cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

A350248 Triangle read by rows: T(n,k) is the number of noncrossing partitions of an n-set into k blocks of size 3 or more, n >= 0, 0 <= k <= floor(n/3).

Original entry on oeis.org

1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 3, 0, 1, 7, 0, 1, 12, 0, 1, 18, 12, 0, 1, 25, 45, 0, 1, 33, 110, 0, 1, 42, 220, 55, 0, 1, 52, 390, 286, 0, 1, 63, 637, 910, 0, 1, 75, 980, 2275, 273, 0, 1, 88, 1440, 4900, 1820, 0, 1, 102, 2040, 9520, 7140, 0, 1, 117, 2805, 17136, 21420, 1428
Offset: 0

Views

Author

Andrew Howroyd and Janaka Rodrigo, Dec 21 2021

Keywords

Examples

			Triangle begins:
  1;
  0;
  0;
  0, 1;
  0, 1;
  0, 1;
  0, 1,   3;
  0, 1,   7;
  0, 1,  12;
  0, 1,  18,   12;
  0, 1,  25,   45;
  0, 1,  33,  110;
  0, 1,  42,  220,   55;
  0, 1,  52,  390,  286;
  0, 1,  63,  637,  910;
  0, 1,  75,  980, 2275,  273;
  0, 1,  88, 1440, 4900, 1820;
  0, 1, 102, 2040, 9520, 7140;
  ...

Links

Crossrefs

Columns k=2..5 are A055998, A350116, A350286, A350303.
Row sums are A114997.
Cf. A001263 (blocks of any size), A108263 (blocks of size 2 or more).

Programs

  • PARI
    T(n)={my(p=1+O(x^3)); for(i=1, n\3, p=1+y*(x*p)^3/(1-x*p)); [Vecrev(t)| t<-Vec(p + O(x*x^n))]}
{my(A=T(12)); for(i=1, #A, print(A[i]))}
  • PARI
    T(n,k) = if(n==0 || k>n\3, k==0, binomial(n+1, n-k+1) * binomial(n-2*k-1, k-1) / (n+1)) \\ Andrew Howroyd, Dec 31 2021

Formula

G.f.: A(x,y) satisfies A(x,y) = 1 + y*(x*A(x,y))^3/(1 - x*A(x,y)).
T(n,k) = binomial(n+1, n-k+1) * binomial(n-2*k-1, k-1) / (n+1) for n > 0.

A350286 Number of different ways to partition the set of vertices of a convex (n+11)-gon into 4 nonintersecting polygons.

Original entry on oeis.org

0, 55, 286, 910, 2275, 4900, 9520, 17136, 29070, 47025, 73150, 110110, 161161, 230230, 322000, 442000, 596700, 793611, 1041390, 1349950, 1730575, 2196040, 2760736, 3440800, 4254250, 5221125, 6363630, 7706286, 9276085, 11102650, 13218400, 15658720, 18462136, 21670495, 25329150
Offset: 0

Views

Author

Janaka Rodrigo, Dec 23 2021

Keywords

Comments

Equivalently, the number of noncrossing set partitions of an (n+11)-set into 4 blocks with 3 or more elements in each block.

Examples

			a(1) = 55; solutions are {1,2,3} {4,5,6} {7,8,9} {10,11,12} with 3 different orientations, {1,2,3} {4,5,6} {11,12,7} {8,9,10} with 12 different orientations, {1,2,3} {12,4,5} {11,6,7} {8,9,10} with 12 different orientations, {1,2,3} {12,4,5} {10,11,6} {7,8,9} with 12 different orientations, {1,2,3} {4,5,6} {12,7,8} {9,10,11} with 12 orientations and {1,2,3} {4,8,12} {5,6,7} {9,10,11} with 4 orientations.
The above numbers can be considered to be the partition of a 12-set into 4 blocks or the partition of the vertices of a convex 12-gon into 4 triangles with vertices labeled 1,2,3,...,12 in order.
a(2) = 286 corresponding to the number of different ways to partition the vertices of a 13-gon into three triangles and one quadrilateral.

Links

Crossrefs

Column k=4 of A350248.
Cf. A350116.

Programs

  • Mathematica
    a[n_] := n*(n + 1)*(n + 2)*(n + 9)*(n + 10)*(n + 11)/144; Array[a, 35, 0] (* Amiram Eldar, Dec 26 2021 *)

Formula

a(n) = (n*(n+1)*(n+2)*(n+9)*(n+10)*(n+11))/144.
G.f.: x*(55 - 99*x + 63*x^2 - 14*x^3)/(1 - x)^7. - Stefano Spezia, Dec 26 2021

A347862 Total number of polygons left out in all partitions of the set of vertices of a convex n-gon into nonintersecting polygons.

Original entry on oeis.org

0, 0, 0, 3, 7, 12, 39, 105, 231, 577, 1482, 3549, 8603, 21340, 52122, 126777, 310859, 761199, 1859014, 4549215, 11141085, 27266225, 66760855, 163567911, 400786617, 982265827, 2408361144, 5906499136, 14489105190, 35553445788, 87264949808, 214241203801
Offset: 3

Views

Author

Janaka Rodrigo, Jan 24 2022

Keywords

Examples

			a(3) = a(4) = a(5) = 0 since the only partition of the vertices of a triangle, quadrilateral or pentagon into polygons is the full polygon so nothing is left out.
a(6) = 3 since the vertices of a hexagon can be partitioned into two non-intersecting triangles in A350248(6,2) = 3 ways and in each of these cases a quadrilateral is left over.
When partitioning the set of vertices of a convex 13-gon into 1 polygon, the number of polygons remaining is 0.
When partitioning it into 2 polygons, the remaining polygons are 52 quadrilaterals.
When partitioning it into 3 polygons, the remaining polygons are 65 hexagons + 650 quadrilaterals.
When partitioning it into 4 polygons, the remaining polygons are 13 octagons + 117 hexagons + 585 quadrilaterals.
This gives the total as 1482 polygons.

Crossrefs

Partitioning into 3 polygons A350116.
Total number of different ways to partition the set of vertices of a convex polygon into nonintersecting polygons A350248.

Programs

  • PARI
    seq(n)={my(p=O(x)); while(serprec(p,x)<=n, p = x + x*y*(1/(1 - x*p^2/(1 - p)) - 1)); Vec(subst(deriv(O(x*x^n) + p^3/(1-p), y), y, 1), 2-n) } \\ Andrew Howroyd, Jan 30 2022

Extensions

More terms from Andrew Howroyd, Jan 30 2022

A350303 a(n) is the number of ways to partition the set of vertices of a convex (n+14)-gon into 5 nonintersecting polygons.

Original entry on oeis.org

0, 273, 1820, 7140, 21420, 54264, 122094, 251370, 482790, 876645, 1519518, 2532530, 4081350, 6388200, 9746100, 14535612, 21244356, 30489585, 43044120, 59865960, 82131896, 111275472, 149029650, 197474550, 259090650, 336817845, 434120778, 555060870, 704375490, 887564720, 1110986184
Offset: 0

Views

Author

Janaka Rodrigo, Dec 24 2021

Keywords

Comments

Equivalently, the number of noncrossing set partitions of an (n+14)-set into 5 blocks with 3 or more elements in each block.

Examples

			The a(1)=273 solutions are {1,2,3} {4,5,6} {7,8,9} {10,11,12} {13,14,15} with its 3 different orientations and each of the following 18 patterns with its 15 orientations:
  {1,2,3} {4,5,15}  {6,7,8}   {9,10,11} {12,13,14}
  {1,2,3} {4,14,15} {5,6,7}   {8,9,10}  {11,12 13}
  {1,2,3} {4,5,6}   {7,8,15}  {9,10,11} {12 13,14}
  {1,2,3} {4,5,15}  {6,7,14}  {8,9,10}  {11,12,13}
  {1,2,3} {4,14,15} {5,12,13} {6,7,8}   {9,10,11}
  {1,2,3} {4,5,15}  {6,13,14} {7,8,9}   {10,11,12}
  {1,2,3} {4,14,15} {5,6,13}  {7,8,9}   {10,11,12}
  {1,2,3} {4,5,15}  {6,7,14}  {8,9,13}  {10,11,12}
  {1,2,3} {4,5,15}  {6,7,14}  {8,12,13} {9,10,11}
  {1,2,3} {4,5,15}  {6,13,14} {7,8,12}  {9,10,11}
  {1,2,3} {4,14,15} {5,12,13} {6,7,11}  {8,9,10}
  {1,2,3} {4,15,8}  {5,6,7}   {9,10,11} {12,13,14}
  {1,2,3} {4,15,8}  {5,6,7}   {9,13,14} {10,11,12}
  {1,2,3} {4,15,8}  {5,6,7}   {9,10,14} {11,12,13}
  {1,2,3} {4,5,15}  {6,7,8}   {9,10,14} {11,12,13}
  {1,2,3} {4,14,15} {5,6,7}   {8,12,13} {9,10,11}
  {1,2,3} {4,14,15} {5,6,7}   {8,9,13}  {10,11,12}
  {1,2,3} {4,5,15}  {6,7,8}   {9,13,14} {10,11,12}
In the above, the numbers can be considered to be the partition of a 15-set into 5 blocks or the partition of the vertices of a convex 15-gon into 5 triangles with vertices labeled 1,2,...,15 in order.
a(2)=1820 corresponding to the number of ways to partition the vertices of a 16-gon into 4 triangles and one quadrilateral.

Links

Crossrefs

Column k=5 of A350248.
Cf. A350116.

Programs

  • Mathematica
    a[n_] := n*(n + 1)*(n + 2)*(n + 3)*(n + 11)*(n + 12)*(n + 13)*(n + 14)/2880; Array[a, 30, 0] (* Amiram Eldar, Dec 26 2021 *)

Formula

a(n) = (1/2880)*n*(n+1)*(n+2)*(n+3)*(n+11)*(n+12)*(n+13)*(n+14).
G.f.: 7*x*(39 - 91*x + 84*x^2 - 36*x^3 + 6*x^4)/(1 - x)^9. - Stefano Spezia, Dec 26 2021

A352477 a(n) is the number of different ways to partition the set of vertices of a convex n-gon into 4 intersecting polygons.

Original entry on oeis.org

15345, 199914, 1610700, 10333050, 57958005, 297787980, 1439757200, 6662364668, 29844152321, 130445708134, 559533869356, 2365296230374, 9885290683829, 40944327268760, 168389163026240, 688631375953560, 280357073972529, 11373212442818370, 46006062638648940
Offset: 12

Views

Author

Janaka Rodrigo, Mar 17 2022

Keywords

Examples

			The set of vertices of a convex 14-gon can be partitioned into 4 polygons in 1611610 different ways:
- 3 triangles and 1 pentagon in (1/3!)*C(14,3)*C(11,3)*C(8,3)*C(5,5) = 560560 different ways, and
- 2 triangles and 2 quadrilaterals in (1/2!)*(1/2!)*C(14,3)*C(11,3)*C(8,4)*C(4,4) = 1051050 ways.
Subtracting the A350286(14-11)=910 nonintersecting partitions leaves a(14)=1610700.

Crossrefs

Cf. A261724, A350116, A350286.

Programs

  • PARI
    a4(n) = (1/12)*(-3^(n - 2)*(n^2 + 5*n + 18) + (1/64)*(2^(2*n + 5) + 3*2^n*(n^4 + 2*n^3 + 19*n^2 + 42*n + 64) - 16*(n^6 - 9*n^5 + 43*n^4 - 91*n^3 + 112*n^2 - 32*n + 8))); \\ A261724
a6(n) = (n*(n+1)*(n+2)*(n+9)*(n+10)*(n+11))/144; \\ A350286
a(n) = a4(n) - a6(n-11); \\ Michel Marcus, Mar 20 2022

Formula

a(n) = A261724(n) - A350286(n - 11), n > 11.

A350599 Number of ways to partition the set of vertices of a convex n-gon into non-intersecting directed polygons.

Original entry on oeis.org

2, 2, 2, 14, 30, 50, 170, 462, 1014, 2810, 7906, 19910, 53278, 148514, 397530, 1073918, 2976390, 8172426, 22413266, 62219830, 172846382, 479683762, 1338281802, 3743620974, 10475828630, 29389158426, 82643684034, 232644515366, 655928162878, 1852640651330, 5239096953274
Offset: 3

Views

Author

Janaka Rodrigo, Jan 08 2022

Keywords

Comments

A directed polygon is a polygon with an associated direction (clockwise or counterclockwise).
Equivalently, the polygons can be colored using two colors. - Andrew Howroyd, Jan 09 2022

Examples

			a(7) = 2 + 28 = 30 since the 7-gon can be given two directions and the 7-gon can also be partitioned into a triangle and a quadrilateral in 7 different ways giving another 7 * 4 = 28 possibilities.

Crossrefs

Cf. A350116, A350248.

Programs

  • Mathematica
    a[n_] := Sum[2^k * Binomial[n + 1, k] * Binomial[n - 2*k - 1, k - 1]/(n + 1), {k, 1, Floor[n/3]}]; Array[a, 30, 3] (* Amiram Eldar, Jan 08 2022 *)
  • PARI
    a(n) = sum(k=1, n\3, 2^k * binomial(n+1, k) * binomial(n-2*k-1, k-1)) / (n+1) \\ Andrew Howroyd, Jan 08 2022

Formula

a(n) = Sum_{k=1..floor(n/3)} 2^k * binomial(n+1, k) * binomial(n-2*k-1, k-1) / (n+1).
a(n) = Sum_{k=1..floor(n/3)} 2^k * A350248(n,k). - Andrew Howroyd, Jan 09 2022
The compositional inverse of x+Sum_{k=1..infinity} a_k x^{k+1} is x(1-x)/(1+x)(1-2x+x^2). Proved at MathOverflow 418996.

A352474 a(n) is the number of different ways to partition the set of vertices of a convex n-gon into 3 intersecting polygons.

Original entry on oeis.org

268, 2055, 10285, 42515, 157911, 548912, 1826846, 5902458, 18679974, 58255005, 179762211, 550473301, 1676299353, 5083919214, 15372833564, 46383749572, 139730014800, 420448279875, 1264071072745, 3798101946855, 11406989330923, 34248214094780
Offset: 9

Views

Author

Janaka Rodrigo, Mar 17 2022

Keywords

Examples

			The set of vertices of a convex 11-gon can be partitioned into 3 polygons in 10395 different ways:
- as 2 triangles and 1 pentagon ((1/2!)*C(11,3)*C(8,3)*C(5,5) = 4620 different ways) or
- as 1 triangle and 2 quadrilaterals ((1/2!)*C(11,3)*C(8,4)*C(4,4) = 5775 different ways).
Subtracting the A350116(11-8) = 110 nonintersecting partitions leaves a(11)=10285.

Links

Crossrefs

Cf. A272982, A350116, A352477.

Programs

  • PARI
    b(n) = if (n==8, 0, 3*b(n-1)+binomial(n-1,2)*(2^(n-4)+2-n-binomial(n-3,2)));
a(n) = b(n) - n*(n-1)*(n-7)*(n-8)/12; \\ Michel Marcus, Mar 19 2022

Formula

a(n) = b(n) - n*(n-1)*(n-7)*(n-8)/12, where b(n) = 3*b(n-1)+C(n-1,2)*(2^(n-4)+2-n-C(n-3,2)) for n > 8 and b(8) = 0. b(n) is given in A272982.
a(n) = A272982(n) - A350116(n-8).
G.f.: x^9*(268 - 1697*x + 4295*x^2 - 5592*x^3 + 4008*x^4 - 1520*x^5 + 240*x^6)/((1 - x)^5*(1 - 2*x)^3*(1 - 3*x)). - Stefano Spezia, Mar 19 2022
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