A024166 -id:A024166 - cp's OEIS frontend

A254869 Seventh partial sums of cubes (A000578).

1, 15, 111, 561, 2211, 7293, 21021, 54483, 129558, 286858, 598026, 1184118, 2242266, 4083366, 7184166, 12257850, 20348031, 32951985, 52179985, 80958735, 123288165, 184562235, 271965915, 394962165, 565884540, 800652996, 1119632580, 1548656956

Offset: 1

Luciano Ancora, Feb 17 2015

			2nd differences:   0,  6,  12,  18,   24,   30, ... (A008588)
1st differences:   1,  7,  19,  37,   61,   91, ... (A003215)
-------------------------------------------------------------------
The cubes:         1,  8,  27,  64,  125,  216, ... (A000578)
-------------------------------------------------------------------
1st partial sums:  1,  9,  36, 100,  225,  441, ... (A000537)
2nd partial sums:  1, 10,  46, 146,  371,  812, ... (A024166)
3rd partial sums:  1, 11,  57, 203,  574, 1386, ... (A101094)
4th partial sums:  1, 12,  69, 272,  846, 2232, ... (A101097)
5th partial sums:  1, 13,  82, 354, 1200, 3432, ... (A101102)
6th partial sums:  1, 14,  96, 450, 1650, 5082, ... (A254469)
7th partial sums:  1, 15, 111, 561, 2211, 7293, ... (this sequence)

Luciano Ancora, Table of n, a(n) for n = 1..1000
Luciano Ancora, Partial sums of m-th powers with Faulhaber polynomials.
Luciano Ancora, Pascal's triangle and recurrence relations for partial sums of m-th powers.
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 15.
Index entries for linear recurrences with constant coefficients, signature (11,-55,165,-330,462,-462,330,-165,55,-11,1).

Cf. A000537, A000578, A003215, A024166, A101094, A101097, A101102, A254469, A254870, A254871, A254872.

[n*(1+n)*(2+n)*(3+n)*(4+n)*(5+n)*(6+n)*(7+n)*(7+7*n+n^2)/604800: n in [1..30]]; // Vincenzo Librandi, Feb 19 2015

Table[n (1 + n) (2 + n) (3 + n) (4 + n) (5 + n) (6 + n) (7 + n) (7 + 7 n + n^2)/604800, {n, 26}] (* or *)
CoefficientList[Series[(- 1 - 4 x - x^2)/(- 1 + x)^11, {x, 0, 25}], x]
Nest[Accumulate,Range[30]^3,7] (* or *) LinearRecurrence[{11,-55,165,-330,462,-462,330,-165,55,-11,1},{1,15,111,561,2211,7293,21021,54483,129558,286858,598026},30] (* Harvey P. Dale, Apr 24 2017 *)

vector(50, n, n*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n)*(6 + n)*(7 + n)*(7 + 7*n + n^2)/604800) \\ Derek Orr, Feb 19 2015

G.f.: x*(1 + 4*x + x^2)/(1 - x)^11.
a(n) = n*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n)*(6 + n)*(7 + n)*(7 + 7*n + n^2)/604800.
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) + n^3.
Sum_{n>=1} 1/a(n) = 1920*sqrt(3/7)*Pi*tan(sqrt(21)*Pi/2) - 251488/49. - Amiram Eldar, Jan 26 2022

A167566 The third left hand column of triangle A167565.

Original entry on oeis.org

2, 16, 67, 202, 497, 1064, 2058, 3684, 6204, 9944, 15301, 22750, 32851, 46256, 63716, 86088, 114342, 149568, 192983, 245938, 309925, 386584, 477710, 585260, 711360, 858312, 1028601, 1224902, 1450087, 1707232, 1999624, 2330768, 2704394

Offset: 3

Johannes W. Meijer, Nov 10 2009

G. C. Greubel, Table of n, a(n) for n = 3..1000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Equals the third left hand column of triangle A167565.

Other left hand columns are A000027, A000292, A167567 and A168304.

Table[(7*n^5 - 30*n^4 + 45*n^3 - 30*n^2 + 8*n)/5!, {n,3,100}] (* or *) LinearRecurrence[{6,-15,20,-15,6,-1}, {2, 16, 67, 202, 497, 1064}, 100] (* G. C. Greubel, Jun 16 2016 *)

Vec((1*z^2 + 4*z + 2)/(1-z)^6 + O(z^50)) \\ Michel Marcus, Jul 05 2017

a(n) = n*(7*n^4 - 30*n^3 + 45*n^2 - 30*n + 8)/120 \\ Charles R Greathouse IV, Jul 14 2017

From Johannes W. Meijer, Nov 23 2009: (Start)
a(n) = (7*n^5 - 30*n^4 + 45*n^3 - 30*n^2 + 8*n)/5!.
G.f.: (1*z^2 + 4*z + 2)/(1-z)^6.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6).
a(n) - 5*a(n-1) + 10*a(n-2) - 10*a(n-3) + 5*a(n-4) - a(n-5) = 7. (End)
a(n) = A024166(n-2) + A000389(n+2). - J. M. Bergot, Jul 04 2017

A212891 Rectangular array: (row n) = bc, where b(h) = h, c(h) = (n-1+h)^2, n>=1, h>=1, and = convolution.

Original entry on oeis.org

1, 6, 4, 20, 17, 9, 50, 46, 34, 16, 105, 100, 84, 57, 25, 196, 190, 170, 134, 86, 36, 336, 329, 305, 260, 196, 121, 49, 540, 532, 504, 450, 370, 270, 162, 64, 825, 816, 784, 721, 625, 500, 356, 209, 81, 1210, 1200, 1164, 1092, 980, 830, 650, 454, 262

Offset: 1

Clark Kimberling, Jun 16 2012

Principal diagonal:  A213436
Antidiagonal sums:  A024166
row 1,  (1,2,3,...)**(1,4,9,...):  A002415(k+1)
row 2,  (1,2,3,...)**(4,9,16,...):  k*(k^3 + 8*k^2 + 23*k + 16)/12
row 3,  (1,2,3,...)**(9,16,25,...):  k*(k^3 + 12*k^2 + 53*k + 42)/12
...
For a guide to related arrays, see A213500.

			Northwest corner (the array is read by falling antidiagonals):
1....6....20....50....105....196...336
4....17...46....100...190....329...532
9....34...84....170...305....504...784
16...57...134...260...450....721...1092
25...86...196...370...625....980...1456
...
T(5,1) = (1)**(25) = 25
T(5,2) = (1,2)**(25,36) = 1*36+2*25 = 86
T(5,3) = (1,2,3)**(25,36,49) = 1*49+2*36+3*25 = 196

Cf. A213500.

b[n_] := n; c[n_] := n^2
t[n_, k_] := Sum[b[k - i] c[n + i], {i, 0, k - 1}]
TableForm[Table[t[n, k], {n, 1, 10}, {k, 1, 10}]]
Flatten[Table[t[n - k + 1, k], {n, 12}, {k, n, 1, -1}]]
r[n_] := Table[t[n, k], {k, 1, 60}]  (* A212891 *)
d = Table[t[n, n], {n, 1, 40}] (* A213436 *)
s[n_] := Sum[t[i, n + 1 - i], {i, 1, n}]
s1 = Table[s[n], {n, 1, 50}] (* A024166  *)

T(n,k) = 5*T(n,k-1) - 10*T(n,k-2) + 10*T(n,k-3) - 5*T(n,k-4) + T(n,k-5).

G.f. for row n: f(x)/g(x), where f(x) = n^2 - (2*n^2 - 2*n - 1)*x + ((n-1)^2)*x^2 and g(x) = (1 - x)^5.

A213555 Rectangular array: (row n) = bc, where b(h) = h^3, c(h) = n-1+h, n>=1, h>=1, and = convolution.

Original entry on oeis.org

1, 10, 2, 46, 19, 3, 146, 82, 28, 4, 371, 246, 118, 37, 5, 812, 596, 346, 154, 46, 6, 1596, 1253, 821, 446, 190, 55, 7, 2892, 2380, 1694, 1046, 546, 226, 64, 8, 4917, 4188, 3164, 2135, 1271, 646, 262, 73, 9, 7942, 6942, 5484, 3948, 2576, 1496, 746

Offset: 1

Clark Kimberling, Jun 17 2012

Principal diagonal: A213556.
Antidiagonal sums: A213547.
Row 1,  (1,8,27,...)**(1,2,3,...):  A024166.
Row 2,  (1,8,27,...)**(2,3,4,...): (3*k^5 + 30*k^4 + 55*k^3 + 30*k^2 + 2*k)/60.
Row 3,  (1,8,27,...)**(3,4,5,...): (3*k^5 + 45*k^4 + 85*k^3 + 45*k^2 + 2*k)/60.
For a guide to related arrays, see A213500.

			Northwest corner (the array is read by falling antidiagonals):
1...10...46....146...371....812
2...19...82....246...596....1253
3...28...118...346...821....1694
4...37...154...446...1046...2135
5...46...190...546...1271...2576
6...55...226...646...1496...3017

Clark Kimberling, Antidiagonals n = 1..60, flattened

Cf. A213500, A213553.

b[n_] := n^3; c[n_] := n
t[n_, k_] := Sum[b[k - i] c[n + i], {i, 0, k - 1}]
TableForm[Table[t[n, k], {n, 1, 10}, {k, 1, 10}]]
Flatten[Table[t[n - k + 1, k], {n, 12}, {k, n, 1, -1}]]
r[n_] := Table[t[n, k], {k, 1, 60}]  (* A213555 *)
d = Table[t[n, n], {n, 1, 40}] (* A213556 *)
s[n_] := Sum[t[i, n + 1 - i], {i, 1, n}]
s1 = Table[s[n], {n, 1, 50}] (* A213547 *)

T(n,k) = 6*T(n,k-1) - 15*T(n,k-2) + 20*T(n,k-3) - 15*T(n,k-4) + 6*T(n,k-5) -T(n,k-6).

G.f. for row n: f(x)/g(x), where f(x) = n + (3*n + 1)*x - (3*n - 4)*x^2 - (n - 1)*x^3 and g(x) = (1 - x)^6.

A257448 a(n) = 13(2^n - 1) - 3n^2 - 9*n.

Original entry on oeis.org

1, 9, 37, 111, 283, 657, 1441, 3051, 6319, 12909, 26149, 52695, 105859, 212265, 425161, 851043, 1702903, 3406725, 6814477, 13630095, 27261451, 54524289, 109050097, 218101851, 436205503, 872412957, 1744828021, 3489658311, 6979319059, 13958640729

Offset: 1

Luciano Ancora, Apr 23 2015

These numbers belong to a family of sequences obtained as follows:
. A000225:   1*(2^n-1);
. A050488:   3*(2^n-1) - 2*n;
.    a(n):  13*(2^n-1) - 3*n^2 -  9*n;
. A257449:  75*(2^n-1) - 4*n^3 - 18*n^2 -  52*n;
. A257450: 541*(2^n-1) - 5*n^4 - 30*n^3 - 130*n^2 - 375*n,
where the sequence 1, 3, 13, 75, 541, ... is A000670 (after the first term), and A208744 gives the triangle of coefficients:
2;
3, 9;
4, 18, 52;
5, 30, 130, 375;
6, 45, 260, 1125, 3246;
7, 63, 455, 2625, 11361, 32781, etc.
Also, the antidiagonal sums in the array are given by the formula (6*n^2 + 6*k*n + (k-1)*k)*(k+n)!/((k+3)!*(n-1)!) for k = 0, 1, 2, 3, 4, ... (see Example field).

			By the second comment, the array begins (antidiagonals in A046902):
k=0: 1,  8, 27,  64,  125,  216, ...  A000578
k=1: 1,  9, 36, 100,  225,  441, ...  A000537
k=2: 1, 10, 46, 146,  371,  812, ...  A024166
k=3: 1, 11, 57, 203,  574, 1386, ...  A101094
k=4: 1, 12, 69, 272,  846, 2232, ...  A101097
k=5: 1, 13, 82, 354, 1200, 3432, ...  A101102
k=6: 1, 14, 96, 450, 1650, 5082, ...  A254469
...
See also A254469 (Example field).

Index entries for linear recurrences with constant coefficients, signature (5, -9, 7, -2).

Cf. A000225, A000670, A050488, A208744, A257449, A257450.

[13*(2^n-1)-3*n^2-9*n: n in [1..30]]; // Vincenzo Librandi, Apr 24 2015

Table[13 (2^n - 1) - 3 n^2 - 9n, {n, 30}]
CoefficientList[Series[x (1 + 4 x + x^2)/((1 - x)^3*(1 - 2 x)), {x, 0, 30}], x] (* Michael De Vlieger, Nov 14 2016 *)

G.f.: x*(1+4*x+x^2)/((1-x)^3*(1-2*x)).

a(n) = 5*a(n-1) - 9*a(n-2) + 7*a(n-3) - 2*a(n-4) for n>4. - Ray Chandler, Jul 25 2015

Edited by Bruno Berselli, Apr 28 2015

A293550 a(n) = Sum_{k=0..n} k^3binomial(2n-k,n).

Original entry on oeis.org

0, 1, 11, 69, 354, 1650, 7293, 31213, 130832, 540702, 2212550, 8989090, 36327810, 146228940, 586823265, 2349424125, 9389012160, 37467344310, 149345215290, 594753416790, 2366845396500, 9413555798556, 37423053793026, 148719333293394, 590842248405024, 2346813893147500

Offset: 0

Ilya Gutkovskiy, Oct 11 2017

nonn

Main diagonal of iterated partial sums array of cubes (starting with the first partial sums). For nonnegative integers see A002054, for squares see A265612.

Seiichi Manyama, Table of n, a(n) for n = 0..1000
Index entries for sequences related to sums of cubes

Cf. A000537, A000578, A002054, A024166, A101094, A101097, A101102, A265612, A302352.

Table[Sum[k^3 Binomial[2 n - k, n], {k, 0, n}], {n, 0, 25}]
Table[SeriesCoefficient[x (1 + 4 x + x^2)/(1 - x)^(n + 5), {x, 0, n}], {n, 0, 25}]
Table[2^(2 n + 1) n^2 (13 n + 7) Gamma[n + 3/2]/(Sqrt[Pi] Gamma[n + 5]), {n, 0, 25}]
CoefficientList[Series[(6 - 6 Sqrt[1 - 4 x] - 36 x + 24 Sqrt[1 - 4 x] x + 55 x^2 - 19 Sqrt[1 - 4 x] x^2 - 15 x^3 + Sqrt[1 - 4 x] x^3)/(2 Sqrt[1 - 4 x] x^4), {x, 0, 25}], x]
CoefficientList[Series[(E^(2 x) (36 - 24 x + 13 x^2) BesselI[0, 2 x])/x^2 + (E^(2 x) (-36 + 24 x - 31 x^2 + 13 x^3) BesselI[1, 2 x])/x^3, {x, 0, 25}], x]* Range[0, 25]!

a(n) = [x^n] x*(1 + 4*x + x^2)/(1 - x)^(n+5).
a(n) = 2^(2*n+1)*n^2*(13*n + 7)*Gamma(n+3/2)/(sqrt(Pi)*Gamma(n+5)).
a(n) ~ 26*4^n/sqrt(Pi*n).

A110197 Number triangle of sums of squared binomial coefficients.

Original entry on oeis.org

1, 2, 1, 3, 5, 1, 4, 14, 10, 1, 5, 30, 46, 17, 1, 6, 55, 146, 117, 26, 1, 7, 91, 371, 517, 251, 37, 1, 8, 140, 812, 1742, 1476, 478, 50, 1, 9, 204, 1596, 4878, 6376, 3614, 834, 65, 1, 10, 285, 2892, 11934, 22252, 19490, 7890, 1361, 82, 1, 11, 385, 4917, 26334, 66352, 82994, 51990, 15761, 2107, 101, 1

Offset: 0

Paul Barry, Jul 15 2005

Alternatively, number square T(n,k) = Sum_{i=0..n} binomial(i+k,k)^2 read by antidiagonals.

			Rows start:
  1;
  2,   1;
  3,   5,   1;
  4,  14,  10,   1;
  5,  30,  46,  17,   1;
  6,  55, 146, 117,  26,   1;
  ...

Columns include A000027, A000330, A024166, A086020, A086023, A086025.
Row sums are A006134.
Antidiagonal sums are A110198.
T(2n,n) gives A112029.

T(n,k) = sum(i=0, n-k, binomial(i+k,k)^2);
tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n, k), ", ")); print();); \\ Michel Marcus, Dec 03 2016

T(n,k) = Sum_{i=0..n-k} binomial(i+k,k)^2.

G.f.: 1/((1-x)*sqrt(x^2*y^2-2*x^2*y-2*x*y+x^2-2*x+1)). - Vladimir Kruchinin, Mar 20 2025

A125166 Triangle R(n,k), companion to A125165, left column n^3.

Original entry on oeis.org

1, 8, 1, 27, 9, 1, 64, 36, 10, 1, 125, 100, 46, 11, 1, 216, 225, 146, 57, 12, 1, 343, 441, 371, 203, 69, 13, 1, 512, 784, 812, 574, 272, 82, 14, 1, 729, 1296, 1596, 1386, 846, 354, 96, 15, 1, 1000, 2025, 2892, 2982, 2232, 1200, 450, 111, 16, 1

Offset: 0

Gary W. Adamson, Nov 22 2006

Riordan array ((1 + 4*x + x^2)/(1 - x)^4, x/(1 - x)). - Philippe Deléham, Dec 09 2013

			With other offset (k >= 1) from first formula: R(5,3) = 146 = R(4,3) + R(4,2) = 46 + 100.
The Riordan triangle R begins:
  n\k|     0    1    2    3    4    5   6   7   8  9
  --------------------------------------------------
   0 |     1
   1 |     8    1
   2 |    27    9    1
   3 |    64   36   10    1
   4 |   125  100   46   11    1
   5 |   216  225  146   57   12    1
   6 |   343  441  371  203   69   13   1
   7 |   512  784  812  574  272   82  14   1
   8 |   729 1296 1596 1386  846  354  96  15   1
   9 |  1000 2025 2892 2982 2232 1200 450 111  16  1
... refomatted and extended by _Wolfdieter Lang_, Mar 25 2025.

Paolo Xausa, Table of n, a(n) for n = 0..11475 (rows 0..150 of triangle, flattened).

Cf. A000578 (column 0), A000537 (column 1), A024166 (column 2), A101094 (column 3).
Cf. A257448 (row sums).
Cf. A008292, A028246, A097805, A125165, A050488, A382057.

A125166[n_, k_] := A125166[n, k] = Switch[k, 0, (n + 1)^3, n, 1, _, A125166[n - 1, k - 1] + A125166[n - 1, k]];
Table[A125166[n, k], {n, 0, 10}, {k, 0, n}] (* Paolo Xausa, Apr 08 2025 *)

y = polygen(QQ, 'y')
x = y.parent()[['x']].gen()
gf = (1 + 4*x + x^2)/((x - 1)^3*(y*x + x - 1))
[list(u) for u in list(gf.O(11))]  # Peter Luschny, Apr 02 2025

Binomial transform of an infinite matrix M with diagonal D, subdiagonal (D-1)..., etc; as follows: (D) = (1,1,1...); (D-1) = (7,7,7...); (D-2) = (12,12,12...); (D-3) = (6,6,6...). Alternatively, given left border n^3: (1, 8, 27, 64...); for k>1, R(n,k) = R(n-1,k) + R(n-1,k-1).
From Wolfdieter Lang, Mar 27 2025: (Start)
Riordan triangle (see a comment above):
R(n, 0) = (n+1)^3, R(n, k) = R(n-1, k-1) + R(n-1, k), for k >= 1. (from the (finite) A-sequence {1, 1} with offset 0),
R(n, 0) = Sum_{k=0..n-1} Z(j)*R(n-1, k), for n >= 1, and R(0, 0) = 1, with the Riordan Z-sequence A382057. For Riordan A- and Z-sequences see the first W. Lang link in A006232.
R(n, k) = Sum{j=0..n} (j+1)^3*A097805(n-j, k) (convolution property).
R(n, k) = Sum{j=0..3} A028246(4, j+1)*binomial(n, k+j). (Proof for k=0 with a standard (n+1)^3  formula, and for k >= 1 using the Pascal type recurrence for the triangle.)
O.g.f. column k (with leading 0s): ((1 + 4*x + x^2)/(1 - x)^4)*(x/(1-x))^k. (Numerator polynomial from row 3 of the triangle A008292.)
O.g.f. row polynomials P(n, y) = Sum_{k=0..n} R(n, k)*y^k:
   G(y, x) = (1 + 4*x + x^2)/((1 - x)^3*(1 - (1+y)*x)). (End)

a(27) corrected by Georg Fischer, Feb 18 2020

A104727 Triangle T(n,k) = (k-1-n)(k-2-n)(k^2+k+2kn+3n^2+5n)/24 read by rows, 1<=k<=n.

Original entry on oeis.org

1, 7, 3, 25, 15, 6, 65, 45, 26, 10, 140, 105, 71, 40, 15, 266, 210, 155, 103, 57, 21, 462, 378, 295, 215, 141, 77, 28, 750, 630, 511, 395, 285, 185, 100, 36, 1155, 990, 826, 665, 510, 365, 235, 126, 45, 1705, 1485, 1266, 1050, 840, 640, 455, 291, 155, 55, 2431, 2145, 1860, 1578, 1302, 1036, 785, 555, 353, 187, 66, 3367, 3003

Offset: 1

Gary W. Adamson, Mar 20 2005

The triangle is created by multiplying the lower triangular matrix A(n,k) = A000217(k) (1<=k<=n) by the lower triangular matrix B(n,k) = n-k+1 (1<=k<=n): T(n,k) = sum_{j=k..n} A(n,j)*B(j,k).

The commuted product B * A generates triangle A098358.

			First few rows of the triangle are:
1;
7, 3;
25, 15, 6;
665, 45, 26, 10;
140, 105, 71, 40, 15;
266, 210, 155, 103, 57, 21;
...

Cf. A098358, A104727, A024166 (row sums).

T(n,1) = A001296(n). - R. J. Mathar, Oct 29 2011

A086755 Sum_{k=1..n} (k(k+1))^2/2.

Original entry on oeis.org

2, 20, 92, 292, 742, 1624, 3192, 5784, 9834, 15884, 24596, 36764, 53326, 75376, 104176, 141168, 187986, 246468, 318668, 406868, 513590, 641608, 793960, 973960, 1185210, 1431612, 1717380, 2047052, 2425502, 2857952, 3349984, 3907552

Offset: 0

Jon Perry, Jul 31 2003

			a(3)=(1*2)^2/2+(2*3)^2/2+(3.4)^2/2=2+18+72=92

Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Table[Sum[(k(k+1))^2/2,{k,n}],{n,40}] (* or *) LinearRecurrence[{6,-15,20,-15,6,-1},{2,20,92,292,742,1624},40] (* Harvey P. Dale, Oct 04 2020 *)

for(i=1,20,print1(","sum(j=1,i,(j*(j+1))^2/2)))

(n+1)(n+2)(n+3)(3n^2+12n+10)/30 = 2*A024166(n+1).

G.f. 2*(1+4*x+x^2) / (x-1)^6 . - R. J. Mathar, Sep 15 2012

More terms from Jason Earls, Aug 01 2003

Definition clarified by Harvey P. Dale, Oct 04 2020

cp's OEIS Frontend

A254869 Seventh partial sums of cubes (A000578).

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A167566 The third left hand column of triangle A167565.

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A212891 Rectangular array: (row n) = b**c, where b(h) = h, c(h) = (n-1+h)^2, n>=1, h>=1, and ** = convolution.

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A213555 Rectangular array: (row n) = b**c, where b(h) = h^3, c(h) = n-1+h, n>=1, h>=1, and ** = convolution.

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Mathematica

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A257448 a(n) = 13*(2^n - 1) - 3*n^2 - 9*n.

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Magma

Mathematica

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Extensions

A293550 a(n) = Sum_{k=0..n} k^3*binomial(2*n-k,n).

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Mathematica

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A110197 Number triangle of sums of squared binomial coefficients.

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PARI

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A125166 Triangle R(n,k), companion to A125165, left column n^3.

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A212891 Rectangular array: (row n) = bc, where b(h) = h, c(h) = (n-1+h)^2, n>=1, h>=1, and = convolution.

A213555 Rectangular array: (row n) = bc, where b(h) = h^3, c(h) = n-1+h, n>=1, h>=1, and = convolution.

A257448 a(n) = 13(2^n - 1) - 3n^2 - 9*n.

A293550 a(n) = Sum_{k=0..n} k^3binomial(2n-k,n).

A104727 Triangle T(n,k) = (k-1-n)(k-2-n)(k^2+k+2kn+3n^2+5n)/24 read by rows, 1<=k<=n.