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 1-5 of 5 results.

A088855 Triangle read by rows: number of symmetric Dyck paths of semilength n with k peaks.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 4, 2, 1, 1, 3, 6, 6, 3, 1, 1, 3, 9, 9, 9, 3, 1, 1, 4, 12, 18, 18, 12, 4, 1, 1, 4, 16, 24, 36, 24, 16, 4, 1, 1, 5, 20, 40, 60, 60, 40, 20, 5, 1, 1, 5, 25, 50, 100, 100, 100, 50, 25, 5, 1, 1, 6, 30, 75, 150, 200, 200, 150, 75, 30, 6, 1, 1, 6, 36, 90, 225, 300, 400, 300, 225, 90, 36, 6, 1
Offset: 1

Views

Author

Emeric Deutsch, Nov 24 2003

Keywords

Comments

Rows 2, 4, 6, ... give A088459.
Diagonal sums are in A088518(n-1). - Philippe Deléham, Jan 04 2009
Row sums are in A001405(n). - Philippe Deléham, Jan 04 2009
Subtriangle (1 <= k <= n) of triangle T(n,k), 0 <= k <= n, read by rows, given by A101455 DELTA A056594 := [0,1,0,-1,0,1,0,-1,0,1,0,-1,0,...] DELTA [1,0,-1,0,1,0,-1,0,1,0,-1,0,1,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Jan 03 2009
Also, number of symmetric noncrossing partitions of an n-set with k blocks. - Andrew Howroyd, Nov 15 2017
From Roger Ford, Oct 17 2018: (Start)
T(n,k) = t(n+2,d) where t(n,d) is the number of different semi-meander arch depth listings with n top arches and with d the depth of the deepest embedded arch.
Examples: /\ semi-meander with 5 top arches
//\\ /\ 2 arches are at depth=0 (no covering arches)
///\\\ //\\ 2 arches are at depth=1 (1 covering arch)
(0)(1)(2) 1 arch is at depth=2 (2 covering arches)
2, 2, 1 is the listing for this t(5,2)
/\ semi-meander with 5 top arches
/ \ (0)(1)
/\ /\ //\/\\ 3, 2 is the listing for this t(5,1)
a(6,5) = t(8,5)= 3 {2,1,1,1,2,1; 2,1,2,1,1,1; 3,1,1,1,1,1} (End)

Examples

			Triangle begins:
  1;
  1,  1;
  1,  1,  1;
  1,  2,  2,   1;
  1,  2,  4,   2,   1;
  1,  3,  6,   6,   3,    1;
  1,  3,  9,   9,   9,    3,    1;
  1,  4, 12,  18,  18,   12,    4,    1;
  1,  4, 16,  24,  36,   24,   16,    4,    1;
  1,  5, 20,  40,  60,   60,   40,   20,    5,    1;
  1,  5, 25,  50, 100,  100,  100,   50,   25,    5,    1;
  1,  6, 30,  75, 150,  200,  200,  150,   75,   30,    6,   1;
  1,  6, 36,  90, 225,  300,  400,  300,  225,   90,   36,   6,   1;
  1,  7, 42, 126, 315,  525,  700,  700,  525,  315,  126,  42,   7,  1;
  1,  7, 49, 147, 441,  735, 1225, 1225, 1225,  735,  441, 147,  49,  7, 1;
  1,  8, 56, 196, 588, 1176, 1960, 2450, 2450, 1960, 1176, 588, 196, 56, 8, 1;
  ...
a(6,2)=3 because we have UUUDDDUUUDDD, UUUUDDUUDDDD, UUUUUDUDDDDD, where
U=(1,1), D=(1,-1).
		

Crossrefs

Cf. A001405 (row sums), A088459, A088518 (diagonal sums).
Column 2 is A008619, column 3 is A002620, column 4 is A028724, column 5 is A028723, column 6 is A028725, column 7 is A331574.

Programs

  • Magma
    [(&*[Binomial(Floor((n-j)/2), Floor((k-j)/2)): j in [0..1]]): k in [1..n], n in [1..15]]; // G. C. Greubel, Apr 08 2022
    
  • Mathematica
    T[n_, k_] := Binomial[Quotient[n-1, 2], Quotient[k-1, 2]]*Binomial[ Quotient[n, 2], Quotient[k, 2]];
    Table[T[n, k], {n,13}, {k,n}]//Flatten (* Jean-François Alcover, Jun 07 2018 *)
  • PARI
    T(n,k) = binomial((n-1)\2, (k-1)\2)*binomial(n\2, k\2); \\ Andrew Howroyd, Nov 15 2017
    
  • Sage
    def A088855(n,k): return product(binomial( (n-j)//2, (k-j)//2 ) for j in (0..1))
    flatten([[A088855(n,k) for k in (1..n)] for n in (1..15)]) # G. C. Greubel, Apr 08 2022

Formula

T(n, k) = binomial(floor(n'), floor(k'))*binomial(ceiling(n'), ceiling(k')), where n' = (n-1)/2, k' = (k-1)/2.
G.f.: 2*u/(u*v + sqrt(x*y*u*v)) - 1, where x = 1+z+t*z, y = 1+z-t*z, u = 1-z+t*z, v = 1-z-t*z.
Triangle T(n,k), 0 <= k <= n, given by A101455 DELTA A056594 begins: 1; 0,1; 0,1,1; 0,1,1,1; 0,1,2,2,1; 0,1,2,4,2,1; 0,1,3,6,6,3,1; 0,1,3,9,9,9,3,1; ... - Philippe Deléham, Jan 03 2009
From G. C. Greubel, Apr 08 2022: (Start)
T(n, n-k+1) = T(n, k).
T(2*n-1, n) = A018224(n-1), n >= 1.
T(2*n, n) = A005566(n-1), n >= 1. (End)

Extensions

Keyword:tabl added Philippe Deléham, Jan 25 2010

A111275 Number of inequivalent non-crossing partitions of n (equally spaced) points on a circle, under rotations and reflections.

Original entry on oeis.org

1, 2, 3, 6, 10, 24, 49, 130, 336, 980, 2904, 9176, 29432, 97356, 326399, 1111770, 3825238, 13293456, 46553116, 164200028, 582706692, 2079517924, 7458493728, 26874412064, 97241528200, 353223728624, 1287668381250, 4709805627484
Offset: 1

Views

Author

David Callan and Len Smiley, Oct 21 2005

Keywords

Comments

These may be viewed as bracelets (able to be turned over in space) designed with n beads on a circle, each of which is a vertex of exactly one of a set of non-touching internal polygons (which may be 1-gons (beads), 2-gons (2 connected beads), etc.).

References

  • S.-C. Chang, J. L. Jacobsen, J. Salas, R. Shrock, "Exact Potts model partition functions for strips of the triangular lattice", J. Statist. Phys. 114, nos.3-4, pp. 763-823 [Corollary 2.1]
  • Motzkin, T. "Relations Between Hypersurface Cross Ratios and a Combinatorial Formula for Partitions of a Polygon for Permanent Preponderance and for Non-Associative Products." Bull. Amer. Math. Soc. 54, page 360, 1948.

Crossrefs

Cf. A209612.

Programs

  • Mathematica
    Table[Length[EquivalenceClasses[NCPartitions[n], groupDihedral[n]]], {n, 9}]

Formula

(A054357(n) + A001405(n))/2.

A303929 Array read by antidiagonals: T(n,k) is the number of noncrossing partitions up to rotation and reflection composed of n blocks of size k.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 3, 1, 1, 1, 1, 3, 5, 6, 1, 1, 1, 1, 3, 8, 13, 12, 1, 1, 1, 1, 4, 11, 34, 49, 27, 1, 1, 1, 1, 4, 16, 60, 169, 201, 65, 1, 1, 1, 1, 5, 20, 109, 423, 1019, 940, 175, 1, 1, 1, 1, 5, 26, 167, 918, 3381, 6710, 4643, 490, 1
Offset: 0

Views

Author

Andrew Howroyd, May 02 2018

Keywords

Examples

			=================================================================
n\k| 1   2    3     4      5       6       7       8        9
---+-------------------------------------------------------------
0  | 1   1    1     1      1       1       1       1        1 ...
1  | 1   1    1     1      1       1       1       1        1 ...
2  | 1   1    1     1      1       1       1       1        1 ...
3  | 1   2    2     3      3       4       4       5        5 ...
4  | 1   3    5     8     11      16      20      26       32 ...
5  | 1   6   13    34     60     109     167     257      359 ...
6  | 1  12   49   169    423     918    1741    3051     4969 ...
7  | 1  27  201  1019   3381    9088   20569   41769    77427 ...
8  | 1  65  940  6710  29335   96315  259431  607696  1280045 ...
9  | 1 175 4643 47104 266703 1072187 3417520 9240444 22066742 ...
...
		

Crossrefs

Columns 2..5 are A006082(n+1), A082938, A303870, A303871.

Programs

  • Mathematica
    u[n_, k_, r_] := (r*Binomial[k*n + r, n]/(k*n + r));
    e[n_, k_] := Sum[ u[j, k, 1 + (n - 2*j)*k/2], {j, 0, n/2}]
    c[n_, k_] := If[n == 0, 1, (DivisorSum[n, EulerPhi[n/#]*Binomial[k*#, #]&] + DivisorSum[GCD[n - 1, k], EulerPhi[#]*Binomial[n*k/#, (n - 1)/#]&])/(k*n) - Binomial[k*n, n]/(n*(k - 1) + 1)];
    T[n_, k_] := (1/2)*(c[n, k] + If[n == 0, 1, If[OddQ[k], If[OddQ[n], 2*u[ Quotient[n, 2], k, (k + 1)/2], u[n/2, k, 1] + u[n/2 - 1, k, k]], e[n, k] + If[OddQ[n], u[Quotient[n, 2], k, k/2]]]/2]) /. Null -> 0;
    Table[T[n - k, k], {n, 1, 12}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Jun 14 2018, translated from PARI *)
  • PARI
    \\ here c(n,k) is A303694
    u(n,k,r) = {r*binomial(k*n + r, n)/(k*n + r)}
    e(n,k) = {sum(j=0, n\2, u(j, k, 1+(n-2*j)*k/2))}
    c(n, k)={if(n==0, 1, (sumdiv(n, d, eulerphi(n/d)*binomial(k*d, d)) + sumdiv(gcd(n-1, k), d, eulerphi(d)*binomial(n*k/d, (n-1)/d)))/(k*n) - binomial(k*n, n)/(n*(k-1)+1))}
    T(n,k)={(1/2)*(c(n,k) + if(n==0, 1, if(k%2, if(n%2, 2*u(n\2,k,(k+1)/2), u(n/2,k,1) + u(n/2-1,k,k)), e(n,k) + if(n%2, u(n\2,k,k/2)))/2))}

A209805 Triangle read by rows: T(n,k) is the number of k-block noncrossing partitions of n-set up to rotations.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 4, 2, 1, 1, 3, 10, 10, 3, 1, 1, 3, 15, 25, 15, 3, 1, 1, 4, 26, 64, 64, 26, 4, 1, 1, 4, 38, 132, 196, 132, 38, 4, 1, 1, 5, 56, 256, 536, 536, 256, 56, 5, 1, 1, 5, 75, 450, 1260, 1764, 1260, 450, 75, 5, 1
Offset: 1

Views

Author

Tilman Piesk, Mar 13 2012

Keywords

Comments

Like the Narayana triangle A001263 (and unlike A152175) this triangle is symmetric.
The diagonal entries are 1, 1, 4, 25, 196, 1764, ... which is probably sequence A001246 - the squares of the Catalan numbers.
The above conjecture about the diagonal entries T(2*n-1, n) is true since gcd(2*n-1, n) = gcd(2*n-1, n-1) = 1 and then T(2*n-1, n) simplifies to A001246(n-1) using the formula given below. - Andrew Howroyd, Nov 15 2017

Examples

			Triangle begins:
  1;
  1,   1;
  1,   1,   1;
  1,   2,   2,   1;
  1,   2,   4,   2,   1;
  1,   3,  10,  10,   3,   1;
  1,   3,  15,  25,  15,   3,   1;
  1,   4,  26,  64,  64,  26,   4,   1;
  1,   4,  38, 132, 196, 132,  38,   4,   1;
  1,   5,  56, 256, 536, 536, 256,  56,   5,   1;
		

Crossrefs

Cf. A054357 (row sums), A001246 (square Catalan numbers).

Programs

  • Mathematica
    b[n_, k_] := Binomial[n-1, n-k] Binomial[n, n-k];
    T[n_, k_] := (DivisorSum[GCD[n, k], EulerPhi[#] b[n/#, k/#]&] + DivisorSum[GCD[n, k - 1], EulerPhi[#] b[n/#, (n + 1 - k)/#]&] - k Binomial[n, k]^2/(n - k + 1))/n;
    Table[T[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jul 01 2018, after Andrew Howroyd *)
  • PARI
    b(n,k)=binomial(n-1,n-k)*binomial(n,n-k);
    T(n,k)=(sumdiv(gcd(n,k), d, eulerphi(d)*b(n/d,k/d)) + sumdiv(gcd(n,k-1), d, eulerphi(d)*b(n/d,(n+1-k)/d)) - k*binomial(n,k)^2/(n-k+1))/n; \\ Andrew Howroyd, Nov 15 2017

Formula

T(n,k) = (1/n)*((Sum_{d|gcd(n,k)} phi(d)*A103371(n/d-1,k/d-1)) + (Sum_{d|gcd(n,k-1)} phi(d)*A103371(n/d-1,(n+1-k)/d-1)) - A132812(n,k)). - Andrew Howroyd, Nov 15 2017

A211359 Triangle read by rows: T(n,k) is the number of noncrossing partitions up to rotation and reflection of an n-set that contain k singleton blocks.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 2, 0, 1, 2, 3, 2, 2, 0, 1, 5, 4, 8, 3, 3, 0, 1, 6, 11, 12, 12, 4, 3, 0, 1, 14, 21, 39, 24, 22, 5, 4, 0, 1, 22, 55, 84, 85, 48, 30, 7, 4, 0, 1, 51, 124, 245, 228, 190, 82, 46, 8, 5, 0, 1, 95, 327, 620, 730, 570, 350, 136, 60
Offset: 0

Views

Author

Tilman Piesk, Apr 12 2012

Keywords

Examples

			From _Andrew Howroyd_, May 02 2018: (Start)
Triangle begins:
   1;
   0,   1;
   1,   0,   1;
   1,   1,   0,   1;
   2,   1,   2,   0,   1;
   2,   3,   2,   2,   0,  1;
   5,   4,   8,   3,   3,  0,  1;
   6,  11,  12,  12,   4,  3,  0, 1;
  14,  21,  39,  24,  22,  5,  4, 0, 1;
  22,  55,  84,  85,  48, 30,  7, 4, 0, 1;
  51, 124, 245, 228, 190, 82, 46, 8, 5, 0, 1;
  ...
(End)
		

Crossrefs

Column k=0 is A303931.
Row sums are A111275.

Programs

  • PARI
    \\ See A303875 for NCPartitionsModDihedral
    { my(rows=Vec(NCPartitionsModDihedral(vector(10, k, if(k==1,y,1)))));
    for(n=1, #rows, for(k=0, n-1, print1(polcoeff(rows[n], k), ", ")); print; ) } \\ Andrew Howroyd, May 02 2018
Showing 1-5 of 5 results.