A033429 -id:A033429 - cp's OEIS frontend

A304659 a(n) = n(n + 1)(16*n - 1)/6.

0, 5, 31, 94, 210, 395, 665, 1036, 1524, 2145, 2915, 3850, 4966, 6279, 7805, 9560, 11560, 13821, 16359, 19190, 22330, 25795, 29601, 33764, 38300, 43225, 48555, 54306, 60494, 67135, 74245, 81840, 89936, 98549, 107695, 117390, 127650, 138491, 149929, 161980, 174660, 187985

Offset: 0

Bruno Berselli, May 22 2018

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A007742, A076455, A139273 (first differences).

First lower diagonal of the rectangular array in A213835.

[n*(n+1)*(16*n-1)/6: n in [0..41]]; // Vincenzo Librandi, May 23 2018

Table[n (n + 1) (16 n - 1)/6, {n, 0, 50}]

concat(0, Vec(x*(5 + 11*x) / (1 - x)^4 + O(x^40))) \\ Colin Barker, May 25 2018

O.g.f.: x*(5 + 11*x)/(1 - x)^4.
E.g.f.: x*(30 + 63*x + 16*x^2)*exp(x)/6.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
a(n) + a(-n) = A033429(n).
a(n) = n*A007742(n) - Sum_{k = 0..n-1} A007742(k) for n > 0.
Also, this sequence is related to A076455 by the same type of recurrence:
A076455(n) = n*a(n) - Sum_{k = 0..n-1} a(k) for n > 0.

A351381 Table read by downward antidiagonals: T(n,k) = n*(k+1)^2.

Original entry on oeis.org

4, 9, 8, 16, 18, 12, 25, 32, 27, 16, 36, 50, 48, 36, 20, 49, 72, 75, 64, 45, 24, 64, 98, 108, 100, 80, 54, 28, 81, 128, 147, 144, 125, 96, 63, 32, 100, 162, 192, 196, 180, 150, 112, 72, 36, 121, 200, 243, 256, 245, 216, 175, 128, 81, 40, 144, 242, 300, 324, 320, 294, 252, 200, 144, 90, 44

Offset: 1

Bernard Schott, Mar 28 2022

When m and k are both positive integers and k | m, with m/k = n, then T(n,k) = S(m,k) = (m+k) + (m-k) + (m*k) + (m/k) = S(n*k,k) = n*(k+1)^2, problem proposed by Yakov Perelman.

All terms are nonsquarefree (A013929).

			Table begins:
  n \ k |   1      2      3      4      5      6      7      8      9     10
  ----------------------------------------------------------------------------
     1  |   4      9     16      25    36     49     64     81    100    121
     2  |   8     18     32      50    72     98    128    162    200    242
     3  |  12     27     48      75   108    147    192    243    300    363
     4  |  16     36     64     100   144    196    256    324    400    484
     5  |  20     45     80     125   180    245    320    405    500    605
     6  |  24     54     96     150   216    294    384    486    600    726
     7  |  28     63    112     175   252    343    448    567    700    847
     8  |  32     72    128     200   288    392    512    648    800    968
     9  |  36     81    144     225   324    441    576    729    900   1089
    10  |  40     90    160     250   360    490    640    810   1000   1210
  ............................................................................
T(3,4) = 75 = 3*(4+1)^2 corresponds to S(3*4,4) = S(12,4) = (12+4) + (12-4) + (12*4) + 12/4 = 75.
S(10,5) = (10+5) + (10-5) + (10*5) + (10/5) = T(10/5,5) = T(2,5) = 72.

I. Perelman, L'Algèbre Récréative, Chapitre IV, Les équations de Diophante, Deux nombres et quatre opérations, Editions en langues étrangères, Moscou, 1959, pp. 101-102.
Ya. I. Perelman, Algebra Can Be Fun, Chapter IV, Diophantine Equations, Two numbers and four operations, Mir Publishers Moscow, 1979, pp. 131-132.

Ya. I. Perelman, Algebra Can Be Fun, Chapter IV, Diophantine Equations, Two numbers and four operations, Mir Publishers Moscow, 1979, pp. 131-132.
Wikipedia, Yakov Perelman.

Cf. A013929.
Cf. A000290 \ {0,1} (row 1), A001105 \ {0,2} (row 2), A033428 \ {0,3} (row 3), A016742 \ {0,4} (row 4), A033429 \ {0,5} (row 5), A033581 \ {0,6} (row 6).
Cf. A008586 \ {0} (column 1), A008591 \ {0} (column 2), A008598 \ {0} (column 3), A008607 \ {0} (column 4), A044102 \ {0} (column 5).
Cf. A045991 \ {0} (diagonal).

T[n_, k_] := n*(k + 1)^2; Table[T[k, n - k + 1], {n, 1, 11}, {k, 1, n}] // Flatten (* Amiram Eldar, Mar 29 2022 *)

T(n,k) = n*(k+1)^2.
T(n,n) = (n+1)^3 - (n+1)^2 = A045991(n+1) for n >= 1.
G.f.: x*(1 + y)/((1 - x)^2*(1 - y)^3). - Stefano Spezia, Mar 31 2022

A103218 Triangle read by rows: T(n, k) = (2k+1)(n+1-k)^2.

Original entry on oeis.org

1, 4, 3, 9, 12, 5, 16, 27, 20, 7, 25, 48, 45, 28, 9, 36, 75, 80, 63, 36, 11, 49, 108, 125, 112, 81, 44, 13, 64, 147, 180, 175, 144, 99, 52, 15, 81, 192, 245, 252, 225, 176, 117, 60, 17, 100, 243, 320, 343, 324, 275, 208, 135, 68, 19, 121, 300, 405, 448, 441, 396, 325, 240

Offset: 0

Lambert Klasen (lambert.klasen(AT)gmx.de) and Gary W. Adamson, Jan 25 2005

The triangle is generated from the product A * B of the infinite lower triangular matrix A =
0 0 0...
1 0 0...
3 1 0...
5 3 1...
... and B =
0 0 0...
3 0 0...
3 5 0...
3 5 7...
...

			Triangle begins:
1,
4,3,
9,12,5,
16,27,20,7,
25,48,45,28,9,

Row sums give A002412 (hexagonal pyramidal numbers).
T(n, 0)=A000290(n+1) (the squares);
T(n, 1)=3*n^2=A033428(n);
T(n, 2)=5*n^2=A033429(n+1);
T(n, 3)=7*n^2=A033582(n+2);
Cf. A103219 (product B*A), A002412, A000290.

T[n_, k_] := (2*k + 1)*(n + 1 - k)^2; Flatten[ Table[ T[n, k], {n, 0, 10}, {k, 0, n}]] (* Robert G. Wilson v, Feb 10 2005 *)

T(n, k) = (2*k+1)*(n+1-k)^2; for(i=0,10, for(j=0,i,print1(T(i,j),","));print())

A247327 Triangle read by rows: T(n,k) = sum of k-th row of n X n square filled with odd numbers 1 through 2*n^2-1 reading across rows left-to-right.

Original entry on oeis.org

1, 4, 12, 9, 27, 45, 16, 48, 80, 112, 25, 75, 125, 175, 225, 36, 108, 180, 252, 324, 396, 49, 147, 245, 343, 441, 539, 637, 64, 192, 320, 448, 576, 704, 832, 960, 81, 243, 405, 567, 729, 891, 1053, 1215, 1377, 100, 300, 500, 700, 900, 1100, 1300, 1500, 1700, 1900, 121, 363

Offset: 1

Kival Ngaokrajang, Sep 13 2014

See illustration in links. Column c(k) = (2*k - 1)*n^2. Diagonal d(m) = (2*n - 2*m + 1)*n^2.

			Triangle begins:
  1
  4   12
  9   27  45
  16  48  80 112
  25  75 125 175 225
  36 108 180 252 324 396
  49 147 245 343 441 539 637

Kival Ngaokrajang, Illustration of initial terms

Column: c(1) = A000290, c(2) = A033428, c(3) = A033429.
Diagonal: d(1) = A015237, d(2) = A015238, d(3) = A015240.
Rows sum: A000538.
Cf. A241016.

trg(nn) = {for (n=1, nn, mm = matrix(n, n, i, j, (2*j-1) + (2*n)*(i-1)); for (i=1, n, print1(sum(j=1, n, mm[i, j]), ", ");); print(););} \\ Michel Marcus, Sep 15 2014

A257093 a(n) = n(n+1)(13*n+2)/6.

Original entry on oeis.org

0, 5, 28, 82, 180, 335, 560, 868, 1272, 1785, 2420, 3190, 4108, 5187, 6440, 7880, 9520, 11373, 13452, 15770, 18340, 21175, 24288, 27692, 31400, 35425, 39780, 44478, 49532, 54955, 60760, 66960, 73568, 80597, 88060, 95970, 104340, 113183, 122512, 132340

Offset: 0

Luce ETIENNE, Apr 16 2015

This sequence gives the number of triangles of all sizes in (5*n^2)-polyiamonds in a tetragonal or hexagonal or heptagonal configuration.
It is the sum of (1/2)*Sum_{j=0..n-1} (n-j)*(5*n+1-j) triangles oriented in one direction and (1/2)*Sum_{j-0..n-1} (n-j)*(5*n-1-3*j) oriented in the opposite direction.
Shäfli's notation: 3.3.3.3.3 for a(1).
The difference between this sequence and A050409(n) equals A000292(n-1).
Also, (1/3)*(A002717(2*n) + A255211(n) - 2*A000330(n)) gives A033994(n): a (5*n^2)-polyiamond in pentagonal configuration that does not belong to this sequence because a(1)=6.
a(n) is odd only when n mod 4 = 1.

			Second comment a(0) = 0; a(1) = 3 + 2; a(2) = 16 + 12; a(3) = 46 + 36; a(4) = 100 + 80; a(5) = 185 + 150; a(6) = 308 + 252.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A002411, A011379, A033429, A050409, A000292, A002717, A000330, A033994, A255211.

[n*(n+1)*(13*n+2)/6: n in [0..40]]; // Vincenzo Librandi, Apr 16 2015

Table[n (n + 1) (13 n + 2)/6, {n, 0, 40}] (* Vincenzo Librandi, Apr 16 2015 *)
CoefficientList[Series[x (5+8x)/(1-x)^4,{x,0,50}],x] (* or *) LinearRecurrence[{4,-6,4,-1},{0,5,28,82},60] (* Harvey P. Dale, Feb 12 2023 *)

a(n) = Sum_{j=0..n-1} (n-j)*(5*n-2*j).
From Vincenzo Librandi, Apr 16 2015: (Start)
G.f.: x*(5+8*x)/(1-x)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). (End)
E.g.f.: exp(x)*x*(30 + 54*x + 13*x^2)/6. - Stefano Spezia, Mar 02 2025

Corrected by Harvey P. Dale, Feb 12 2023

A303302 a(n) = 34*n^2.

Original entry on oeis.org

0, 34, 136, 306, 544, 850, 1224, 1666, 2176, 2754, 3400, 4114, 4896, 5746, 6664, 7650, 8704, 9826, 11016, 12274, 13600, 14994, 16456, 17986, 19584, 21250, 22984, 24786, 26656, 28594, 30600, 32674, 34816, 37026, 39304, 41650, 44064, 46546, 49096, 51714, 54400, 57154, 59976, 62866, 65824, 68850, 71944

Offset: 0

Omar E. Pol, May 13 2018

Sequence found by reading the line from 0, in the direction 0, 34, ..., in the square spiral whose vertices are the generalized 19-gonal numbers A303813.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A005843, A008599, A303813.

Cf. similar sequences of the type k*n^2: A000290 (k=1), A001105 (k=2), A033428 (k=3), A016742 (k=4), A033429 (k=5), A033581 (k=6), A033582 (k=7), A139098 (k=8), A016766 (k=9), A033583 (k=10), A033584 (k=11), A135453 (k=12), A152742 (k=13), A144555 (k=14), A064761 (k=15), A016802 (k=16), A244630 (k=17), A195321 (k=18), A244631 (k=19), A195322 (k=20), A064762 (k=21), A195323 (k=22), A244632 (k=23), A195824 (k=24), A016850 (k=25), A244633 (k=26), A244634 (k=27), A064763 (k=28), A244635 (k=29), A244636 (k=30), A244082 (k=32), this sequence (k=34), A016910 (k=36), A016982 (k=49), A017066 (k=64), A017162 (k=81), A017270 (k=100), A017390 (k=121), A017522 (k=144).

[34*n^2: n in [0..50]]; // Vincenzo Librandi Jun 07 2018

Table[34 n^2, {n, 0, 40}]
LinearRecurrence[{3,-3,1},{0,34,136},50] (* Harvey P. Dale, Jul 23 2018 *)

PARI
```
a(n) = 34*n^2;
```

concat(0, Vec(34*x*(1 + x) / (1 - x)^3 + O(x^40))) \\ Colin Barker, Jun 12 2018

a(n) = 34*A000290(n) = 17*A001105(n) = 2*A244630(n).
G.f.: 34*x*(1 + x)/(1 - x)^3. - Vincenzo Librandi, Jun 07 2018
From Elmo R. Oliveira, Dec 02 2024: (Start)
E.g.f.: 34*x*(1 + x)*exp(x).
a(n) = A005843(n)*A008599(n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A330082 a(n) = 5*A064038(n+1).

Original entry on oeis.org

0, 5, 15, 15, 25, 75, 105, 70, 90, 225, 275, 165, 195, 455, 525, 300, 340, 765, 855, 475, 525, 1155, 1265, 690, 750, 1625, 1755, 945, 1015, 2175, 2325, 1240, 1320, 2805, 2975, 1575, 1665, 3515, 3705, 1950, 2050, 4305, 4515, 2365, 2475, 5175, 5405, 2820, 2940

Offset: 0

Paul Curtz, Dec 01 2019

Main column of a pentagonal spiral for A026741:
                       (25)
                    49 (15) 31
                24  29 (15)  8  16
            47  14   7 ( 5)  3  17  33
        23  27  13   2 ( 0)  1   7   9  17
          45  13   6   3   1   4  19  35
            22  25  11   5   9  10  18
              43  12  23  11  21  37
                21  41  20  39  19
a(n) = 5 * A064038(n+1) from a pentagonal spiral.
Compare to A319127 =  6 * A002620 in the hexagonal spiral:
                22  23  23  22 (24)
              20  12  13  13 (12) 25
            21  10   5   4 ( 6) 14  25
          21  11   5   1 ( 0)  7  15  24
        20  11   4   1 ( 0)  2   7  15  26
          18  10   2   3   3   6  14  27
            19   8   9   9   8  16  27
              19  18  16  17  17  26
                30  28  29  29  28

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-6,10,-12,12,-10,6,-3,1).

Cf. A026741, A028895, A064038. A033429, A062786, A087348, A147874, A158447, A168668 are in the spiral.

A330082[n_]:=5Numerator[n(n+1)/4];Array[A330082,100,0] (* Paolo Xausa, Dec 04 2023 *)

concat(0, Vec(5*x*(1 + 4*x^3 + x^6) / ((1 - x)^3*(1 + x^2)^3) + O(x^50))) \\ Colin Barker, Dec 08 2019

a(n) = A026741(A028895(n)).
From Colin Barker, Dec 08 2019: (Start)
G.f.: 5*x*(1 + 4*x^3 + x^6) / ((1 - x)^3*(1 + x^2)^3).
a(n) = 3*a(n-1) - 6*a(n-2) + 10*a(n-3) - 12*a(n-4) + 12*a(n-5) - 10*a(n-6) + 6*a(n-7) - 3*a(n-8) + a(n-9) for n>8.
a(n) = (-5/16 + (5*i)/16)*(((-3-3*i) + (-i)^n + i^(1+n))*n*(1+n)) where i=sqrt(-1).
(End)

More terms from Colin Barker, Dec 22 2019

Name corrected by Paolo Xausa, Dec 04 2023

A332495 a(n-2) = a(n-6) + 5(1+2n) with a(0)=0, a(1)=2, a(2)=7, a(3)=15 for n>=4.

Original entry on oeis.org

0, 2, 7, 15, 25, 37, 52, 70, 90, 112, 137, 165, 195, 227, 262, 300, 340, 382, 427, 475, 525, 577, 632, 690, 750, 812, 877, 945, 1015, 1087, 1162, 1240, 1320, 1402, 1487, 1575, 1665, 1757, 1852, 1950, 2050, 2152, 2257

Offset: 0

Paul Curtz, Feb 14 2020

a(-2)=2, a(-1)=0. 4 evens followed by 4 odds.
Last digit is only 0, 2, 5, 7.
The vertical spoke S-N of the pentagonal spiral for A004526.
                          37
                      37  25  25
                  36  24  15  15  26
              36  24  14   7   8  16  26
          35  23  14   7   2   3   8  16  27
      35  23  13   6   2   0   0   3   9  17  27
        34  22  13   6   1   1   4   9  17  28
          34  22  12   5   5   4  10  18  28
            33  21  12  11  11  10  18  29
              33  21  20  20  19  19  29
                32  32  31  31  30  30
Rank of multiples of 10: 0, 7, 8, 15, 16, ... = A047521. Compare to A154260 in the formula.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-4,4,-3,1).

Cf. A004526, A033429, A062786, A168668, A135706, A147874, 2*A147875 (all in the spiral).

Cf. A005408, A047521, A154260

CoefficientList[Series[x (2 + x + 2 x^2)/((1 - x)^3*(1 + x^2)), {x, 0, 42}], x] (* Michael De Vlieger, Feb 14 2020 *)

concat(0, Vec(x*(2 + x + 2*x^2) / ((1 - x)^3*(1 + x^2)) + O(x^40))) \\ Colin Barker, Feb 14 2020

a(-1-n) = a(n).
a(2*n) + a(1+2*n) = 2, 22, 62, ... = A273366(n).
Second differences give the sequence of period 4: repeat [3, 3, 2, 2].
From Colin Barker, Feb 14 2020: (Start)
G.f.: x*(2 + x + 2*x^2) / ((1 - x)^3*(1 + x^2)).
a(n) = 3*a(n-1) - 4*a(n-2) + 4*a(n-3) - 3*a(n-4) + a(n-5) for n>4.
(End)
Multiples of 10: 10*(0, 7, 9, 30, 34, ... = A154260).
4*a(n) = A087960(n) +5*n -1 +5*n^2. - R. J. Mathar, Feb 28 2020

A362885 Array read by ascending antidiagonals: A(n, k) = (1 + 2n)k^n.

Original entry on oeis.org

1, 0, 1, 0, 3, 1, 0, 5, 6, 1, 0, 7, 20, 9, 1, 0, 9, 56, 45, 12, 1, 0, 11, 144, 189, 80, 15, 1, 0, 13, 352, 729, 448, 125, 18, 1, 0, 15, 832, 2673, 2304, 875, 180, 21, 1, 0, 17, 1920, 9477, 11264, 5625, 1512, 245, 24, 1, 0, 19, 4352, 32805, 53248, 34375, 11664, 2401, 320, 27, 1

Offset: 0

Stefano Spezia, May 08 2023

			The array begins:
    1,  1,   1,    1,     1,     1, ...
    0,  3,   6,    9,    12,    15, ...
    0,  5,  20,   45,    80,   125, ...
    0,  7,  56,  189,   448,   875, ...
    0,  9, 144,  729,  2304,  5625, ...
    0, 11, 352, 2673, 11264, 34375, ...
    ...

Cf. A000007 (k=0), A000012 (n=0), A004248, A005408 (k=1), A008585 (n=1), A014480 (k=2), A033429 (n=2), A058962 (k=4), A124647 (k=3), A155988 (k=9), A171220 (k=5), A176043, A199299 (k=6), A199300 (k=7), A199301 (k=8), A244727 (n=3), A362886 (antidiagonal sums).

A[n_,k_]:=(1+2n)k^n; Join[{1}, Table[A[n-k,k],{n,10},{k,0,n}]]//Flatten (* or *)
A[n_,k_]:=SeriesCoefficient[(1+k*x)/(1-k*x)^2,{x,0,n}]; Table[A[n-k,k],{n,0,10},{k,0,n}]//Flatten (* or *)
A[n_,k_]:=n!SeriesCoefficient[Exp[k*x](1+2k*x),{x,0,n}]; Table[A[n-k,k],{n,0,10},{k,0,n}]//Flatten

A(n, k) = A005408(n)*A004248(n, k).
O.g.f. of column k: (1 + k*x)/(1 - k*x)^2.
E.g.f. of column k: exp(k*x)*(1 + 2*k*x).
A(n, n) = A176043(n+1).

A363436 Array read by ascending antidiagonals: A(n, k) = k*n^2, with k >= 0.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 4, 2, 0, 0, 9, 8, 3, 0, 0, 16, 18, 12, 4, 0, 0, 25, 32, 27, 16, 5, 0, 0, 36, 50, 48, 36, 20, 6, 0, 0, 49, 72, 75, 64, 45, 24, 7, 0, 0, 64, 98, 108, 100, 80, 54, 28, 8, 0, 0, 81, 128, 147, 144, 125, 96, 63, 32, 9, 0, 0, 100, 162, 192, 196, 180, 150, 112, 72, 36, 10, 0

Offset: 0

Stefano Spezia, Jul 08 2023

			The array begins:
  0,  0,  0,   0,   0,   0,   0, ...
  0,  1,  2,   3,   4,   5,   6, ...
  0,  4,  8,  12,  16,  20,  24, ...
  0,  9, 18,  27,  36,  45,  54, ...
  0, 16, 32,  48,  64,  80,  96, ...
  0, 25, 50,  75, 100, 125, 150, ...
  0, 36, 72, 108, 144, 180, 216, ...
  ...

Cf. A000290 (k = 1), A001105 (k = 2), A033428 (k = 3), A016742 (k = 4), A033429 (k = 5), A033581 (k = 6), A033582 (k = 7), A139098 (k = 8), A016766 (k = 9), A033583 (k = 10), A033584 (k = 11), A135453 (k = 12),  A152742 (k = 13), A144555 (k = 14), A064761 (k = 15), A016802 (k = 16), A244630 (k = 17), A195321 (k = 18), A244631 (k = 19), A195322 (k = 20), A064762 (k = 21), A195323 (k = 22), A244632 (k = 23), A195824 (k = 24), A016850 (k = 25), A244633 (k = 26), A244634 (k = 27), A064763 (k = 28), A244635 (k = 29), A244636 (k = 30).
Cf. A001477 (n = 1), A008586 (n = 2), A008591 (n = 3), A008598 (n = 4), A008607 (n = 5), A044102 (n = 6), A152691 (n = 8).
Cf. A000007 (n = 0 or k = 0), A000578 (main diagonal), A002415 (antidiagonal sums), A004247.

A[n_,k_]:=k n^2; Table[A[n-k,k],{n,0,11},{k,0,n}]//Flatten

O.g.f.: x*y*(1 + x)/((1 - x)^3*(1 - y)^2).
E.g.f.: x*y*(1 + x)*exp(x + y).
A(n, k) = n*A004247(n, k).

cp's OEIS Frontend

A304659 a(n) = n*(n + 1)*(16*n - 1)/6.

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A351381 Table read by downward antidiagonals: T(n,k) = n*(k+1)^2.

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A103218 Triangle read by rows: T(n, k) = (2*k+1)*(n+1-k)^2.

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A247327 Triangle read by rows: T(n,k) = sum of k-th row of n X n square filled with odd numbers 1 through 2*n^2-1 reading across rows left-to-right.

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A257093 a(n) = n*(n+1)*(13*n+2)/6.

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A303302 a(n) = 34*n^2.

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A330082 a(n) = 5*A064038(n+1).

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A304659 a(n) = n(n + 1)(16*n - 1)/6.

A103218 Triangle read by rows: T(n, k) = (2k+1)(n+1-k)^2.

A257093 a(n) = n(n+1)(13*n+2)/6.

A332495 a(n-2) = a(n-6) + 5(1+2n) with a(0)=0, a(1)=2, a(2)=7, a(3)=15 for n>=4.

A362885 Array read by ascending antidiagonals: A(n, k) = (1 + 2n)k^n.