A256859 -id:A256859 - cp's OEIS frontend

A257055 a(n) = n(n + 1)(n^2 - n + 3)/6.

0, 1, 5, 18, 50, 115, 231, 420, 708, 1125, 1705, 2486, 3510, 4823, 6475, 8520, 11016, 14025, 17613, 21850, 26810, 32571, 39215, 46828, 55500, 65325, 76401, 88830, 102718, 118175, 135315, 154256, 175120, 198033, 223125, 250530, 280386, 312835, 348023, 386100

Offset: 0

Bruno Berselli, Apr 15 2015

Partial sums of A037235.
After 0, this sequence is the 2nd diagonal of the square array in A080851.
For n > 2, a(n)-n is the 4th column of the triangular array in A208657.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000332, A037235, A080851, A208657.

Cf. similar sequences listed in A256859.

Magma
```
[n*(n+1)*(n^2-n+3)/6: n in [0..40]];
```

Table[n (n + 1) (n^2 - n + 3)/6, {n, 40}]

vector(40, n, n--; n*(n+1)*(n^2-n+3)/6)

Sage
```
[n*(n+1)*(n^2-n+3)/6 for n in (0..40)]
```

G.f.: x*(1 + 3*x^2)/(1 - x)^5.
a(n) = 3*A000332(n+1) + A000332(n+3).
a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5). - Wesley Ivan Hurt, May 27 2021

A261721 Fourth-dimensional figurate numbers.

Original entry on oeis.org

1, 1, 5, 1, 6, 15, 1, 7, 20, 35, 1, 8, 25, 50, 70, 1, 9, 30, 65, 105, 126, 1, 10, 35, 80, 140, 196, 210, 1, 11, 40, 95, 175, 266, 336, 330, 1, 12, 45, 110, 210, 336, 462, 540, 495, 1, 13, 50, 125, 245, 406, 588, 750, 825, 715, 1, 14, 55, 140, 280, 476, 714, 960, 1155, 1210, 1001, 1

Offset: 1

Gary W. Adamson, Aug 30 2015

Generating polygons for the sequences are: Triangle, Square, Pentagon, Hexagon, Heptagon, Octagon, ... .
n-th row sequence is the binomial transform of the fourth row of Pascal's triangle (1,n) followed by zeros; and the fourth partial sum of (1, n, n, n, ...).
n-th row sequence is the binomial transform of:
((n-1) * (0, 1, 3, 3, 1, 0, 0, 0) + (1, 4, 6, 4, 1, 0, 0, 0)).
Given the n-th row of the array (1, b, c, d, ...), the next row of the array is (1, b, c, d, ...) + (0, 1, 5, 15, 35, ...)

			The array as shown in A257200:
  1,  5, 15,  35,  70, 126, 210,  330, ... A000332
  1,  6, 20,  50, 105, 196, 336,  540, ... A002415
  1,  7, 25,  65, 140, 266, 462,  750, ... A001296
  1,  8, 30,  80, 175, 336, 588,  960, ... A002417
  1,  9, 35,  95, 210, 406, 714, 1170, ... A002418
  1, 10, 40, 110, 245, 476, 840, 1380, ... A002419
  ...
(1, 7, 25, 65, 140, ...) is the third row of the array and is the binomial transform of the fourth row of Pascal's triangle (1,3) followed by zeros: (1, 6, 12, 10, 3, 0, 0, 0, ...); and the fourth partial sum of (1, 3, 3, 3, 0, 0, 0).
(1, 7, 25, 65, 140, ...) is the third row of the array and is the binomial transform of: (2 * (0, 1, 3, 3, 1, 0, 0, 0, ...) + (1, 4, 6, 4, 1, 0, 0, 0, ...)); that is, the binomial transform of (1, 6, 12, 10, 3, 0, 0, 0, ...).
Row 2 of the array is (1, 5, 15, 35, 70, ...) + (0, 1, 5, 15, 35, ...), = (1, 6, 20, 50, 105, ...).

Albert H. Beiler, "Recreations in the Theory of Numbers"; Dover, 1966, p. 195 (Table 80).

Alois P. Heinz, Antidiagonals n = 1..141, flattened
Index to sequences related to pyramidal numbers

Cf. A257200, A261720 (pyramidal numbers), A000332, A002415, A001296, A002417, A002418, A002419.
Similar to A080852 but without row n=0.
Main diagonal gives A256859.

A:= (n, k)-> binomial(k+3, 3) + n*binomial(k+3, 4):
seq(seq(A(d-k, k), k=0..d-1), d=1..13);  # Alois P. Heinz, Aug 31 2015

row[1] = LinearRecurrence[{5, -10, 10, -5, 1}, {1, 5, 15, 35, 70}, m = 10];
row1 = Join[{0}, row[1] // Most]; row[n_] := row[n] = row[n-1] + row1;
Table[row[n-k+1][[k]], {n, 1, m}, {k, 1, n}] // Flatten (* Jean-François Alcover, May 26 2016 *)

A(n, k) = binomial(k+3, 3) + n*binomial(k+3, 4)
table(n, k) = for(x=1, n, for(y=0, k-1, print1(A(x, y), ", ")); print(""))
/* Print initial 6 rows and 8 columns as follows: */
table(6, 8) \\ Felix Fröhlich, Jul 28 2016

G.f. for row n: (1 + (n-1)*x)/(1 - x)^5.
A(n,k) = C(k+3,3) + n * C(k+3,4) = A080852(n,k).
E.g.f. as array: exp(y)*(exp(x)*(24 + 24*(3 + x)*y + 36*(1 + x)*y^2 + 4*(1 + 3*x)*y^3 + x*y^4) - 4*(6 + 18*y + 9*y^2 + y^3))/24. - Stefano Spezia, Aug 15 2024

A256860 a(n) = n(n + 1)(n + 2)(n + 3)(n^2 - n + 5)/120.

Original entry on oeis.org

1, 7, 33, 119, 350, 882, 1974, 4026, 7623, 13585, 23023, 37401, 58604, 89012, 131580, 189924, 268413, 372267, 507661, 681835, 903210, 1181510, 1527890, 1955070, 2477475, 3111381, 3875067, 4788973, 5875864, 7161000, 8672312, 10440584, 12499641, 14886543

Offset: 1

Luciano Ancora, Apr 14 2015

This is the case k = n of b(n,k) = n*(n+1)*(n+2)*(n+3)*(k*(n-1)+5)/120, where b(n,k) is the n-th hypersolid number in 5 dimensions generated from an arithmetical progression with the first term 1 and common difference k (see Sardelis et al. paper).

D. A. Sardelis and T. M. Valahas, On Multidimensional Pythagorean Numbers, arXiv:0805.4070v1 [math.GM], 2008.
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A000579.

Cf. similar sequences listed in A256859.

Table[n (n + 1) (n + 2) (n + 3) (n^2 - n + 5)/120, {n, 40}]

vector(40, n, n*(n+1)*(n+2)*(n+3)*(n^2-n+5)/120) \\ Bruno Berselli, Apr 15 2015

G.f.: x*(1 + 5*x^2)/(1 - x)^7.

a(n) = 5*A000579(n+3) + A000579(n+5). [Bruno Berselli, Apr 15 2015]

A256861 a(n) = n(n + 1)(n + 2)(n + 3)(n + 4)*(n^2 - n + 6)/720.

Original entry on oeis.org

1, 8, 42, 168, 546, 1512, 3696, 8184, 16731, 32032, 58058, 100464, 167076, 268464, 418608, 635664, 942837, 1369368, 1951642, 2734424, 3772230, 5130840, 6888960, 9140040, 11994255, 15580656, 20049498, 25574752, 32356808, 40625376, 50642592, 62706336

Offset: 1

Luciano Ancora, Apr 14 2015

This is the case k = n of b(n,k) = n*(n+1)*(n+2)*(n+3)*(n+4)*(k*(n-1)+6)/120, where b(n,k) is the n-th hypersolid number in 6 dimensions generated from an arithmetical progression with the first term 1 and common difference k (see Sardelis et al. paper).

D. A. Sardelis and T. M. Valahas, On Multidimensional Pythagorean Numbers, arXiv:0805.4070 [math.GM], 2008.
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Cf. A000580.

Cf. similar sequences listed in A256859.

Table[n (1 + n) (2 + n) (3 + n) (4 + n) (6 - n + n^2)/720, {n, 40}]
Table[Times@@(n+Range[0,4])(n^2-n+6)/720,{n,40}] (* or *) LinearRecurrence[ {8,-28,56,-70,56,-28,8,-1},{1,8,42,168,546,1512,3696,8184},40] (* Harvey P. Dale, Sep 25 2019 *)

vector(40, n, n*(n+1)*(n+2)*(n+3)*(n+4)*(n^2-n+6)/720) \\ Bruno Berselli, Apr 15 2015

G.f.: x*(1 + 6*x^2)/(1 - x)^8.

a(n) = 6*A000580(n+4) + A000580(n+6). [Bruno Berselli, Apr 15 2015]

A301973 a(n) = (n^2 - 3n + 6)binomial(n+2,3)/4.

Original entry on oeis.org

0, 1, 4, 15, 50, 140, 336, 714, 1380, 2475, 4180, 6721, 10374, 15470, 22400, 31620, 43656, 59109, 78660, 103075, 133210, 170016, 214544, 267950, 331500, 406575, 494676, 597429, 716590, 854050, 1011840, 1192136, 1397264, 1629705, 1892100, 2187255, 2518146, 2887924, 3299920, 3757650, 4264820

Offset: 0

Ilya Gutkovskiy, Mar 29 2018

For n > 2, a(n) is the n-th term of the partial sums of n-gonal pyramidal numbers (in other words, a(n) is the n-th 4-dimensional n-gonal number).

Index to sequences related to pyramidal numbers
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A000292, A000332, A001296, A002415, A006484, A060354, A080852, A256859, A291288 (partial sums), A292551.

Table[(n^2 - 3 n + 6) Binomial[n + 2, 3]/4, {n, 0, 40}]
nmax = 40; CoefficientList[Series[x (1 - 2 x + 6 x^2)/(1 - x)^6, {x, 0, nmax}], x]
nmax = 40; CoefficientList[Series[Exp[x] x (24 + 24 x + 24 x^2 + 10 x^3 + x^4)/24, {x, 0, nmax}], x] Range[0, nmax]!
Table[SeriesCoefficient[x (1 - 3 x + n x)/(1 - x)^5, {x, 0, n}], {n, 0, 40}]
LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 1, 4, 15, 50, 140}, 41]

O.g.f.: x*(1 - 2*x + 6*x^2)/(1 - x)^6.
E.g.f.: exp(x)*x*(24 + 24*x + 24*x^2 + 10*x^3 + x^4)/24.
a(n) = [x^n] x*(1 - 3*x + n*x)/(1 - x)^5.
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).

cp's OEIS Frontend

A257055 a(n) = n*(n + 1)*(n^2 - n + 3)/6.

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A261721 Fourth-dimensional figurate numbers.

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A256860 a(n) = n*(n + 1)*(n + 2)*(n + 3)*(n^2 - n + 5)/120.

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A256861 a(n) = n*(n + 1)*(n + 2)*(n + 3)*(n + 4)*(n^2 - n + 6)/720.

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A301973 a(n) = (n^2 - 3*n + 6)*binomial(n+2,3)/4.

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A257055 a(n) = n(n + 1)(n^2 - n + 3)/6.

A256860 a(n) = n(n + 1)(n + 2)(n + 3)(n^2 - n + 5)/120.

A256861 a(n) = n(n + 1)(n + 2)(n + 3)(n + 4)*(n^2 - n + 6)/720.

A301973 a(n) = (n^2 - 3n + 6)binomial(n+2,3)/4.