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 1-8 of 8 results.

A054514 Number of ways to place non-crossing diagonals in convex (n+4)-gon so as to create no triangles or quadrilaterals.

Original entry on oeis.org

1, 1, 1, 5, 10, 16, 45, 109, 222, 540, 1341, 3065, 7328, 18112, 43530, 105390, 260254, 639244, 1570257, 3893805, 9669236, 24014264, 59903650, 149806494, 374982790, 940835404, 2365679689, 5955973237, 15018854005, 37935575685, 95942896837, 242954350457, 616034170069, 1563810857705, 3974000543475
Offset: 1

Views

Author

Len Smiley, Apr 08 2000

Keywords

Examples

			a(4)=5 because the octagon has the null placement and four ways to place a single diagonal.
		

Crossrefs

Programs

  • Mathematica
    f[x_] = InverseSeries[Series[(y - y^2 - y^4)/(1 - y), {y, 0, 38}], x];
    CoefficientList[(f[x] - x)/x^4, x]
    (* Second program: *)
    a[n_] := Sum[Binomial[n-2j-1, n-3j-1] Binomial[n+3+j, n+2]/(n+3), {j, 0, (n-1)/3}]; Array[a, 35] (* Jean-François Alcover, Dec 08 2018, after David Callan *)
    Table[HypergeometricPFQ[{1/3 - n/3, 2/3 - n/3, 1 - n/3, 4 + n}, {2, 1/2 - n/2, 1 - n/2}, -27/4], {n, 1, 40}] (* Vaclav Kotesovec, Sep 16 2023 *)

Formula

a(n) = Sum_{j=0..(n-1)/3} binomial[n-2j-1, n-3j-1] binomial[n+3+j, n+2]/(n+3). This counts the polygon dissections above by number j of diagonals. - David Callan, Jul 15 2004

Extensions

More terms from Joerg Arndt, Jan 28 2014

A348479 Number of interval posets of permutations with n minimal elements.

Original entry on oeis.org

1, 1, 3, 12, 52, 240, 1160, 5795, 29681, 155025, 822563, 4421458, 24025518, 131759106, 728330062, 4053823980, 22699853940, 127790656040, 722835069984, 4106096464006, 23414579166050, 133984343279790, 769124367124594, 4427878983496972, 25559244203741228
Offset: 1

Views

Author

Mathilde Bouvel, Oct 21 2021

Keywords

Crossrefs

For interval posets which are in addition trees, see A054515.

Programs

  • Mathematica
    Join[{1},Table[Sum[Sum[Binomial[n+i-1,i]Binomial[i,k]Binomial[n-2k-2,i-1],{k,0,Min[i,(n-i-1)/2]}],{i,n-1}]/n,{n,2,25}]] (* Stefano Spezia, Oct 23 2021 *)
  • PARI
    a(n) = if (n==1, 1, (1/n) * sum(i=1, n-1, sum(k=0, min(i,(n-i-1)/2), binomial(n+i-1,i)* binomial(i,k)*binomial(n-2*k-2,i-1)))); \\ Michel Marcus, Oct 21 2021

Formula

a(n) = (1/n) * Sum_{i=1..(n-1)} Sum_{k=0..Min(i,(n-i-1)/2)} binomial(n+i-1,i)* binomial(i,k)*binomial(n-2k-2,i-1) if n>1. Proved in M. Bouvel, L. Cioni, B. Izart (Theorem 18).
G.f. A(z) = Sum_{n>=0} a(n)*z^n satisfies the equation A(z) = z + (A(z)^2 + A(z)^4)/(1-A(z)). Proved in M. Bouvel, L. Cioni, B. Izart (Equation (1) page 14).
Asymptotic behavior of a(n) is c*n^(-3/2)*r^n with c approximately 0.0622 and r approximately 6.1403. Proved in M. Bouvel, L. Cioni, B. Izart (Theorem 19).
D-finite with recurrence 177*n*(n-1)*(n-2) *(1884*n-6797)*a(n) -(n-1) *(n-2) *(2079652*n^2-10492117*n+10802220) *a(n-1) +6*(n-2) *(98404*n^3-611787*n^2+893503*n+124240) *a(n-2) +2*(-1206916*n^4+13262653*n^3-52943063*n^2+90096428*n-54243072) *a(n-3) +(-16564*n^4+1171171*n^3-12487565*n^2+47878166*n-62441016) *a(n-4) +3 *(3*n-14) *(n-5) *(388*n-1861) *(3*n-16)*a(n-5)=0. - R. J. Mathar, Nov 04 2021

A367317 Expansion of (1/x) * Series_Reversion( x * (1-x-x^4/(1-x)) ).

Original entry on oeis.org

1, 1, 2, 5, 15, 50, 177, 649, 2436, 9307, 36080, 141610, 561732, 2248709, 9073415, 36863549, 150676275, 619169360, 2556446520, 10600160707, 44121921044, 184291848864, 772204252280, 3244999395406, 13672564904027, 57749354647408, 244469827514066
Offset: 0

Views

Author

Seiichi Manyama, Jan 26 2024

Keywords

Crossrefs

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec(serreverse(x*(1-x-x^4/(1-x)))/x)
    
  • PARI
    a(n) = sum(k=0, n\4, binomial(n+k, k)*binomial(2*n-2*k, n-4*k))/(n+1);

Formula

a(n) = (1/(n+1)) * Sum_{k=0..floor(n/4)} binomial(n+k,k) * binomial(2*n-2*k,n-4*k).

A253194 Number of ways to place non-intersecting diagonals in a convex (n+2)-gon so as to create no pentagons.

Original entry on oeis.org

1, 3, 10, 39, 162, 707, 3190, 14766, 69719, 334481, 1625846, 7989908, 39631204, 198151579, 997623275, 5053274850, 25734158411, 131680565544, 676693557574, 3490897656337, 18071699948492, 93851485181749, 488815126122166
Offset: 1

Views

Author

Michael D. Weiner, Mar 24 2015

Keywords

Examples

			a(3)=10 because the pentagon allows all but the null placement, i.e., 5 placements of 1 diagonal and 5 placements of two diagonals.
		

Crossrefs

Programs

  • Mathematica
    Rest[CoefficientList[(InverseSeries[Series[(y-2*y^2+y^4-y^5)/(1-y),{y,0,24}],x]-x)/x,x]]
  • PARI
    A253194(n)=sum(i=0,(n-1)\3,sum(k=i+1,n-2*i, (-1)^i*binomial(n+k,k)*binomial(k,i)*binomial(n-3*i-1,k-i-1)),if(n%3==0,(-1)^(n/3)*binomial(4*n/3,n/3)))/(n+1) \\ M. F. Hasler, Apr 07 2015

Formula

a(n) = (1/(n+1))*Sum_{i=0..floor(n/3)} Sum_{k=i+1..n-2*i} (-1)^i*binomial(n+k,k)*binomial(k,i)*binomial(n-3*i-1,k-i-1), n !== 0 (mod 3),
a(n) = ((-1)^(n/3)/(n+1))*binomial(4*n/3,n/3) + (1/(n+1))*Sum_{i=0..(n/3)-1} Sum_{k=i+1..n-2*i} (-1)^i*binomial(n+k,k)*binomial(k,i)*binomial(n-3*i-1,k-i-1), n == 0 (mod 3).
Recurrence: 275*(n-2)*(n-1)*n*(n+1)*(13962464*n^5 - 196202616*n^4 + 1069508732*n^3 - 2802358002*n^2 + 3480787751*n - 1597000860)*a(n) = 900*(n-2)*(n-1)*n*(27924928*n^6 - 406367696*n^5 + 2336399896*n^4 - 6678345644*n^3 + 9735406192*n^2 - 6526643891*n + 1424056473)*a(n-1) - 8*(n-2)*(n-1)*(1870970176*n^7 - 30033090896*n^6 + 197840216728*n^5 - 682911269612*n^4 + 1297104157966*n^3 - 1273486799084*n^2 + 486871358313*n + 21712608900)*a(n-2) - 16*(n-2)*(2569093376*n^8 - 47662201536*n^7 + 375012676176*n^6 - 1627682459628*n^5 + 4239503473896*n^4 - 6734585440155*n^3 + 6299789310412*n^2 - 3112752665481*n + 598681926090)*a(n-3) + 16*(2457393664*n^9 - 54190809728*n^8 + 518749193184*n^7 - 2816319789288*n^6 + 9492888047100*n^5 - 20388222826734*n^4 + 27407291375141*n^3 - 21462176121217*n^2 + 8117426803296*n - 745665750648)*a(n-4) - 8*(n-4)*(2*n - 7)*(4*n - 17)*(4*n - 11)*(13962464*n^5 - 126390296*n^4 + 424322908*n^3 - 631422862*n^2 + 369599799*n - 31302531)*a(n-5). - Vaclav Kotesovec, Mar 30 2015

A367413 Expansion of (1/x) * Series_Reversion( x * (1-x-x^3/(1-x)^2) ).

Original entry on oeis.org

1, 1, 2, 6, 22, 87, 356, 1493, 6398, 27936, 123906, 556734, 2528668, 11590555, 53545932, 249065874, 1165482126, 5482782933, 25914899804, 123009541412, 586121731150, 2802470267460, 13441993044464, 64660400422341, 311861855749484, 1507802756171072, 7306422899878394
Offset: 0

Views

Author

Seiichi Manyama, Jan 26 2024

Keywords

Crossrefs

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec(serreverse(x*(1-x-x^3/(1-x)^2))/x)
    
  • PARI
    a(n) = sum(k=0, n\3, binomial(n+k, k)*binomial(2*n, n-3*k))/(n+1);

Formula

a(n) = (1/(n+1)) * Sum_{k=0..floor(n/3)} binomial(n+k,k) * binomial(2*n,n-3*k).
From Seiichi Manyama, Nov 27 2024: (Start)
G.f.: exp( Sum_{k>=1} A378464(k) * x^k/k ).
a(n) = (1/(n+1)) * [x^n] 1/(1 - x - x^3/(1 - x)^2)^(n+1). (End)

A373456 Number of tree interval posets of permutations of size n, considered up to isomorphism.

Original entry on oeis.org

1, 1, 1, 3, 6, 14, 32, 79, 192, 488, 1244, 3240, 8497, 22561, 60309, 162541, 440598, 1201377, 3291426, 9058464, 25027797, 69401101, 193071153, 538724060, 1507288378, 4227824974, 11886150870, 33488522111, 94539554742, 267383598840, 757539956852, 2149698586706, 6109515731611
Offset: 1

Views

Author

Mathilde Bouvel, Jun 06 2024

Keywords

Comments

See Remark 24 in [Bouvel-Cioni-Izart].

References

  • Bridget E. Tenner. Interval Posets of Permutations. Order, 39(3):523-536, 2022.

Crossrefs

For the same posets but not considered up to isomorphism, see A054515.
For interval posets that are not necessarily trees, see A373455 (for posets also considered up to isomorphism) and A348479 (otherwise).

Formula

Asymptotic behavior of a(n) is c*n^(-3/2)*r^n with c approximately 0.2597 and r approximately 2.9784. See M. Bouvel, L. Cioni, B. Izart (Remark 24).

A378463 a(n) = Sum_{k=0..floor(n/3)} binomial(n+k-1,k) * binomial(2*n-k-1,n-3*k).

Original entry on oeis.org

1, 1, 3, 13, 59, 266, 1203, 5489, 25259, 117022, 545038, 2549592, 11970035, 56372460, 266194295, 1259910113, 5975382699, 28390616727, 135108035502, 643891031826, 3072604703774, 14679493913048, 70206875750168, 336103001918788, 1610476039036259, 7723148579525441
Offset: 0

Views

Author

Seiichi Manyama, Nov 27 2024

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n\3, binomial(n+k-1, k)*binomial(2*n-k-1, n-3*k));

Formula

a(n) = [x^n] 1/(1 - x - x^3/(1 - x))^n.

A256752 Number of ways to place non-intersecting diagonals in a convex (n+2)-gon so as to create no hexagons.

Original entry on oeis.org

1, 3, 11, 44, 190, 859, 4015, 19248, 94117, 467575, 2353443, 11975568, 61505088, 318406927, 1659801852, 8704865907, 45898065978, 243163198928, 1293769867676, 6910165762943, 37036898772008, 199140325574519, 1073849938338566
Offset: 1

Views

Author

Michael D. Weiner, Apr 09 2015

Keywords

Examples

			a(3)=11 because all 11 dissections of the pentagon are allowed, i.e., the null placement, 5 placements of 1 diagonal and 5 placements of two diagonals.
		

Crossrefs

Programs

  • Mathematica
    Rest[CoefficientList[(InverseSeries[Series[(y-2*y^2+y^5-y^6)/(1-y), {y, 0, 24}], x]-x)/x, x]]

Formula

a(n) = (1/(n+1))*Sum_{i=0..floor(n/4)} Sum_{k=i+1..n-3*i} (-1)^i*binomial(n+k,k)*binomial(k,i)*binomial(n-4*i-1,k-i-1), n !== 0 (mod 4),
a(n) = ((-1)^(n/4)/(n+1))*binomial(5*n/4,n/4) + (1/(n+1))*Sum_{i=0..(n/4)-1} Sum_{k=i+1..n-3*i} (-1)^i*binomial(n+k,k)*binomial(k,i)*binomial(n-4*i-1,k-i-1), n == 0 (mod 4).
Showing 1-8 of 8 results.