A099903 -id:A099903 - cp's OEIS frontend

A163102 a(n) = n^2*(n+1)^2/2.

0, 2, 18, 72, 200, 450, 882, 1568, 2592, 4050, 6050, 8712, 12168, 16562, 22050, 28800, 36992, 46818, 58482, 72200, 88200, 106722, 128018, 152352, 180000, 211250, 246402, 285768, 329672, 378450, 432450, 492032, 557568, 629442, 708050, 793800, 887112, 988418

Offset: 0

Omar E. Pol, Jul 24 2009

Row sums of triangle A163282.
Also, the number of nonattacking placements of 2 rooks on an (n+1) X (n+1) board. - Thomas Zaslavsky, Jun 26 2010
If P_{k}(n) is the n-th k-gonal number, then a(n) = P_{s}(n+1)*P_{t}(n+1) - P_{s+1}(n+1)*P_{t-1}(n+1) for s=t+1. - Bruno Berselli, Sep 05 2014
Subsequence of A000982, see formula. - David James Sycamore, Jul 31 2018
Number of edges in the (n+1) X (n+1) rook complement graph. - Freddy Barrera, May 02 2019
Number of paths from (0,0) to (n+2,n+2) consisting of exactly three forward horizontal steps and three upward vertical steps. - Greg Dresden and Snezhana Tuneska, Aug 24 2023

Seth Chaiken, Christopher R. H. Hanusa, and Thomas Zaslavsky, A q-queens problem, in preparation. - Thomas Zaslavsky, Jun 26 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Seth Chaiken, Christopher R. H. Hanusa, and Thomas Zaslavsky, A q-Queens Problem. IV. Queens, Bishops, Nightriders (and Rooks), arXiv preprint arXiv:1609.00853 [math.CO], 2016-2020.
A. Umar, Combinatorial Results for Semigroups of Orientation-Preserving Partial Transformations, J. Int. Seq., Vol. 14 (2011), Article 11.7.5.
Eric Weisstein's World of Mathematics, Rook Complement Graph.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Column k=2 of A144084.
Cf. A000217, A000537, A000982, A035287.
Cf. A006002, A099903, A163274, A163275, A163276, A163277, A163282.
Cf. A060300, A254371.

List([0..40],n->(n*(n+1))^2/2); # Muniru A Asiru, Aug 02 2018

[n^2*(n+1)^2/2: n in [0..40]]; // Vincenzo Librandi, Mar 26 2012

CoefficientList[Series[2*x*(1+4*x+x^2)/(1-x)^5,{x,0,40}],x] (* Vincenzo Librandi, Mar 26 2012 *)

a(n)=n^2*(n+1)^2/2 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 2*A000537(n) = A035287(n+1)/2. - Omar E. Pol, Nov 29 2011
G.f.: 2*x*(1+4*x+x^2)/(1-x)^5. - R. J. Mathar, Nov 30 2011
Let t(n) = A000217(n). Then a(n) = (t(n-1)*(t(n)+t(n+1)) + t(n)*(t(n-1)+t(n+1)) + t(n+1)*(t(n-1)+t(n)))/3. - J. M. Bergot, Jun 21 2012
a(n) = A000982(n*(n+1)). - David James Sycamore, Jul 31 2018
From Amiram Eldar, Nov 02 2021: (Start)
Sum_{n>=1} 1/a(n) = 2*Pi^2/3 - 6.
Sum_{n>=1} (-1)^(n+1)/a(n) = 6 - 8*log(2). (End)
Another identity: ..., a(4) = 200 = 1*(2+4+6+8) + 3*(4+6+8) + 5*(6+8) + 7*(8), a(5) = 450 = 1*(2+4+6+8+10) + 3*(4+6+8+10) + 5*(6+8+10) + 7*(8+10) + 9*(10) = 30+84+120+126+90, and so on. - J. M. Bergot, Aug 25 2022
From Elmo R. Oliveira, Aug 14 2025: (Start)
E.g.f.: x*(2 + x)*(2 + 6*x + x^2)*exp(x)/2.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = A254371(n)/4 = A060300(n)/8. (End)

A163274 a(n) = n^4*(n+1)^2/2.

Original entry on oeis.org

0, 2, 72, 648, 3200, 11250, 31752, 76832, 165888, 328050, 605000, 1054152, 1752192, 2798978, 4321800, 6480000, 9469952, 13530402, 18948168, 26064200, 35280000, 47064402, 61960712, 80594208, 103680000, 132031250, 166567752, 208324872

Offset: 0

Omar E. Pol, Jul 24 2009

Row sums of triangle A163284.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A006002, A099903, A163102, A163275, A163276, A163277, A163284.

Table[(n^4 (n+1)^2)/2,{n,0,30}] (* or *) LinearRecurrence[ {7,-21,35,-35,21,-7,1},{0,2,72,648,3200,11250,31752},30] (* Harvey P. Dale, May 07 2012 *)

a(n)=n^4*(n+1)^2/2 \\ Charles R Greathouse IV, Oct 07 2015

From R. J. Mathar, Jul 29 2009: (Start)
a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7).
G.f.: -2*x*(1 + 29*x + 93*x^2 + 53*x^3 + 4*x^4)/(x-1)^7. (End)
From Amiram Eldar, May 14 2022: (Start)
Sum_{n>=1} 1/a(n) = 4*Pi^2/3 + Pi^4/45 - 4*zeta(3) - 10.
Sum_{n>=1} (-1)^(n+1)/a(n) = 10 + Pi^2/3 + 7*Pi^4/360 - 16*log(2) - 3*zeta(3). (End)

More terms from R. J. Mathar, Jul 29 2009

A163275 a(n) = n^5*(n+1)^2/2.

Original entry on oeis.org

0, 2, 144, 1944, 12800, 56250, 190512, 537824, 1327104, 2952450, 6050000, 11595672, 21026304, 36386714, 60505200, 97200000, 151519232, 230016834, 341067024, 495219800, 705600000, 988352442, 1363135664, 1853666784

Offset: 0

Omar E. Pol, Jul 24 2009

Row sums of triangle A163285.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Cf. A006002, A099903, A163102, A163274, A163276, A163277, A163285.

A163275 := proc(n) n^5*(n+1)^2/2 ; end proc: seq(A163275(n),n=0..60) ; # R. J. Mathar, Feb 05 2010

Table[(1/2)*n^5*(n + 1)^2, {n,0,50}] (* or *) LinearRecurrence[{8,-28,56, -70,56,-28,8,-1}, {0,2,144,1944,12800,56250,190512,537824}, 50] (* G. C. Greubel, Dec 12 2016 *)

concat([0], Vec(2*x*(1+64*x+424*x^2+584*x^3+179*x^4+8*x^5)/(x-1)^8 + O(x^50))) \\ G. C. Greubel, Dec 12 2016

From R. J. Mathar, Feb 05 2010: (Start)
a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 70*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8).
G.f.: 2*x*(1 + 64*x + 424*x^2 + 584*x^3 + 179*x^4  +8*x^5)/(x-1)^8. (End)
From Amiram Eldar, May 14 2022: (Start)
Sum_{n>=1} 1/a(n) = 12 -5*Pi^2/3 - 2*Pi^4/45 + 6*zeta(3) + 2*zeta(5).
Sum_{n>=1} (-1)^(n+1)/a(n) = 20*log(2) + 9*zeta(3)/2 + 15*zeta(5)/8 - 12 - Pi^2/2 - 7*Pi^4/180. (End)

Extended by R. J. Mathar, Feb 05 2010

A163276 a(n) = n^6*(n+1)^2/2.

Original entry on oeis.org

0, 2, 288, 5832, 51200, 281250, 1143072, 3764768, 10616832, 26572050, 60500000, 127552392, 252315648, 473027282, 847072800, 1458000000, 2424307712, 3910286178, 6139206432, 9409176200, 14112000000, 20755401282, 29988984608

Offset: 0

Omar E. Pol, Jul 24 2009

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Cf. A006002, A099903, A163102, A163274, A163275, A163277.

[n^6*(n+1)^2/2: n in [0..30]]; // Vincenzo Librandi, Dec 13 2016

seq((1/2)*n^6*(n+1)^2, n = 0 .. 25); # Emeric Deutsch, Aug 01 2009

Table[(1/2)*n^6*(n + 1)^2, {n,0,50}] (* or *) LinearRecurrence[{9,-36,84,-126,126,-84,36,-9,1}, {0, 2, 288, 5832, 51200, 281250, 1143072, 3764768, 10616832}, 50] (* G. C. Greubel, Dec 12 2016 *)

concat([0], Vec(2*x*(1 + 135*x +1656*x^2 +4456*x^3 +3231*x^4 +585*x^5 +16*x^6)/(1-x)^9 + O(x^50))) \\ G. C. Greubel, Dec 12 2016

G.f.: 2*x*(1+135*x+1656*x^2+4456*x^3+3231*x^4+585*x^5+16*x^6)/(1-x)^9. - Colin Barker, May 05 2012
From Amiram Eldar, May 14 2022: (Start)
Sum_{n>=1} 1/a(n) = 2*Pi^2 + Pi^4/15 + 2*Pi^6/945 - 14 - 8*zeta(3) - 4*zeta(5).
Sum_{n>=1} (-1)^(n+1)/a(n) = 14 + 2*Pi^2/3 + 7*Pi^4/120 + 31*Pi^6/15120 - 24*log(2) - 6*zeta(3) - 15*zeta(5)/4. (End)

Extended by Emeric Deutsch, Aug 01 2009

A163277 a(n) = n^7*(n+1)^2/2.

Original entry on oeis.org

0, 2, 576, 17496, 204800, 1406250, 6858432, 26353376, 84934656, 239148450, 605000000, 1403076312, 3027787776, 6149354666, 11859019200, 21870000000, 38788923392, 66474865026, 110505715776, 178774347800, 282240000000

Offset: 0

Omar E. Pol, Jul 24 2009

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

Cf. A006002, A099903, A163102, A163274, A163275, A163276.

[n^7*(n+1)^2/2: n in [0..30]]; // Vincenzo Librandi, Dec 13 2016

A163277 := proc(n) n^7*(n+1)^2/2 ; end proc: seq(A163277(n),n=0..60) ; \\ R. J. Mathar, Feb 05 2010

Table[(1/2)*n^7*(n + 1)^2, {n,0,50}] (* G. C. Greubel, Dec 12 2016 *)

concat([0], Vec(2*x*(1 +278*x +5913*x^2 +27760*x^3 +38435*x^4 +16434*x^5 +1867*x^6 +32*x^7)/(x-1)^10 + O(x^50))) \\ G. C. Greubel, Dec 12 2016

From R. J. Mathar, Feb 05 2010: (Start)
a(n) = n^2*A163275(n).
G.f.: 2*x*(1 +278*x +5913*x^2 +27760*x^3 +38435*x^4 +16434*x^5 +1867*x^6 +32*x^7)/(x-1)^10. (End)
From Amiram Eldar, May 14 2022: (Start)
Sum_{n>=1} 1/a(n) = 16 - 7*Pi^2/3 - 4*Pi^4/45 - 4*Pi^6/945 + 10*zeta(3) + 6*zeta(5) + 2*zeta(7).
Sum_{n>=1} (-1)^(n+1)/a(n) = 28*log(2) + 15*zeta(3)/2 + 45*zeta(5)/8 + 63*zeta(7)/32 - 16 - 5*Pi^2/6 - 7*Pi^4/90 - 31*Pi^6/7560. (End)

Extended by R. J. Mathar, Feb 05 2010

A163283 Triangle read by rows in which row n lists n+1 terms, starting with n^3 and ending with n^4, such that the difference between successive terms is equal to n^3 - n^2.

Original entry on oeis.org

0, 1, 1, 8, 12, 16, 27, 45, 63, 81, 64, 112, 160, 208, 256, 125, 225, 325, 425, 525, 625, 216, 396, 576, 756, 936, 1116, 1296, 343, 637, 931, 1225, 1519, 1813, 2107, 2401, 512, 960, 1408, 1856, 2304, 2752, 3200, 3648, 4096, 729, 1377, 2025, 2673, 3321, 3969

Offset: 0

Omar E. Pol, Jul 24 2009

The first term of row n is A000578(n) and the last term of row n is A000583(n).

			Triangle begins:
0;
1,    1;
8,    12,   16;
27,   45,   63,   81;
64,   112,  160,  208,  256;
125,  225,  325,  425,  525,  625;
216,  396,  576,  756,  936,  1116, 1296;
343,  637,  931,  1225, 1519, 1813, 2107, 2401;
512,  960,  1408, 1856, 2304, 2752, 3200, 3648, 4096;
729,  1377, 2025, 2673, 3321, 3969, 4617, 5265, 5913, 6561;
1000, 1900, 2800, 3700, 4600, 5500, 6400, 7300, 8200, 9100, 10000;
...

G. C. Greubel, Table of n, a(n) for the first 50 rows

Row sums: A099903.

Cf. A000578, A000583, A045991, A159797, A162611, A162614, A163282, A163284, A163285.

Table[n^3 + k*(n^3 - n^2), {n, 0, 5}, {k, 0, n}]//Flatten (* G. C. Greubel, Dec 13 2016 *)

A163283(n, k)=n^3 +k*(n^3 -n^2) \\ G. C. Greubel, Dec 13 2016

T(n, k) = n^3 + k*(n^3 - n^2), for 0 <= k <= n, n >= 0. - G. C. Greubel, Dec 13 2016

Edited by Omar E. Pol, Jul 25 2009

A019584 a(n) = n^2*(n-1)^3/4.

Original entry on oeis.org

0, 0, 1, 18, 108, 400, 1125, 2646, 5488, 10368, 18225, 30250, 47916, 73008, 107653, 154350, 216000, 295936, 397953, 526338, 685900, 882000, 1120581, 1408198, 1752048, 2160000, 2640625, 3203226, 3857868, 4615408, 5487525, 6486750, 7626496, 8921088, 10385793

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..600
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A099903.

[n^2*(n-1)^3/4: n in [0..60]]; // Vincenzo Librandi, Apr 26 2011

Mathematica
```
Table[n^2*(n-1)^3/4, {n,0,100}]
```

a(n)=n^2*(n-1)^3/4 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = Sum_{j=1..n-2} Sum_{i=1..n-2} (i^3 + j^3)/2. - Alexander Adamchuk, Oct 24 2004
G.f.: x^2*(1 + 12*x + 15*x^2 + 2*x^3)/(1 - x)^6. - Colin Barker, May 04 2012
a(n) = Sum_{i=0..n-1} (n-1)*(n-1-i)^3 for n>0. - Bruno Berselli, Oct 31 2017
From Amiram Eldar, Feb 13 2023: (Start)
a(n) = A099903(n-1)/2.
Sum_{n>=2} 1/a(n) = 16 - 2*Pi^2 + 4*zeta(3).
Sum_{n>=2} (-1)^n/a(n) = 24*log(2) - 16 - Pi^2/3 + 3*zeta(3). (End)

A099904 Numerator of sum of all matrix elements of N X N matrix M(i,j) = i^3+j^3, (i,j = 1..n) divided by n!.

Original entry on oeis.org

2, 18, 36, 100, 75, 147, 98, 18, 45, 605, 121, 169, 1183, 7, 1, 289, 289, 361, 361, 1, 11, 5819, 529, 1, 13, 13, 1, 841, 841, 961, 961, 1, 17, 17, 1, 1369, 26011, 19, 1, 1681, 1681, 1849, 1849, 1, 23, 50807, 2209, 1, 1, 1, 1, 2809, 2809, 1, 1, 1, 29, 100949, 3481, 3721

Offset: 1

Alexander Adamchuk, Oct 29 2004

nonn

Sum M(i,j) (i,j = 1..n) is A099903(n). a(n) is an irregular sequence with highest champions belonging to Pentagonal pyramidal numbers n^2*(n+1)/2 (A002411) and n/2*(n+1)^2 (A006002).

			A099903(n)/n! begins 2, 18, 36, 100/3, 75/4, 147/20, 98/45, 18/35, 45/448, ... So a(6) = 147.

Cf. A099903, A019584, A098077, A002411, A006002.

Table[ Numerator[ Sum[(i^3 + j^3), {i, n}, {j, n}]/n! ], {n, 60}]

a(n) = Numerator[1/n!*Sum[Sum[(i^3+j^3), {i, 1, n}], {j, 1, n}]] a(n) = Numerator[1/2 * (n^3)*(n+1)^2 /n! ].

cp's OEIS Frontend

A163102 a(n) = n^2*(n+1)^2/2.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

GAP

Magma

Mathematica

PARI

Formula

A163274 a(n) = n^4*(n+1)^2/2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A163275 a(n) = n^5*(n+1)^2/2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

A163276 a(n) = n^6*(n+1)^2/2.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

Extensions

A163277 a(n) = n^7*(n+1)^2/2.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

Extensions

A163283 Triangle read by rows in which row n lists n+1 terms, starting with n^3 and ending with n^4, such that the difference between successive terms is equal to n^3 - n^2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A019584 a(n) = n^2*(n-1)^3/4.