A163275 -id:A163275 - 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

A163285 Triangle read by rows in which row n lists n+1 terms, starting with n^5 and ending with n^6, such that the difference between successive terms is equal to n^5 - n^4.

Original entry on oeis.org

0, 1, 1, 32, 48, 64, 243, 405, 567, 729, 1024, 1792, 2560, 3328, 4096, 3125, 5625, 8125, 10625, 13125, 15625, 7776, 14256, 20736, 27216, 33696, 40176, 46656, 16807, 31213, 45619, 60025, 74431, 88837, 103243, 117649, 32768, 61440, 90112, 118784, 147456

Offset: 0

Omar E. Pol, Jul 24 2009

The first term of row n is A000584(n) and the last term of row n is A001014(n).
The main entry for this sequence is A159797. See also A163282, A163283 and A163284.
Row sums give A163275. - Omar E. Pol, Mar 18 2012

			Triangle begins:
0;
1,1;
32,48,64;
243,405,567,729;
1024,1792,2560,3328,4096;
3125,5625,8125,10625,13125,15625;
7776,14256,20736,27216,33696,40176,46656;
16807,31213,45619,60025,74431,88837,103243,117649;
32768,61440,90112,118784,147456,176128,204800,233472,262144;
59049,111537,164025,216513,269001,321489,373977,426465,478953,531441;
100000,190000,280000,370000,460000,550000,640000,730000,820000,910000,1000000;

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

Cf. A000584, A001014, A085538, A159797, A162611, A162614, A162622, A163282, A163283, A163284.

rw[n_]:=Range[n^5,n^6,n^5-n^4]; Join[{0,1},Flatten[Array[rw,10]]] (* Harvey P. Dale, Mar 18 2012 *)

A163285(n, k)=n^5 +k*(n^5 -n^4) \\ G. C. Greubel, Dec 17 2016

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

cp's OEIS Frontend

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

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

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A163285 Triangle read by rows in which row n lists n+1 terms, starting with n^5 and ending with n^6, such that the difference between successive terms is equal to n^5 - n^4.

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

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

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