A033329(k),
T(4,k) = A033336(k),
T(5,k) = A033345(k),
T(6,k) = A033356(k),
T(7,k) = A033369(k),
T(8,k) = A033384(k),
T(9,k) = A033401(k),
T(10,k) = A033420(k),
T(100,k) = A033422(k),
T(10^3,k) = A033426(k),
T(10^4,k) = A033424(k).
Columns of this triangle:
T(n,1) = A000290(n),
T(n,2) = A007590(n),
T(n,3) = A000212(n),
T(n,4) = A002620(n),
T(n,5) = A118015(n),
T(n,6) = A056827(n),
T(n,7) = A056834(n),
T(n,8) = A130519(n+1),
T(n,9) = A056838(n),
T(n,10)= A056865(n),
T(n,12)= A174709(n+2).

A277646:=func;
[A277646(n,k):k in[1..n^2],n in[1..7]];

Table[Floor[n^2/k], {n, 7}, {k, n^2}] // Flatten (* Michael De Vlieger, Nov 24 2016 *)

T(n,k) = A010766(n^2,k).

A194222 a(n) = floor(Sum_{k=1..n} frac(k/5)), where frac() = fractional part.

Original entry on oeis.org

0, 0, 1, 2, 2, 2, 2, 3, 4, 4, 4, 4, 5, 6, 6, 6, 6, 7, 8, 8, 8, 8, 9, 10, 10, 10, 10, 11, 12, 12, 12, 12, 13, 14, 14, 14, 14, 15, 16, 16, 16, 16, 17, 18, 18, 18, 18, 19, 20, 20, 20, 20, 21, 22, 22, 22, 22, 23, 24, 24, 24, 24, 25, 26, 26, 26, 26, 27, 28, 28, 28, 28, 29, 30

Offset: 1

Clark Kimberling, Aug 19 2011

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

Cf. A118015.

seq(floor((n+1)/5)+floor((n+2)/5), n=1..80); # Ridouane Oudra, Dec 14 2021

r = 1/5;
a[n_] := Floor[Sum[FractionalPart[k*r], {k, 1, n}]]
Table[a[n], {n, 1, 90}]    (* A194222 *)
s[n_] := Sum[a[k], {k, 1, n}]
Table[s[n], {n, 1, 100}]   (* A118015 *)
LinearRecurrence[{1,0,0,0,1,-1},{0,0,1,2,2,2},80] (* Harvey P. Dale, Jun 06 2024 *)

From Chai Wah Wu, Jun 10 2020: (Start)
a(n) = a(n-1) + a(n-5) - a(n-6) for n > 6.
G.f.: x^3*(x + 1)/((x-1)^2*(1+x+x^2+x^3+x^4)). (End)
a(n) = floor((n+1)/5) + floor((n+2)/5). - Ridouane Oudra, Dec 14 2021
a(n) = A002266(n+1)+A002266(n+2). - R. J. Mathar, Nov 21 2023

A279169 a(n) = floor( 4*n^2/5 ).

Original entry on oeis.org

0, 0, 3, 7, 12, 20, 28, 39, 51, 64, 80, 96, 115, 135, 156, 180, 204, 231, 259, 288, 320, 352, 387, 423, 460, 500, 540, 583, 627, 672, 720, 768, 819, 871, 924, 980, 1036, 1095, 1155, 1216, 1280, 1344, 1411, 1479, 1548, 1620, 1692, 1767, 1843, 1920, 2000, 2080, 2163, 2247

Offset: 0

Bruno Berselli, Dec 07 2016

Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,1,-2,1).

Cf. A090223: floor(4*n/5).
Subsequence of A008728, A014601, A118015, A131242.
Cf. similar sequences with closed form floor(k*n^2/5): A118015 (k=1), A033437 (k=2), A184535 (k=3).

Magma
```
[4*n^2 div 5: n in [0..60]];
```

Table[Floor[4 n^2/5], {n, 0, 60}]
LinearRecurrence[{2,-1,0,0,1,-2,1},{0,0,3,7,12,20,28},60] (* Harvey P. Dale, Nov 07 2020 *)

PARI
```
vector(60, n, n--; floor(4*n^2/5))
```
Python
```
[int(4*n**2/5) for n in range(60)]
```
Sage
```
[floor(4*n^2/5) for n in range(60)]
```

O.g.f.: x^2*(3 + x + x^2 + 3*x^3)/((1 - x)^3*(1 + x + x^2 + x^3 + x^4)).
a(n) = a(-n) = 2*a(n-1) - a(n-2) + a(n-5) - 2*a(n-6) + a(n-7).
a(5*m+r) = 4*m*(5*m + 2*r) + a(r), where m >= 0 and 0 <= r < 5. Example: for m=4 and r=3, a(5*4+3) = a(23) = 4*4*(5*4 + 2*3) + a(3) = 416 + 7 = 423.
a(n) = A118015(2*n) = A008728(4*n+2) = A131242(4*n+4) = A014601(floor(2*n^2/5)).
Sum_{n>=2} 1/a(n) = Pi^2/120 + sqrt(29 - 62/sqrt(5))*Pi/8 + 5/16. - Amiram Eldar, Sep 26 2022

A194200 [sum{(k*e) : 1<=k<=n}], where [ ]=floor, ( )=fractional part.

Original entry on oeis.org

0, 1, 1, 2, 2, 3, 3, 3, 4, 4, 5, 6, 6, 6, 7, 7, 7, 8, 9, 9, 9, 10, 11, 11, 12, 13, 13, 13, 14, 15, 15, 16, 16, 17, 17, 18, 18, 19, 19, 19, 20, 20, 21, 22, 22, 22, 23, 23, 23, 24, 25, 25, 25, 26, 27, 27, 28, 28, 29, 29, 30, 30, 31, 32, 32, 33, 33, 34, 34, 34, 35, 36, 37

Offset: 1

Clark Kimberling, Aug 19 2011

nonn

The defining [sum] is equivalent to
...
a(n)=[n(n+1)r/2]-sum{[k*r] : 1<=k<=n},
...
where []=floor and r=sqrt(2).  Let s(n) denote the n-th partial sum of the sequence a; then the difference sequence d defined by d(n)=s(n+1)-s(n) gives the runlengths of a.
...
Examples:
...
r...........a........s....
1/2......A002265...A001972
1/3......A002264...A001840
2/3......A002264...A001840
1/4......A194220...A194221
1/5......A194222...A118015
2/5......A057354...A011858
3/5......A194222...A118015
4/5......A057354...A011858
1/6......A194223...A194224
3/7......A057357...A194229
1/8......A194235...A194236
3/8......A194237...A194238
sqrt(2)..A194161...A194162
sqrt(3)..A194163...A194164
sqrt(5)..A194169...A194170
sqrt(6)..A194173...A194174
tau......A194165...A194166; tau=(1+sqrt(5))/2
e........A194200...A194201
2e.......A194202...A194203
e/2......A194204...A194205
pi.......A194206...A194207

			a(5)=[(e)+(2e)+(3e)+4(e)+5(e)]
    =[.718+.436+.154+.873+.591]
    =[2.77423]=2.

G. C. Greubel, Table of n, a(n) for n = 1..5000

Cf. A194201.

r = E;
a[n_] := Floor[Sum[FractionalPart[k*r], {k, 1, n}]]
Table[a[n], {n, 1, 90}]  (* A194200 *)
s[n_] := Sum[a[k], {k, 1, n}]
Table[s[n], {n, 1, 100}] (* A194201 *)

A126696 Tenth-squares: floor(n/10)*ceiling(n/10).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 4, 6, 6, 6, 6, 6, 6, 6, 6, 6, 9, 12, 12, 12, 12, 12, 12, 12, 12, 12, 16, 20, 20, 20, 20, 20, 20, 20, 20, 20, 25, 30, 30, 30, 30, 30, 30, 30, 30, 30, 36, 42, 42, 42, 42, 42, 42, 42, 42, 42, 49, 56, 56, 56, 56, 56, 56, 56, 56, 56

Offset: 0

Jonathan Vos Post, May 27 2007

Cf. A002620, A024206, A072280, A002984, A007590, A000212, A118015, A056827, A118013.

[ Floor(n/10)*Ceiling(n/10) : n in [0..100]];

f[n_]:=Module[{c=n/10},Floor[c]Ceiling[c]];f[Range[0,90]] (* Harvey P. Dale, Apr 04 2011 *)

Equivalently, floor(n^2/100).

cp's OEIS Frontend

A147874 a(n) = (5*n-7)*(n-1).

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A056838 a(n) = floor(n^2/9).

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A118013 Triangle read by rows: T(n,k) = floor(n^2/k), 1<=k<=n.

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A277646 Triangle T(n,k) = floor(n^2/k) for 1 <= k <= n^2, read by rows.

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A194222 a(n) = floor(Sum_{k=1..n} frac(k/5)), where frac() = fractional part.

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A279169 a(n) = floor( 4*n^2/5 ).

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A194200 [sum{(k*e) : 1<=k<=n}], where [ ]=floor, ( )=fractional part.

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A147874 a(n) = (5n-7)(n-1).