A103532 -id:A103532 - cp's OEIS frontend

A262000 a(n) = n^2(7n - 5)/2.

0, 1, 18, 72, 184, 375, 666, 1078, 1632, 2349, 3250, 4356, 5688, 7267, 9114, 11250, 13696, 16473, 19602, 23104, 27000, 31311, 36058, 41262, 46944, 53125, 59826, 67068, 74872, 83259, 92250, 101866, 112128, 123057, 134674, 147000, 160056, 173863, 188442, 203814, 220000

Offset: 0

Bruno Berselli, Sep 08 2015

Also, structured enneagonal prism numbers.

			For n=8, a(8) = 8*(7*0+1)+8*(7*1+1)+8*(7*2+1)+8*(7*3+1)+8*(7*4+1)+8*(7*5+1)+8*(7*6+1)+8*(7*7+1) = 1632.

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

Cf. similar sequences with the formula n^2*(k*n - k + 2)/2: A000290 (k=0), A002411 (k=1), A000578 (k=2), A050509 (k=3), A015237 (k=4), A006597 (k=5), A100176 (k=6), this sequence (k=7), A103532 (k=8).

Cf. A001106, A069125.

Magma
```
[n^2*(7*n-5)/2: n in [0..40]];
```

Table[n^2 (7 n - 5)/2, {n, 0, 40}]
LinearRecurrence[{4,-6,4,-1},{0,1,18,72},50] (* Harvey P. Dale, Oct 04 2016 *)

PARI
```
vector(40, n, n--; n^2*(7*n-5)/2)
```
Sage
```
[n^2*(7*n-5)/2 for n in (0..40)]
```

G.f.: x*(1 + 14*x + 6*x^2)/(1 - x)^4.
a(n) = Sum_{i=0..n-1} n*(7*i+1) for n>0, a(0)=0.
a(n+1) + a(-n) = A069125(n+1).
Sum_{i>0} 1/a(i) = 1.082675669875907610300284768825... = (42*(log(14) + 2*(cos(Pi/7)*log(cos(3*Pi/14)) + log(sin(Pi/7))*sin(Pi/14) - log(cos(Pi/14)) * sin(3*Pi/14))) + 21*Pi*tan(3*Pi/14))/75 - Pi^2/15. - Vaclav Kotesovec, Oct 04 2016
From Elmo R. Oliveira, Aug 06 2025: (Start)
E.g.f.: exp(x)*x*(2 + 16*x + 7*x^2)/2.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). (End)

A257238 Triangle T(n, k) = n^3 - k^3, 0 <= k < = n.

Original entry on oeis.org

0, 1, 0, 8, 7, 0, 27, 26, 19, 0, 64, 63, 56, 37, 0, 125, 124, 117, 98, 61, 0, 216, 215, 208, 189, 152, 91, 0, 343, 342, 335, 316, 279, 218, 127, 0, 512, 511, 504, 485, 448, 387, 296, 169, 0, 729, 728, 721, 702, 665, 604, 513, 386, 217, 0, 1000, 999, 992, 973, 936, 875, 784, 657, 488, 271, 0

Offset: 0

Wolfdieter Lang, May 12 2015

See the comments in A025581 and A079904 on a problem by François Viète (Vieta)(1593). Regarding that problem, note the simple identity: n^3 - k^3 = (n - k)^3 + 3*n*k*(n - k), for n > = k >= 0.
Row sums give A126274(n-1) for n >= 1, and 0 for n=0.
Alternating row sums are ars(2*n) = ars(2*n-1) = (4*n-3)*n^2 = A103532(n-1), for n >= 1, and ars(0) = 0.

			The triangle T(n, k) begins:
  n\k    0   1   2   3   4   5   6   7   8   9  10
   0:    0
   1:    1   0
   2:    8   7   0
   3:   27  26  19   0
   4:   64  63  56  37   0
   5:  125 124 117  98  61   0
   6:  216 215 208 189 152  91   0
   7:  343 342 335 316 279 218 127   0
   8:  512 511 504 485 448 387 296 169   0
   9:  729 728 721 702 665 604 513 386 217   0
  10: 1000 999 992 973 936 875 784 657 488 271   0
  ...

Robert Israel, Table of n, a(n) for n = 0..10010 (rows 0 to 140, flattened)

Cf. A126274, A103532, A025581, A079904, A335821.

for n from 0 to 10 do seq(n^3-k^3,k=0..n) od; # Robert Israel, May 10 2018

Table[n^3-k^3,{n,0,10},{k,0,n}]//Flatten (* Harvey P. Dale, Jan 02 2021 *)

T(n, k) =  A025581(n, k)*(A025581(n, k)^2 + 3* A079904(n, k)) (see the identity mentioned in a comment).
Columns (with one leading zero and offset 0): k=0: l^3 = A000578(l), k=1: (l+1)^3 - 1 = A068601(l+1), k=2: l*(l^2 + 6*l + 12), k=3: l*(l^2 + 9*l + 27), k=4: l*(l^2 + 12*l + 48), k=5: l*(l^2 + 15*l + 75), ...
G.f. for T(n,k): (1+4*x+4*x*y+x^2-14*x^2*y+x^2*y^2-2*x^3*y-2*x^3*y^2+7*x^4*y^2)*x/((1-x*y)^3*(1-x)^4). - Robert Israel, May 10 2018

A340242 Square array read by upward antidiagonals: T(n,k) is the number of n-ary strings of length k containing 000.

Original entry on oeis.org

1, 1, 3, 1, 5, 8, 1, 7, 21, 20, 1, 9, 40, 81, 47, 1, 11, 65, 208, 295, 107, 1, 13, 96, 425, 1021, 1037, 238, 1, 15, 133, 756, 2621, 4831, 3555, 520, 1, 17, 176, 1225, 5611, 15569, 22276, 11961, 1121, 1, 19, 225, 1856, 10627, 40091, 90085, 100768, 39667, 2391

Offset: 2

Robert P. P. McKone, Jan 01 2021

			For n = 4 and k = 5, there are 40 strings: {00000, 00001, 00002, 00003, 00010, 00011, 00012, 00013, 00020, 00021, 00022, 00023, 00030, 00031, 00032, 00033, 01000, 02000, 03000, 10000, 10001, 10002, 10003, 11000, 12000, 13000, 20000, 20001, 20002, 20003, 21000, 22000, 23000, 30000, 30001, 30002, 30003, 31000, 32000, 33000}.
Square table T(n,k):
      k=3: k=4:  k=5:   k=6:    k=7:     k=8:
n=2:    1    3     8     20      47      107
n=3:    1    5    21     81     295     1037
n=4:    1    7    40    208    1021     4831
n=5:    1    9    65    425    2621    15569
n=6:    1   11    96    756    5611    40091
n=7:    1   13   133   1225   10627    88717
n=8:    1   15   176   1856   18425   175967
n=9:    1   17   225   2673   29881   321281

Robert P. P. McKone, Antidiagonals n = 2..100, flattened

Rows: A050231 (n=2), A231430 (n=3).
Columns: A000567 (k=5), A103532 (k=6).
Cf. A340156 (containing 00).
Cf. A341050.
Cf. A000073, A068601.

m[r_] := Normal[With[{p = 1/n}, SparseArray[{Band[{1, 2}] -> p, {i_, 1} /; i <= r -> 1 - p, {r + 1, r + 1} -> 1}]]];
T[n_, k_, r_] := MatrixPower[m[r], k][[1, r + 1]]*n^k;
Reverse[Table[T[n, k - n + 3, 3], {k, 2, 11}, {n, 2, k}], 2] // Flatten

my(x2='x^2+'x+1); T(n,k) = n^k - polcoeff(lift(x2*Mod('x, 'x^3-(n-1)*x2)^k), 2); \\ Kevin Ryde, Jan 02 2021

m(3) = [1 - 1/n, 1/n, 0, 0; 1 - 1/n, 0, 1/n, 0; 1 - 1/n, 0, 0, 1/n; 0, 0, 0, 1], is the probability/transition matrix for three consecutive "0" -> "containing 000".

A341050 Cube array read by upward antidiagonals ignoring zero and empty terms: T(n, k, r) is the number of n-ary strings of length k, containing r consecutive 0's.

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 3, 1, 5, 8, 1, 1, 3, 1, 5, 8, 1, 7, 21, 19, 1, 1, 3, 1, 5, 8, 1, 7, 21, 20, 1, 9, 40, 81, 43, 1, 1, 3, 1, 5, 8, 1, 7, 21, 20, 1, 9, 40, 81, 47, 1, 11, 65, 208, 295, 94, 1, 1, 3, 1, 5, 8, 1, 7, 21, 20, 1, 9, 40, 81, 48, 1, 11, 65, 208, 297, 107, 1, 13, 96, 425, 1024, 1037, 201

Offset: 2

Robert P. P. McKone, Feb 04 2021

			For n = 5, k = 6 and r = 4, there are 65 strings: {000000, 000001, 000002, 000003, 000004, 000010, 000011, 000012, 000013, 000014, 000020, 000021, 000022, 000023, 000024, 000030, 000031, 000032, 000033, 000034, 000040, 000041, 000042, 000043, 000044, 010000, 020000, 030000, 040000, 100000, 100001, 100002, 100003, 100004, 110000, 120000, 130000, 140000, 200000, 200001, 200002, 200003, 200004, 210000, 220000, 230000, 240000, 300000, 300001, 300002, 300003, 300004, 310000, 320000, 330000, 340000, 400000, 400001, 400002, 400003, 400004, 410000, 420000, 430000, 440000}
The first seven slices of the tetrahedron (or pyramid) are:
-----------------Slice 1-----------------
  1
-----------------Slice 2-----------------
    1
  1  3
-----------------Slice 3-----------------
      1
    1  3
  1  5  8
-----------------Slice 4-----------------
        1
      1  3
    1  5   8
  1  7  21  19
-----------------Slice 5-----------------
          1
        1  3
      1  5   8
    1  7  21  20
  1  9  40  81  43
-----------------Slice 6-----------------
              1
           1    3
        1    5     8
      1   7    21    20
    1   9   40    81    47
  1  11  65   208   295   94
-----------------Slice 7-----------------
                 1
              1     3
           1     5     8
         1    7     21    20
      1    9    40     81      48
    1   11   65    208     297     107
  1  13   96   425    1024    1037    201

Robert P. P. McKone, Antidiagonals n = 2..50, flattened

Cf. A340156 (r=2), A340242 (r=3).
Cf. A008466 (n=2, r=2), A186244 (n=3, r=2), A050231 (n=2, r=3), A231430 (n=3, r=3).
Cf. A005408, A003215, A005917, A022521, A022522, A022523, A022524, A022525, A022526, A022527, A022528, A022529, A022530, A022531, A022532, A022533, A022534, A022535, A022536, A022537, A022538, A022539, A022540 (k=x, r=1, where x is the x-th Nexus Number).
Cf. A000567 [(k=4, r=2),(k=5, r=3),(k=6, r=4),...,(k=x, r=x-2)].
Cf. A103532 [(k=6, r=3),(k=7, r=4),(k=8, r=5),...,(k=x, r=x-3)].

m[r_, n_] := Normal[With[{p = 1/n}, SparseArray[{Band[{1, 2}] -> p, {i_, 1} /; i <= r -> 1 - p, {r + 1, r + 1} -> 1}]]]; T[n_, k_, r_] := MatrixPower[m[r, n], k][[1, r + 1]]*n^k; DeleteCases[Transpose[PadLeft[Reverse[Table[T[n, k, r], {k, 2, 8}, {r, 2, k}, {n, 2, r}], 2]], 2 <-> 3], 0, 3] // Flatten

A268644 a(n) = 4n^3 - 3n^2 - 2*n - 1.

Original entry on oeis.org

-1, -2, 15, 74, 199, 414, 743, 1210, 1839, 2654, 3679, 4938, 6455, 8254, 10359, 12794, 15583, 18750, 22319, 26314, 30759, 35678, 41095, 47034, 53519, 60574, 68223, 76490, 85399, 94974, 105239, 116218, 127935, 140414, 153679, 167754, 182663, 198430, 215079, 232634, 251119

Offset: 0

Ilya Gutkovskiy, Feb 09 2016

In general, the ordinary generating function for the values of cubic polynomial p*n^3 + q*n^2 + k*n + m is (m + (p + q + k - 3*m)*x + (4*p - 2*k + 3*m)*x^2 + (p - q + k - m)*x^3)/(1 - x)^4.
Primes in this sequence: 199, 743, 15583, 105239, 435359, 620999, 770239, 853079, 1738423, 3511103, 7580119, 8737039, 10006063, ...
If a(n) is a positive prime then n is congruent to 0 or 4 (mod 6).

Ilya Gutkovskiy, Examples of the ordinary generating function for the values of cubic polynomialK
Eric Weisstein's World of Mathematics, Cubic Polynomial
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A005408, A056578, A103532, A131464.

[4*n^3-3*n^2-2*n-1: n in [0..40]]; // Vincenzo Librandi, Feb 10 2016

Table[4 n^3 - 3 n^2 - 2 n - 1, {n, 0, 40}]
LinearRecurrence[{4, -6, 4, -1}, {-1, -2, 15, 74}, 41]
CoefficientList[Series[(-1 + 2 x + 17 x^2 + 6 x^3) / (1 - x)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Feb 10 2016 *)

a(n)=4*n^3-3*n^2-2*n-1 \\ Charles R Greathouse IV, Jul 26 2016

G.f.: (-1 + 2*x + 17*x^2 + 6*x^3)/(1 - x)^4.
a(n) = A103532(n - 1) - A005408(n), n>0.
a(n) = 4*a(n - 1) - 6*a(n - 2) + 4*a(n - 3) - a(n - 4).
Sum_{n>=0} 1/a(n) = -1.407823506818026589265...
E.g.f.: exp(x)*(-1 - x + 9*x^2 + 4*x^3). - Stefano Spezia, Nov 17 2024

A380747 Array read by ascending antidiagonals: A(n,k) = [x^n] (1 - x)/(1 - k*x)^2.

Original entry on oeis.org

1, -1, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 8, 5, 1, 0, 1, 20, 21, 7, 1, 0, 1, 48, 81, 40, 9, 1, 0, 1, 112, 297, 208, 65, 11, 1, 0, 1, 256, 1053, 1024, 425, 96, 13, 1, 0, 1, 576, 3645, 4864, 2625, 756, 133, 15, 1, 0, 1, 1280, 12393, 22528, 15625, 5616, 1225, 176, 17, 1

Offset: 0

Stefano Spezia, Jan 31 2025

			The array begins as:
   1, 1,   1,    1,     1,     1, ...
  -1, 1,   3,    5,     7,     9, ...
   0, 1,   8,   21,    40,    65, ...
   0, 1,  20,   81,   208,   425, ...
   0, 1,  48,  297,  1024,  2625, ...
   0, 1, 112, 1053,  4864, 15625, ...
   0, 1, 256, 3645, 22528, 90625, ...
   ...

Cf. A000012 (k=1 or n=0), A000567 (n=2), A001792 (k=2), A007778, A060747 (n=1), A081038 (k=3), A081039 (k=4), A081040 (k=5), A081041 (k=6), A081042 (k=7), A081043 (k=8), A081044 (k=9), A081045 (k=10), A103532, A154955, A380748 (antidiagonal sums).

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

A(n,k) = ((k - 1)*n + k)*k^(n-1) with A(0,0) = 1.
A(n,k) = n! * [x^n] exp(k*x)*(1 + (k - 1)*x).
A(n,0) = A154955(n+1).
A(3,n) = A103532(n-1) for n > 0.
A(n,n) = A007778(n) for n > 0.

cp's OEIS Frontend

A262000 a(n) = n^2*(7*n - 5)/2.

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A257238 Triangle T(n, k) = n^3 - k^3, 0 <= k < = n.

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A340242 Square array read by upward antidiagonals: T(n,k) is the number of n-ary strings of length k containing 000.

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Mathematica

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A341050 Cube array read by upward antidiagonals ignoring zero and empty terms: T(n, k, r) is the number of n-ary strings of length k, containing r consecutive 0's.

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Mathematica

A268644 a(n) = 4*n^3 - 3*n^2 - 2*n - 1.

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Magma

Mathematica

PARI

Formula

A380747 Array read by ascending antidiagonals: A(n,k) = [x^n] (1 - x)/(1 - k*x)^2.

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Author

Keywords

Examples

Crossrefs

Programs

Mathematica

Formula

A262000 a(n) = n^2(7n - 5)/2.

A268644 a(n) = 4n^3 - 3n^2 - 2*n - 1.