A033488 -id:A033488 - cp's OEIS frontend

A343200 Expansion of Product_{k>=1} (1 + x^k)^binomial(k+3,3).

1, 4, 16, 64, 221, 736, 2338, 7132, 21093, 60652, 170172, 467140, 1257571, 3325824, 8654576, 22189340, 56116043, 140122760, 345769094, 843827436, 2038017983, 4874329024, 11550814704, 27134195608, 63215468883, 146120097736, 335227455982, 763592477104, 1727482413548

Offset: 0

Ilya Gutkovskiy, May 09 2021

nonn

Vaclav Kotesovec, Table of n, a(n) for n = 0..7650

Cf. A000292, A033488, A219555, A255050, A258343, A338645, A344097, A344098.

nmax = 28; CoefficientList[Series[Product[(1 + x^k)^Binomial[k + 3, 3], {k, 1, nmax}], {x, 0, nmax}], x]
a[n_] := a[n] = If[n == 0, 1, (1/n) Sum[Sum[(-1)^(k/d + 1) d Binomial[d + 3, 3], {d, Divisors[k]}] a[n - k], {k, 1, n}]]; Table[a[n], {n, 0, 28}]

a(0) = 1; a(n) = (1/n) * Sum_{k=1..n} ( Sum_{d|k} (-1)^(k/d+1) * A033488(d) ) * a(n-k).

a(n) ~ (3*zeta(5))^(1/10) / (2^(7/10) * 5^(2/5) * sqrt(Pi) * n^(3/5)) * exp(-469*log(2)/720 - 2401*Pi^16 / (656100000000*zeta(5)^3) + 539*Pi^8*zeta(3) / (8100000*zeta(5)^2) - 7*Pi^6 / (27000*zeta(5)) - 121*zeta(3)^2 / (600*zeta(5)) + (343*Pi^12 / (303750000 * 2^(3/5) * 15^(1/5) * zeta(5)^(11/5)) - 77*Pi^4*zeta(3) / (4500 * 2^(3/5) * 15^(1/5) * zeta(5)^(6/5)) + Pi^2 / (6*2^(3/5) * (15*zeta(5))^(1/5))) * n^(1/5) + (-49*Pi^8 / (270000 * 2^(1/5) * 15^(2/5) * zeta(5)^(7/5)) + 11*zeta(3) / (4*2^(1/5) * (15*zeta(5))^(2/5))) * n^(2/5) + (7*Pi^4 / (90*2^(4/5) * (15*zeta(5))^(3/5))) * n^(3/5) + (5*(15*zeta(5))^(1/5) / (4*2^(2/5))) * n^(4/5)). - Vaclav Kotesovec, May 12 2021

A174002 a(n) = n*binomial(n+4, 4).

Original entry on oeis.org

0, 5, 30, 105, 280, 630, 1260, 2310, 3960, 6435, 10010, 15015, 21840, 30940, 42840, 58140, 77520, 101745, 131670, 168245, 212520, 265650, 328900, 403650, 491400, 593775, 712530, 849555, 1006880, 1186680, 1391280, 1623160, 1884960, 2179485

Offset: 0

Reinhard Zumkeller, Mar 05 2010, Mar 17 2010

This sequence can be computed from Pascal's triangle. Find the fifth number in a row and multiply it by the second number of the next row. - Alonso del Arte, Jan 21 2018

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

Cf. A033488, A027480, A003506.

[ (n^5+10*n^4+35*n^3+50*n^2+24*n)/24: n in [0..40] ]; // Vincenzo Librandi, Dec 28 2010

Table[n Binomial[n + 4, 4], {n, 0, 40}] (* or *) LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 5, 30, 105, 280, 630}, 40] (* Harvey P. Dale, Dec 03 2011 *)

a(n) = (n^5 + 10*n^4 + 35*n^3 + 50*n^2 + 24*n) / 24.
For n > 0: a(n) = A003506(n+4, 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), with a(0)=0, a(1)=5, a(2)=30, a(3)=105, a(4)=280, a(5)=630. - Harvey P. Dale, Dec 03 2011
G.f.: 5*x/(1-x)^6. - Colin Barker, Mar 18 2012

Title switched with first Formula section entry, at the suggestion of Alonso del Arte, by Jon E. Schoenfield, Jan 28 2018

A363174 Array read by rows: T(n,k) is the number of triangles inside a regular n-gon formed by intersecting line segments, considering all configurations of 3 line segments from k distinct vertices, with n >= 3, 3 <= k <= 6.

Original entry on oeis.org

1, 0, 0, 0, 4, 4, 0, 0, 10, 20, 5, 0, 20, 60, 30, 0, 35, 140, 105, 7, 56, 280, 280, 16, 84, 504, 630, 84, 120, 840, 1260, 180, 165, 1320, 2310, 462, 220, 1980, 3960, 796, 286, 2860, 6435, 1716, 364, 4004, 10010, 2856, 455, 5460, 15015, 5005, 560, 7280, 21840, 7744

Offset: 3

Paolo Xausa, May 19 2023

See Sommars and Sommars (1998) for a complete analysis of the problem.

			Array begins:
  n\k|     3     4     5     6
  ---+---------------------------
   3 |     1,    0,    0,    0;
   4 |     4,    4,    0,    0;
   5 |    10,   20,    5,    0;
   6 |    20,   60,   30,    0;
   7 |    35,  140,  105,    7;
   8 |    56,  280,  280,   16;
   9 |    84,  504,  630,   84;
  10 |   120,  840, 1260,  180;
  ...

Paolo Xausa, Table of n, a(n) for n = 3..10002 (rows 3..2502 of array, flattened).
Bjorn Poonen and Michael Rubinstein, The number of intersection points made by the diagonals of a regular polygon, arXiv:math/9508209 [math.MG], 1995-2006.
Steven E. Sommars and Tim Sommars, The Number of Triangles Formed by Intersecting Diagonals of a Regular Polygon, Journal of Integer Sequences, Vol. 1 (1998), Article 98.1.5.
Paolo Xausa, Illustration of T(9,6).
Sequences formed by drawing all diagonals in regular polygon

Cf. A000579, A006561, A006600 (row sums), A260417.

Cf. A000292 (column k = 3), A033488 (column k = 4), A174002 (column k = 5), A363173 (column k = 6).

A363174list[rowmax_]:=Module[{d},d[m_,n_]:=Boole[Divisible[n,m]];Table[Binomial[n,k]If[4<=k<=5,k,1]-If[k==6&&EvenQ[n],((1/8n^2-9/8n+7/4)d[2,n]+3/4d[4,n]+(6n-106/3)d[6,n]-33d[12,n]-36d[18,n]-24d[24,n]+96d[30,n]+72d[42,n]+264d[60,n]+96d[84,n]+48d[90,n]+96d[120,n]+48d[210,n])n,0],{n,3,rowmax},{k,3,6}]];A363174list[20]

T(n,3) = binomial(n,3) = A000292(n-2).
T(n,4) = 4*binomial(n,4) = A033488(n-3).
T(n,5) = 5*binomial(n,5) = A174002(n-4), for n >= 4.
T(n,6) = binomial(n,6) = A000579(n) if n is odd, A000579(n) - A260417(n/2) if n is even.
Sum_{k=3..6} T(n,k) = A006600(n).

A121547 Fourth slice along the 1-2-plane in the cube a(m,n,o) = a(m-1,n,o) + a(m,n-1,o) + a(m,n,o-1) for which the first slice is Pascal's triangle (slice read by antidiagonals).

Original entry on oeis.org

0, 0, 1, 0, 4, 4, 0, 10, 20, 10, 0, 20, 60, 60, 20, 0, 35, 140, 210, 140, 35, 0, 56, 280, 560, 560, 280, 56, 0, 84, 504, 1260, 1680, 1260, 504, 84, 0, 120, 840, 2520, 4200, 4200, 2520, 840, 120, 0, 165, 1320, 4620, 9240, 11550, 9240, 4620, 1320, 165, 0, 220, 1980, 7920, 18480, 27720, 27720, 18480, 7920, 1980, 220

Offset: 0

Thomas Wieder, Aug 06 2006

Essentially the same as triangle A178820 with an additional column of zeros at the left border. - Georg Fischer, Jul 31 2023

			The second row is 1, 4, 10, 20, 35, 56, 84, 120, 165, 220 = A000292, i.e., Tetrahedral (or pyramidal) numbers: binomial(n+2,3) = n(n+1)(n+2)/6 (core).
The third row is 4, 20, 60, 140, 280, 504, 840, 1320, 1980, 2860 = A033488 = n*(n+1)*(n+2)*(n+3)/6.
The main diagonal is 0, 4, 60, 560, 4200, 27720, 168168, 960960, 5250960, 27713400 = {0} U A002803*4.
Triangle starts:
  0
  0, 1
  0, 4, 4
  0, 10, 20, 10
  0, 20, 60, 60, 20
  0, 35, 140, 210, 140, 35
  0, 56, 280, 560, 560, 280, 56
  0, 84, 504, 1260, 1680, 1260, 504, 84

Cf. A000292, A003506, A007318, A094305, A121306, A119800, A178820.

T:=(n, k)->binomial(n+3, 3)*binomial(n, k): seq(print(seq(T(n-1, k-1), k=0..n)), n=0..10); # Georg Fischer, Jul 31 2023

a(m-1,n,o) + a(m,n-1,o) + a(m,n,o-1) with initialization values a(1,0,0) = 1 and a(m<>1=0, n>=0, 0>=o) = 0.

a(55)-a(56) corrected and more terms from Georg Fischer, Jul 31 2023

A291980 Triangle read by rows, T(n, k) = n![t^k] ([x^n] exp(xt)/(1 - log(1+x))) for 0<=k<=n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 3, 3, 1, 4, 8, 6, 4, 1, 14, 20, 20, 10, 5, 1, 38, 84, 60, 40, 15, 6, 1, 216, 266, 294, 140, 70, 21, 7, 1, 600, 1728, 1064, 784, 280, 112, 28, 8, 1, 6240, 5400, 7776, 3192, 1764, 504, 168, 36, 9, 1

Offset: 0

Peter Luschny, Sep 15 2017

			Triangle starts:
[1]
[1,      1]
[1,      2,    1]
[2,      3,    3,   1]
[4,      8,    6,   4,   1]
[14,    20,   20,  10,   5,   1]
[38,    84,   60,  40,  15,   6,  1]
[216,  266,  294, 140,  70,  21,  7, 1]
[600, 1728, 1064, 784, 280, 112, 28, 8, 1]

Row sums: A291981.
Columns: A006252 (c=1), A108125 (c=2).
Diagonals: A000217 (d=3), A007290 (d=4), A033488 (d=5).
Cf. A291978.

T_row := proc(n) exp(x*t)/(1 - log(1+x)): series(%, x, n+1):
seq(n!*coeff(coeff(%,x,n), t, k), k=0..n) end:
seq(T_row(n), n=0..10);

T[n_, k_] := Binomial[n, n - k]*Sum[j!*StirlingS1[n - k, j], {j, 0, n - k}]; Flatten[Table[T[n, k], {n, 0, 9}, {k, 0, n}]] (* Detlef Meya, May 12 2024 *)

T(n, k) = binomial(n, n - k)*Sum_{j=0..n - k} j!*Stirling1(n - k, j). - Detlef Meya, May 12 2024

cp's OEIS Frontend

A343200 Expansion of Product_{k>=1} (1 + x^k)^binomial(k+3,3).

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A174002 a(n) = n*binomial(n+4, 4).

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A363174 Array read by rows: T(n,k) is the number of triangles inside a regular n-gon formed by intersecting line segments, considering all configurations of 3 line segments from k distinct vertices, with n >= 3, 3 <= k <= 6.

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A121547 Fourth slice along the 1-2-plane in the cube a(m,n,o) = a(m-1,n,o) + a(m,n-1,o) + a(m,n,o-1) for which the first slice is Pascal's triangle (slice read by antidiagonals).

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A291980 Triangle read by rows, T(n, k) = n!*[t^k] ([x^n] exp(x*t)/(1 - log(1+x))) for 0<=k<=n.

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Maple

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A291980 Triangle read by rows, T(n, k) = n![t^k] ([x^n] exp(xt)/(1 - log(1+x))) for 0<=k<=n.