cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A290131 Number of regions in a regular drawing of the complete bipartite graph K_{n,n}.

Original entry on oeis.org

0, 2, 12, 40, 96, 204, 368, 634, 1012, 1544, 2236, 3186, 4360, 5898, 7764, 10022, 12712, 16026, 19844, 24448, 29708, 35756, 42604, 50602, 59496, 69650, 80940, 93600, 107540, 123316, 140428, 159642, 180632, 203618, 228556, 255822, 285080, 317326, 352020, 389498
Offset: 1

Views

Author

R. J. Mathar, Jul 20 2017

Keywords

Crossrefs

For K_n see A007569, A007678, A135563.
The following eight sequences are all essentially the same. The simplest is A115004(n), which we denote by z(n). Then A088658(n) = 4*z(n-1); A114043(n) = 2*z(n-1)+2*n^2-2*n+1; A114146(n) = 2*A114043(n); A115005(n) = z(n-1)+n*(n-1); A141255(n) = 2*z(n-1)+2*n*(n-1); A290131(n) = z(n-1)+(n-1)^2; A306302(n) = z(n)+n^2+2*n. - N. J. A. Sloane, Feb 04 2020

Programs

  • Maple
    A290131 := proc(n)
        A115004(n-1)+(n-1)^2 ;
    end proc:
    seq(A290131(n),n=1..30) ;
  • Mathematica
    z[n_] := Sum[(n - i + 1)(n - j + 1) Boole[GCD[i, j] == 1], {i, n}, {j, n}];
    a[n_] := z[n - 1] + (n - 1)^2;
    Array[a, 40] (* Jean-François Alcover, Mar 24 2020 *)
  • Python
    from math import gcd
    def a115004(n):
        r=0
        for a in range(1, n + 1):
            for b in range(1, n + 1):
                if gcd(a, b)==1:r+=(n + 1 - a)*(n + 1 - b)
        return r
    def a(n): return a115004(n - 1) + (n - 1)**2
    print([a(n) for n in range(1, 51)]) # Indranil Ghosh, Jul 20 2017, after Maple code
    
  • Python
    from sympy import totient
    def A290131(n): return 2*(n-1)**2 + sum(totient(i)*(n-i)*(2*n-i) for i in range(2,n)) # Chai Wah Wu, Aug 16 2021

Formula

a(n) = A115004(n-1) + (n-1)^2.
a(n) = 2*(n-1)^2 + Sum_{i=2..n-1} (n-i)*(2n-i)*phi(i). - Chai Wah Wu, Aug 16 2021

A114043 Take an n X n square grid of points in the plane; a(n) = number of ways to divide the points into two sets using a straight line.

Original entry on oeis.org

1, 7, 29, 87, 201, 419, 749, 1283, 2041, 3107, 4493, 6395, 8745, 11823, 15557, 20075, 25457, 32087, 39725, 48935, 59457, 71555, 85253, 101251, 119041, 139351, 161933, 187255, 215137, 246691, 280917, 319347, 361329, 407303
Offset: 1

Views

Author

Ugo Merlone (merlone(AT)econ.unito.it) and N. J. A. Sloane, Feb 22 2006

Keywords

Comments

Also, half of the number of two-dimensional threshold functions (A114146).
The line may not pass through any point. This is the "labeled" version - rotations and reflections are not taken into account (cf. A116696).
The number of ways to divide a (2n) X (2n) grid into two sets of equal size is given by 2*A099957(n). - David Applegate, Feb 23 2006
All terms are odd: the line that misses the grid contributes 1 to the total and all other lines contribute 2, 4 or 8, so the total must be odd.
What can be said about the 3-D generalization? - Max Alekseyev, Feb 27 2006

Examples

			Examples: the two sets are indicated by X's and o's.
a(2) = 7:
XX oX Xo XX XX oo oX
XX XX XX Xo oX XX oX
--------------------
a(3) = 29:
XXX oXX ooX ooo ooX ooo
XXX XXX XXX XXX oXX oXX
XXX XXX XXX XXX XXX XXX
-1- -4- -8- -4- -4- -8- Total = 29
--------------------
a(4)= 87:
XXXX XXXX XXXX XXXX XXXX XXXX XXXX XXXX XXXX XXXX
XXXX XXXX XXXX XXXX XXXX XXXX XXXX XXXX XXXX XXXX
XXXX XXXX XXXX XXXX XXXX XXXo XXXo XXXo XXoo XXoo
XXXX XXXo XXoo Xooo oooo XXoo Xooo oooo Xooo oooo
--1- --4- --8- --8- --4- --4- --8- --8- --8- --8-
XXXX XXXX XXXX XXXX XXXX
XXXo XXXX XXXX XXXo XXXo
XXoo Xooo oooo Xooo XXoo
Xooo oooo oooo oooo oooo
--4- --8- --2- --4- --8- Total = 87.
--------------------
		

Crossrefs

Cf. A114499, A115004, A115005, A116696 (unlabeled case), A114531, A114146.
Cf. A099957.
The following eight sequences are all essentially the same. The simplest is A115004(n), which we denote by z(n). Then A088658(n) = 4*z(n-1); A114043(n) = 2*z(n-1)+2*n^2-2*n+1; A114146(n) = 2*A114043(n); A115005(n) = z(n-1)+n*(n-1); A141255(n) = 2*z(n-1)+2*n*(n-1); A290131(n) = z(n-1)+(n-1)^2; A306302(n) = z(n)+n^2+2*n. - N. J. A. Sloane, Feb 04 2020

Programs

  • Mathematica
    a[n_] := 2*Sum[(n - i)*(n - j)*Boole[CoprimeQ[i, j]], {i, 1, n - 1}, {j, 1, n - 1}] + 2*n^2 - 2*n + 1; Array[a, 40] (* Jean-François Alcover, Apr 25 2016, after Max Alekseyev *)
  • Python
    from sympy import totient
    def A114043(n): return 4*n**2-6*n+3 + 2*sum(totient(i)*(n-i)*(2*n-i) for i in range(2,n)) # Chai Wah Wu, Aug 15 2021

Formula

Let V(m,n) = Sum_{i=1..m, j=1..n, gcd(i,j)=1} (m+1-i)*(n+1-j); then a(n+1) = 2*(n^2 + n + V(n,n)) + 1. - Max Alekseyev, Feb 22 2006
a(n) ~ (3/Pi^2) * n^4. - Max Alekseyev, Feb 22 2006
a(n) = A141255(n) + 1. - T. D. Noe, Jun 17 2008
a(n) = 4*n^2 - 6*n + 3 + 2*Sum_{i=2..n-1} (n-i)*(2n-i)*phi(i). - Chai Wah Wu, Aug 15 2021

Extensions

More terms from Max Alekseyev, Feb 22 2006

A141255 Total number of line segments between points visible to each other in a square n X n lattice.

Original entry on oeis.org

0, 6, 28, 86, 200, 418, 748, 1282, 2040, 3106, 4492, 6394, 8744, 11822, 15556, 20074, 25456, 32086, 39724, 48934, 59456, 71554, 85252, 101250, 119040, 139350, 161932, 187254, 215136, 246690, 280916, 319346, 361328, 407302, 457180, 511714, 570232
Offset: 1

Views

Author

T. D. Noe, Jun 17 2008

Keywords

Comments

A line segment joins points (a,b) and (c,d) if the points are distinct and gcd(c-a,d-b)=1.

Examples

			The 2 x 2 square lattice has a total of 6 line segments: 2 vertical, 2 horizontal and 2 diagonal.
		

References

  • D. M. Acketa, J. D. Zunic: On the number of linear partitions of the (m,n)-grid. Inform. Process. Lett., 38 (3) (1991), 163-168. See Table A.1.
  • Jovisa Zunic, Note on the number of two-dimensional threshold functions, SIAM J. Discrete Math. Vol. 25 (2011), No. 3, pp. 1266-1268. See Eq. (1.2).

Crossrefs

Cf. A141224.
The following eight sequences are all essentially the same. The simplest is A115004(n), which we denote by z(n). Then A088658(n) = 4*z(n-1); A114043(n) = 2*z(n-1)+2*n^2-2*n+1; A114146(n) = 2*A114043(n); A115005(n) = z(n-1)+n*(n-1); A141255(n) = 2*z(n-1)+2*n*(n-1); A290131(n) = z(n-1)+(n-1)^2; A306302(n) = z(n)+n^2+2*n. - N. J. A. Sloane, Feb 04 2020

Programs

  • Mathematica
    Table[cnt=0; Do[If[GCD[c-a,d-b]<2, cnt++ ], {a,n}, {b,n}, {c,n}, {d,n}]; (cnt-n^2)/2, {n,20}]
    (* This recursive code is much more efficient. *)
    a[n_]:=a[n]=If[n<=1,0,2*a1[n]-a[n-1]+R1[n]]
    a1[n_]:=a1[n]=If[n<=1,0,2*a[n-1]-a1[n-1]+R2[n]]
    R1[n_]:=R1[n]=If[n<=1,0,R1[n-1]+4*EulerPhi[n-1]]
    R2[n_]:=(n-1)*EulerPhi[n-1]
    Table[a[n],{n,1,37}]
    (* Seppo Mustonen, May 13 2010 *)
    a[n_]:=2 Sum[(n-i) (n-j) Boole[CoprimeQ[i,j]], {i,1,n-1}, {j,1,n-1}] + 2 n^2 - 2 n; Array[a, 40] (* Vincenzo Librandi, Feb 05 2020 *)
  • Python
    from sympy import totient
    def A141255(n): return 2*(n-1)*(2*n-1) + 2*sum(totient(i)*(n-i)*(2*n-i) for i in range(2,n)) # Chai Wah Wu, Aug 16 2021

Formula

a(n) = A114043(n) - 1.
a(n) = 2*(n-1)*(2n-1) + 2*Sum_{i=2..n-1} (n-i)*(2n-i)*phi(i). - Chai Wah Wu, Aug 16 2021

A288187 Triangle read by rows: T(n,m) (n >= m >= 1) = number of chambers (or regions) formed by drawing the line segments connecting any two of the (n+1) X (m+1) lattice points in an n X m lattice polygon.

Original entry on oeis.org

4, 16, 56, 46, 176, 520, 104, 388, 1152, 2584, 214, 822, 2502, 5700, 12368, 380, 1452, 4392, 9944, 21504, 37400, 648, 2516, 7644, 17380, 37572, 65810, 115532, 1028, 3952, 12120, 27572, 59784, 105128, 184442, 294040, 1562, 6060, 18476, 42066, 91654, 161352, 282754, 450864, 690816
Offset: 1

Views

Author

Hugo Pfoertner, Jun 06 2017

Keywords

Comments

Chambers are counted regardless of their numbers of vertices.
The n X m lattice polygon mentioned in the definition is an n X m grid of square cells, formed using a grid of n+1 X m+1 points. - N. J. A. Sloane, Feb 07 2019

Examples

			The diagonals of the 1 X 1 lattice polygon, i.e. the square, cut it into 4 triangles. Therefore T(1,1)=4.
Triangle begins
4,
16, 56,
46, 176, 520,
104, 388, 1152, 2584,
214, 822, 2502, 5700, 12368,
...
		

Crossrefs

The first column is A306302. For column 2 see A333279, A333280, A333281.
If the initial points are arranged around a circle rather than a square we get A006533 and A007678.

Extensions

T(4,1) added from A306302. - N. J. A. Sloane, Feb 07 2019
T(3,3) corrected and rows for n=4..9 added by Max Alekseyev, Apr 05 2019.

A331757 Number of edges in a figure made up of a row of n adjacent congruent rectangles upon drawing diagonals of all possible rectangles.

Original entry on oeis.org

8, 28, 80, 178, 372, 654, 1124, 1782, 2724, 3914, 5580, 7626, 10352, 13590, 17540, 22210, 28040, 34670, 42760, 51962, 62612, 74494, 88508, 104042, 121912, 141534, 163664, 187942, 215636, 245490, 279260, 316022, 356456, 399898, 447612, 498698, 555352
Offset: 1

Views

Author

N. J. A. Sloane, Feb 04 2020

Keywords

Crossrefs

A306302 gives number of regions in the figure.
This is column 1 of A331454.

Programs

  • Mathematica
    Table[n^2 + 4n + 1 + Sum[Sum[(2 * Boole[GCD[i, j] == 1] - Boole[GCD[i, j] == 2]) * (n + 1 - i) * (n + 1 - j), {j, 1, n}], {i, 1, n}], {n, 1, 37}] (* Joshua Oliver, Feb 05 2020 *)
  • Python
    from sympy import totient
    def A331757(n): return 8 if n == 1 else 2*(n*(n+3) + sum(totient(i)*(n+1-i)*(n+1+i) for i in range(2,n//2+1)) + sum(totient(i)*(n+1-i)*(2*n+2-i) for i in range(n//2+1,n+1))) # Chai Wah Wu, Aug 16 2021

Formula

a(n) = (2*n + 2 + 3*A324042(n) + 4*A324043(n))/2 [Corrected by Chai Wah Wu, Aug 16 2021]
For n > 1, a(n) = 2*(n*(n+3) + Sum_{i=2..floor(n/2)} (n+1-i)*(n+1+i)*phi(i) + Sum_{i=floor(n/2)+1..n} (n+1-i)*(2*n+2-i)*phi(i)). - Chai Wah Wu, Aug 16 2021

A335678 Array read by antidiagonals: T(m,n) (m>=1, n>=1) = number of cells in figure formed by taking m equally spaced points on a line and n equally spaced points on a parallel line, and joining each of the m points to each of the n points by a line segment.

Original entry on oeis.org

0, 1, 1, 2, 4, 2, 3, 8, 8, 3, 4, 13, 16, 13, 4, 5, 19, 27, 27, 19, 5, 6, 26, 40, 46, 40, 26, 6, 7, 34, 56, 69, 69, 56, 34, 7, 8, 43, 74, 98, 104, 98, 74, 43, 8, 9, 53, 95, 130, 149, 149, 130, 95, 53, 9, 10, 64, 118, 168, 198, 214, 198, 168, 118, 64, 10, 11, 76, 144, 210, 257, 285, 285, 257, 210, 144, 76, 11
Offset: 1

Views

Author

Keywords

Comments

The case m=n (the main diagonal) is dealt with in A306302, where there are illustrations for m = 1 to 15.

Examples

			The initial rows of the array are:
  0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...
  1, 4, 8, 13, 19, 26, 34, 43, 53, 64, 76, 89, ...
  2, 8, 16, 27, 40, 56, 74, 95, 118, 144, 172, 203, ...
  3, 13, 27, 46, 69, 98, 130, 168, 210, 257, 308, 365, ...
  4, 19, 40, 69, 104, 149, 198, 257, 322, 395, 474, 563, ...
  5, 26, 56, 98, 149, 214, 285, 371, 466, 573, 688, 818, ...
  6, 34, 74, 130, 198, 285, 380, 496, 624, 768, 922, 1097, ...
  7, 43, 95, 168, 257, 371, 496, 648, 816, 1005, 1207, 1437, ...
  8, 53, 118, 210, 322, 466, 624, 816, 1028, 1267, 1522, 1813, ...
  9, 64, 144, 257, 395, 573, 768, 1005, 1267, 1562, 1877, 2237, ...
  10, 76, 172, 308, 474, 688, 922, 1207, 1522, 1877, 2256, 2690, ...
  ...
The initial antidiagonals are:
  0
  1, 1
  2, 4, 2
  3, 8, 8, 3
  4, 13, 16, 13, 4
  5, 19, 27, 27, 19, 5
  6, 26, 40, 46, 40, 26, 6
  7, 34, 56, 69, 69, 56, 34, 7
  8, 43, 74, 98, 104, 98, 74, 43, 8
  9, 53, 95, 130, 149, 149, 130, 95, 53, 9
  10, 64, 118, 168, 198, 214, 198, 168, 118, 64, 10
  ...
		

Crossrefs

This is one of a set of five arrays: A335678, A335679, A335680, A335681, A335682.
For the diagonal see A306302.
See also A114999.

Formula

Euler's formula implies that A335679[m,n] = A335678[m,n] + A335680[m,n] - 1 for all m,n.
Comment from Max Alekseyev, Jun 28 2020 (Start):
T(m,n) = A114999(m-1,n-1) + m*n - 1 for all m, n >= 1. This follows from the Alekseyev-Basova-Zolotykh (2015) article.
Proof: Here is the appropriate modification of the corresponding comment in A306302: Assuming that K(m,n) has vertices at (i,0) and (j,1), for i=0..m-1 and j=0..n-1, the projective map (x,y) -> ((1-y)/(x+1), y/(x+1)) maps K(m,n) to the partition of the right triangle described by Alekseyev et al. (2015), for which Theorem 13 gives the number of regions, line segments, and intersection points. (End)
Max Alekseyev's formula is an analog of Theorem 3 of Griffiths (2010), and gives an explicit formula for this array. - N. J. A. Sloane, Jun 30 2020

A115005 a(n) = (A114043(n) - 1)/2.

Original entry on oeis.org

0, 3, 14, 43, 100, 209, 374, 641, 1020, 1553, 2246, 3197, 4372, 5911, 7778, 10037, 12728, 16043, 19862, 24467, 29728, 35777, 42626, 50625, 59520, 69675, 80966, 93627, 107568, 123345, 140458, 159673, 180664, 203651, 228590, 255857, 285116, 317363, 352058
Offset: 1

Views

Author

N. J. A. Sloane, Feb 23 2006

Keywords

Crossrefs

The following eight sequences are all essentially the same. The simplest is A115004(n), which we denote by z(n). Then A088658(n) = 4*z(n-1); A114043(n) = 2*z(n-1)+2*n^2-2*n+1; A114146(n) = 2*A114043(n); A115005(n) = z(n-1)+n*(n-1); A141255(n) = 2*z(n-1)+2*n*(n-1); A290131(n) = z(n-1)+(n-1)^2; A306302(n) = z(n)+n^2+2*n. - N. J. A. Sloane, Feb 04 2020

Programs

  • Mathematica
    a[n_]:=2 Sum[(n-i) (n-j) Boole[CoprimeQ[i,j]], {i, 1, n-1}, {j, 1, n-1}] / 2 + n^2 - n; Array[a, 40] (* Vincenzo Librandi, Feb 05 2020 *)
  • Python
    from sympy import totient
    def A115005(n): return (n-1)*(2*n-1) + sum(totient(i)*(n-i)*(2*n-i) for i in range(2,n)) # Chai Wah Wu, Aug 15 2021

Formula

a(n) = (n-1)*(2n-1) + Sum_{i=2..n-1} (n-i)*(2n-i)*phi(i). - Chai Wah Wu, Aug 15 2021

Extensions

Offset corrected by Max Alekseyev, Apr 10 2019

A088658 Number of triangles in an n X n unit grid that have minimal possible area (of 1/2).

Original entry on oeis.org

0, 4, 32, 124, 320, 716, 1328, 2340, 3792, 5852, 8544, 12260, 16864, 22916, 30272, 39188, 49824, 62948, 78080, 96348, 117232, 141260, 168480, 200292, 235680, 276100, 321056, 371484, 427024, 489900, 558112, 634724, 718432, 810116, 909600, 1018388, 1135136, 1263828, 1402304, 1551908
Offset: 1

Views

Author

Yuval Dekel (dekelyuval(AT)hotmail.com), Nov 21 2003

Keywords

Examples

			a(2)=4 because 4 (isosceles right) triangles with area 1/2 can be placed on a 2 X 2 grid.
		

Crossrefs

Cf. A045996.
The following eight sequences are all essentially the same. The simplest is A115004(n), which we denote by z(n). Then A088658(n) = 4*z(n-1); A114043(n) = 2*z(n-1)+2*n^2-2*n+1; A114146(n) = 2*A114043(n); A115005(n) = z(n-1)+n*(n-1); A141255(n) = 2*z(n-1)+2*n*(n-1); A290131(n) = z(n-1)+(n-1)^2; A306302(n) = z(n)+n^2+2*n. - N. J. A. Sloane, Feb 04 2020

Programs

  • Mathematica
    z[n_] := Sum[(n - i + 1)(n - j + 1) Boole[GCD[i, j] == 1], {i, n}, {j, n}];
    a[n_] := 4 z[n - 1];
    Array[a, 40] (* Jean-François Alcover, Mar 24 2020 *)
  • Python
    from sympy import totient
    def A088658(n): return 4*(n-1)**2 + 4*sum(totient(i)*(n-i)*(2*n-i) for i in range(2,n)) # Chai Wah Wu, Aug 15 2021

Formula

a(n+1) = 4*A115004(n).
a(n) = 4*(n-1)^2 + 4*Sum_{i=2..n-1} (n-i)*(2n-i)*phi(i). - Chai Wah Wu, Aug 15 2021

Extensions

a(7)-a(28) from Ray Chandler, May 03 2011
Corrected and extended by Ray Chandler, May 18 2011

A114146 Number of threshold functions on n X n grid.

Original entry on oeis.org

1, 2, 14, 58, 174, 402, 838, 1498, 2566, 4082, 6214, 8986, 12790, 17490, 23646, 31114, 40150, 50914, 64174, 79450, 97870, 118914, 143110, 170506, 202502, 238082, 278702, 323866, 374510, 430274, 493382, 561834, 638694, 722658, 814606, 914362, 1023430, 1140466
Offset: 0

Views

Author

N. J. A. Sloane, Feb 22 2006

Keywords

Comments

Also, number of intersections of a halfspace with an n X n grid. While A114043 counts cuts, this sequence counts sides of cuts. The only difference between this and twice A114043 is that this makes sense for the empty grid. This is the "labeled" version - rotations and reflections are not taken into account. - David Applegate, Feb 24 2006
In the terminology of Koplowitz et al., this is the number of linear dichotomies on a square grid. - N. J. A. Sloane, Mar 14 2020

Crossrefs

The following eight sequences are all essentially the same. The simplest is A115004(n), which we denote by z(n). Then A088658(n) = 4*z(n-1); A114043(n) = 2*z(n-1)+2*n^2-2*n+1; A114146(n) = 2*A114043(n); A115005(n) = z(n-1)+n*(n-1); A141255(n) = 2*z(n-1)+2*n*(n-1); A290131(n) = z(n-1)+(n-1)^2; A306302(n) = z(n)+n^2+2*n. - N. J. A. Sloane, Feb 04 2020

Programs

  • Mathematica
    a[0] = 1; a[n_] := 4 Sum[(n-i)(n-j) Boole[CoprimeQ[i, j]], {i, 1, n-1}, {j, 1, n-1}] + 4 n^2 - 4 n + 2;
    Array[a, 38, 0] (* Jean-François Alcover, Sep 04 2018, after Max Alekseyev in A114043 *)
  • Python
    from sympy import totient
    def A114146(n): return 1 if n == 0 else 8*n**2-12*n+6 + 4*sum(totient(i)*(n-i)*(2*n-i) for i in range(2,n)) # Chai Wah Wu, Aug 15 2021

Formula

For n>0, a(n) = 2*A114043(n).
For n>0, a(n) = 8*n^2 - 12*n + 6 + 4*Sum_{i=2..n-1} (n-i)*(2n-i)*phi(i). - Chai Wah Wu, Aug 15 2021

Extensions

Definition corrected by Max Alekseyev, Oct 23 2008
a(0)=1 prepended by Max Alekseyev, Jan 23 2015

A324043 Number of quadrilateral regions into which a figure made up of a row of n adjacent congruent rectangles is divided upon drawing diagonals of all possible rectangles.

Original entry on oeis.org

0, 2, 14, 34, 90, 154, 288, 462, 742, 1038, 1512, 2074, 2904, 3774, 4892, 6154, 7864, 9662, 12022, 14638, 17786, 20998, 25024, 29402, 34672, 40038, 46310, 53038, 61090, 69454, 79344, 89890, 101792, 113854, 127476, 141866, 158428, 175182, 193760, 213274, 235444, 258182, 283858, 310750, 339986
Offset: 1

Views

Author

Jinyuan Wang, May 01 2019

Keywords

Comments

A row of n adjacent congruent rectangles can only be divided into triangles (cf. A324042) or quadrilaterals when drawing diagonals. Proof is given in Alekseyev et al. (2015) under the transformation described in A306302.

Examples

			For k adjacent congruent rectangles, the number of quadrilateral regions in the j-th rectangle is:
k\j|  1   2   3   4   5   6   7  ...
---+--------------------------------
1  |  0,  0,  0,  0,  0,  0,  0, ...
2  |  1,  1,  0,  0,  0,  0,  0, ...
3  |  3,  8,  3,  0,  0,  0,  0, ...
4  |  5, 12, 12,  5,  0,  0,  0, ...
5  |  7, 22, 32, 22,  7,  0,  0, ...
6  |  9, 28, 40, 40, 28,  9,  0, ...
7  | 11, 38, 58, 74, 58, 38, 11, ...
...
a(4) = 5 + 12 + 12 + 5 = 34.
		

Crossrefs

Programs

  • Maple
    See Robert Israel link.
    There are also Maple programs for both A306302 and A324042. Then a := n -> A306302(n) - A324042(n); # N. J. A. Sloane, Mar 04 2020
  • Mathematica
    Table[Sum[Sum[(Boole[GCD[i, j] == 1] - 2 * Boole[GCD[i, j] == 2]) * (n + 1 - i) * (n + 1 - j), {j, 1, n}], {i, 1, n}] - n^2, {n, 1, 45}] (* Joshua Oliver, Feb 05 2020 *)
  • PARI
    { A324043(n) = sum(i=1, n, sum(j=1, n, ( (gcd(i, j)==1) - 2*(gcd(i,j)==2) ) * (n+1-i) * (n+1-j) )) - n^2; } \\ Max Alekseyev, Jul 08 2019
    
  • Python
    from sympy import totient
    def A324043(n): return 0 if n==1 else -2*(n-1)**2 + sum(totient(i)*(n+1-i)*(7*i-2*n-2) for i in range(2,n//2+1)) + sum(totient(i)*(n+1-i)*(2*n+2-i) for i in range(n//2+1,n+1)) # Chai Wah Wu, Aug 16 2021

Formula

a(n) = A115005(n+1) - A177719(n+1) - n - 1 = Sum_{i,j=1..n; gcd(i,j)=1} (n+1-i)*(n+1-j) - 2*Sum_{i,j=1..n; gcd(i,j)=2} (n+1-i)*(n+1-j) - n^2. - Max Alekseyev, Jul 08 2019
a(n) = A306302(n) - A324042(n).
For n>1, a(n) = -2(n-1)^2 + Sum_{i=2..floor(n/2)} (n+1-i)*(7i-2n-2)*phi(i) + Sum_{i=floor(n/2)+1..n} (n+1-i)*(2n+2-i)*phi(i). - Chai Wah Wu, Aug 16 2021

Extensions

a(8)-a(23) from Robert Israel, Jul 07 2019
Terms a(24) onward from Max Alekseyev, Jul 08 2019
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