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.

A137593 Triangle read by rows: matrix product A000012 * A136717.

Original entry on oeis.org

1, 1, 2, 1, 4, 4, 1, 4, 16, 12, 1, 4, 16, 84, 48, 1, 4, 16, 156, 504, 192, 1, 4, 16, 156, 1464, 3312, 960, 1, 4, 16, 156, 1464, 14112, 24720, 5760, 1, 4, 16, 156, 1464, 24912, 158640, 189360, 34560, 1, 4, 16, 156, 1464, 24912, 400560, 1761840, 1607040, 241920
Offset: 1

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Author

Gary W. Adamson, Jan 28 2008

Keywords

Comments

Right border = A004527: (1, 2, 4, 12, 48, 192, ...).
Row sums = A007489 starting (1, 3, 9, 33, 153, ...).

Examples

			First few rows of the triangle:
  1;
  1, 2;
  1, 4,  4;
  1, 4, 16,  12;
  1, 4, 16,  84,   48;
  1, 4, 16, 156,  504,  192;
  1, 4, 16, 156, 1464, 3312, 960;
  ...
		

Crossrefs

Formula

A000012 * A136717 as infinite lower triangular matrices.

Extensions

Typo in eighth row corrected by Olivier Gérard, Oct 30 2012

A137594 Triangle read by rows: A007318 * A136717.

Original entry on oeis.org

1, 1, 2, 1, 6, 4, 1, 12, 24, 12, 1, 20, 72, 120, 48, 1, 30, 160, 552, 696, 192, 1, 42, 300, 1752, 4416, 4272, 960, 1, 56, 504, 4452, 17976, 36672, 30480, 5760, 1, 72, 784, 9744, 55776, 195312, 350880, 229680, 34560
Offset: 0

Views

Author

Gary W. Adamson, Jan 28 2008

Keywords

Comments

Row sums = A001339: (1, 3, 11, 49, 261, ...).

Examples

			First few rows of the triangle:
  1;
  1,  2;
  1,  6,   4;
  1, 12,  24,   12;
  1, 20,  72,  120,   48;
  1, 30, 160,  552,  696,  192;
  1, 42, 300, 1752, 4416, 4272, 960;
  ...
		

Crossrefs

Formula

Binomial transform of A136717.

A136715 Triangle T(n,k), 1 <= k <= n, read by rows: T(n,k) is the number of permutations of the set {2,4,6,...,2n} with k excedances. Equivalently, T(n,k) is the number of permutations in the symmetric group S_n having k multiplicative 2-excedances.

Original entry on oeis.org

1, 1, 1, 0, 4, 2, 0, 4, 16, 4, 0, 0, 36, 72, 12, 0, 0, 36, 324, 324, 36, 0, 0, 0, 576, 2592, 1728, 144, 0, 0, 0, 576, 9216, 20736, 9216, 576, 0, 0, 0, 0, 14400, 115200, 172800, 57600, 2880, 0, 0, 0, 0, 14400, 360000, 1440000, 1440000, 360000, 14400
Offset: 1

Views

Author

Peter Bala, Jan 18 2008

Keywords

Comments

Let E_n denote the set {2,4,6,...,2n} and let p denote a bijection p:E_n -> E_n. We say the permutation p has an excedance at position i, 1 <= i <= n, if p(2i) > i. For example, if we represent p in one line notation by the vector (p(2),p(4),...,p(2n)), then the permutation (6,2,8,4,10) of E_5 has three excedances in total (at positions 1, 3 and 5). This array gives the number of permutations of the set E_n with k excedances. This is the viewpoint taken in [Jansson]. See A136716 for the corresponding array when the set E_n is replaced by the set O_n := {1,3,5,...,2n-1}.
Alternatively, we can work with permutations (p(1),p(2),...,p(n)) in the symmetric group S_n and define p to have a multiplicative 2-excedance at position i, 1 <= i <= n, if 2*p(i) > i. Then the (n,k)-th entry of this array gives the number of permutations in S_n with k multiplicative 2-excedances. Compare with A008292, the triangle of Eulerian numbers, which enumerates permutations by the usual excedance number and A136717 which enumerates permutations by multiplicative 3-excedances.
Let e(p)= |{i | 1 <= i < = n, 2*p(i) > i}| denote the number of multiplicative 2-excedances in the permutation p of S_n. This 2-excedance statistic e(p) on the symmetric group S_n is related to a descent statistic as follows.
Define a permutation p in S_n to have a descent from even at position i, 1 <= i <= n-1, if p(i) is even and p(i) > p(i+1). For example, the permutation (2,1,3,5,6,4) in S_6 has two descents from even (at position 1 and position 5). Array A134434 records the number of permutations of S_n with k descents from even.
Let d(p) = |{i | 1 <= i <= n-1, p(i) is even & p(i) > p(i+1)}| count the descents from even in the permutation p. Comparison of the formulas for the entries of this table with the formulas for the entries of A134434 shows that e(p) and d(p) are related by sum {p in S_n} x^e(p) = x^ceiling(n/2)* sum {p in S_n} x^d(p). Thus the shifted multiplicative 2-excedance statistic e(p) - ceiling(n/2) and the descent statistic d(p) are equidistributed on the symmetric group S_n.

Examples

			T(4,2) = 4; the four permutations in S_4 with two multiplicative 2-excedances are (3,4,1,2), (4,3,1,2), (3,1,4,2) and (4,1,3,2). Alternatively, the four permutations (6,8,2,4), (8,6,2,4), (6,2,8,4) and (8,2,6,4) of the set E_4 each have 2 excedances.
Triangle starts
n\k|..1....2....3....4....5....6
--------------------------------
1..|..1
2..|..1....1
3..|..0....4....2
4..|..0....4...16....4
5..|..0....0...36...72...12
6..|..0....0...36..324..324...36
		

Crossrefs

Formula

Recurrence relations:
T(2n,k) = (k+1-n)*T(2n-1,k) + (3n-k)*T(2n-1,k-1) for n >= 1;
T(2n+1,k) = (k-n)*T(2n,k) + (3n+2-k)*T(2n,k-1) for n >= 0. Boundary conditions: T(0,k) = 0 all k; T(n,0) = 0 all n; T(1,1) = 1.
The recurrence relations have the explicit solution T(2n,n+k) = [n!* C(n,k)]^2 and T(2n+1,n+k+1) = 1/(k+1)*[(n+1)!*C(n,k)]^2 = n!*(n+1)!*C(n,k)*C(n+1,k+1); or as a single formula, T(n,k) = floor(n/2)! * floor((n+1)/2)! * C(floor(n/2),k-floor((n+1)/2)) * C(floor((n+1)/2),k-floor(n/2)). Also T(2n,n+k) = n!^2 * A008459 (n,k); T(2n+1,n+k+1) = n!*(n+1)!* A103371 (n,k).
For the even numbered rows, define the shifted row polynomials F(2n,x) := x^(1-n)* sum {k = n..2n} T(2n,k)*x^k = n!^2 * x * (1 + C(n,1)^2*x + C(n,2)^2*x^2 + ... + C(n,n)^2*x^n). For the odd numbered rows, define the shifted row polynomials F(2n+1,x) := x^(-n)* sum {k = n+1..2n+1} T(2n+1,k)*x^k = n!*(n+1)!* ((n+1)*N(n+1,1)*x + n*N(n+1,2)*x^2 +(n-1)* N(n+1,3)*x^3 + ... + N(n+1,n+1)*x^(n+1)), where N(n,k) denotes the Narayana numbers. The first few values are F(1,x) = x, F(2,x) =x+x^2, F(3,x) = 4x+2x^2 and F(4,x) = 4x+16x^2+4x^2.
The recurrence relations yield the identities x*d/dx(F(2n-1,x)/(1-x)^(2n)) = F(2n,x)/(1-x)^(2n+1) and x*d/dx(1/x*F(2n,x)/(1-x)^(2n+1)) = F(2n+1,x)/(1-x)^(2n+2), for n = 1,2,3,... . An easy induction argument now gives the Taylor series expansions: F(2n,x)/(1-x)^(2n+1) = sum {m = 1..inf} (m*(m+1)*...*(m+n-1))^2*x^m; F(2n+1,x)/(1-x)^(2n+2) = sum {m = 1..inf} m*((m+1)*(m+2)*...*(m+n))^2*x^m.
For example, when n = 3 we have for row 6 the expansion (36x + 324x^2 + 324x^3 + 36x^4)/(1-x)^7 = 36x + 576x^2 + 3600x^3 + ... = (1.2.3)^2*x + (2.3.4)^2*x^2 + (3.4.5)^2*x^3 + ... and for row 7 the expansion (576x + 2592x^2 + 1728x^3 + 144x^4)/(1-x)^8 = 576x + 7200x^2 + 43200x^3 + ... = 1*(2.3.4)^2*x + 2*(3.4.5)^2*x^2 + 3*(4.5.6)^2*x^3 + ... .
Relation with the Jacobi polynomials P_n(a,b,x): F(2n,x) = n!^2*x*(1-x)^n *P_n(0,0,(1+x)/(1-x)), F(2n+1,x) = n!*(n+1)!*x*(1-x)^n *P_n(1,0,(1+x)/(1-x)).
Worpitzky-type identities:
Sum {k = n..2n} T(2n,k)*C(x+k,2n) = ((x+1)*(x+2)*...*(x+n))^2;
sum {k = n+1..2n+1} T(2n+1,k)*C(x+k,2n+1) = ((x+1)*(x+2)*...*(x+n))^2*(x+n+1);
and for the odd numbered rows read in reverse order, sum {k = n+1..2n+1} T(2n+1,3n+2-k)*C(x+k,2n+1) = (x+1)*((x+2)*(x+3)*...*(x+n+1))^2.

A136716 Triangle T(n,k), 0 <= k < n, read by rows: T(n,k) is the number of permutations of the set O_n = {1,3,5,...,2n-1} with k excedances.

Original entry on oeis.org

1, 0, 2, 0, 2, 4, 0, 0, 12, 12, 0, 0, 12, 72, 36, 0, 0, 0, 144, 432, 144, 0, 0, 0, 144, 1728, 2592, 576, 0, 0, 0, 0, 2880, 17280, 17280, 2880, 0, 0, 0, 0, 2880, 57600, 172800, 115200, 14400, 0, 0, 0, 0, 0, 86400, 864000, 1728000, 864000, 86400
Offset: 1

Views

Author

Peter Bala, Jan 22 2008

Keywords

Comments

A permutation p of the set O_n := {1,3,5,...,2n-1} is a bijection p:O_n -> O_n. We say the permutation p has an excedance at position i, 1 <= i <= n, if p(2i-1) > i. For example, if we represent permutations in one line notation by the vector (p(1),p(3),...,p(2n-1)) then the permutation (1,3,7,9,5) of O_5 has three excedances in total (at positions 2, 3 and 4).
This array gives the number of permutations of the set O_n with k excedances. This is the viewpoint taken in [Jansson]. See A136715 for the corresponding array when O_n is replaced by the set E_n := {2,4,6,...,2n}.
Alternatively, the bijection f:O_n -> [n] defined by f(i) = (i+1)/2 allows us to work with permutations in the symmetric group S_n, rather than permutations of O_n. The excedance condition p(2i-1) > i on permutations of O_n becomes the equivalent condition 2*q(i)-1 > i on permutations q of [n].
Let now q be a permutation in the group S_n and define e(q):= |{i | 1 <= i < = n, 2*q(i)-1 > i}|. Then the (n,k)-th entry in this array gives the number of permutations q in S_n such that e(q) = k.
This excedance type statistic e(q) on the symmetric group S_n is related to a descent statistic as follows. Define a permutation q in S_n to have a descent from odd at position i, 1 <= i <= n-1, if q(i) is odd and q(i) > q(i+1).
For example, the permutation (2,6,5,3,1,4) in the group S_6 has two descents from odd (at position 3 and position 4). Array A134435 records the number of permutations in S_n with k descents from odd. Let d(q) = |{i | 1 <= i <= n-1, q(i) is odd & q(i) > q(i+1)}| count the descents from odd in the permutation q.
Comparison of the formulas for the entries of this table with the formulas for the entries of A134435 shows that e(q) and d(q) are related by sum {q in S_n} x^e(q) = x^floor(n/2)* sum {q in S_n} x^d(q). Thus the shifted excedance statistic e(q) - floor(n/2) and the descent statistic d(q) are equidistributed on the symmetric group S_n.

Examples

			T(3,2) = 4; the four permutations of the set {1,3,5} with 2 excedances are (1,3,5), (3,1,5), (3,5,1) and (5,3,1).
Triangle starts
n\k|..0....1....2....3....4....5
--------------------------------
1..|..1
2..|..0....2
3..|..0....2....4
4..|..0....0...12...12
5..|..0....0...12...72...36
6..|..0....0....0..144..432..144
		

Crossrefs

Formula

Recurrence relations: T(2n,k) = (k+1-n)*T(2n-1,k) + (3n-k)*T(2n-1,k-1) for n >= 1; T(2n+1,k) = (k+1-n)*T(2n,k) + (3n+1-k)*T(2n,k-1) for n >= 0. Boundary conditions: T(0,k) = 0 all k; T(1,0) = 1; T(n,0) = 0 for n > 1.
The recurrence relations have the explicit solution T(2n,n+k) = n!^2 * C(n-1,k) * C(n+1,k+1) and T(2n+1,n+k) = n!*(n+1)!*C(n,k)*C(n+1,k), or as a single formula, T(n,k) = floor(n/2)! * floor((n+1)/2)! * C(floor((n-1)/2),k-floor(n/2)) * C(floor(n/2)+1,k-floor((n-1)/2)).
Also T(2n,n+k) = n!*(n+1)!*N(n,k+1), where N(n,k) denotes the Narayana number A001263 (n,k).
For the even numbered rows, define the shifted row polynomials G(2n,x) := x^(1-n)* sum {k = n..2n-1} T(2n,k)*x^k . For the odd numbered rows, define the shifted row polynomials G(2n+1,x) := x^(1-n)* sum {k = n..2n} T(2n+1,k)*x^k.
The first few values are G(1,x) = x, G(2,x) = 2x, G(3,x) = 2x+4x^2 and G(4,x) = 12x+12x^2. The recurrence relations yield the identities x*d/dx(1/x*G(2n-1,x)/(1-x)^(2n)) = G(2n,x)/(1-x)^(2n+1) and x*d/dx(G(2n,x)/(1-x)^(2n+1)) = G(2n+1,x)/(1-x)^(2n+2), for n = 1,2,3,... .
An easy induction argument now gives the Taylor series expansions: G(2n,x)/(1-x)^(2n+1) = sum {m = 1..inf} m*((m+1)*(m+2)*...*(m+n-1))^2*(m+n)*x^m (note: there is a typo in proposition 3.3 of [Jansson]); G(2n+1,x)/(1-x)^(2n+2) = sum {m = 1..inf} (m*(m+1)*...*(m+n-1))^2*(m+n)*x^m.
For example, when n = 3 we have for row 6 the expansion (144x+432x^2+144x^3)/(1-x)^7 = 144x + 1440x^2+ 7200x^3 + ... = 1*(2*3)^2*4*x + 2*(3*4)^2*5*x^2 + 3*(4*5)^2*6*x^3 + ... and for row 7 the expansion (144x + 1728x^2 + 2592x^3 + 576x^4)/(1-x)^8 = 144x + 2880x^2 + 21600x^3 + ... = (1.2.3)^2*4*x + (2.3.4)^2*5*x^2 + (3.4.5)^2*6*x^3 + ... .
Relation with the Jacobi polynomials P_n(a,b,x): G(2n+2,x) = n!*(n+2)!*x*(1-x)^n*P_n(1,1,(1+x)/(1-x)), G(2n+1,x) = n!*(n+1)!*x*(1-x)^n *P_n(0,1,(1+x)/(1-x)).
Worpitzky-type identities:
Sum {k = n..2n-1} T(2n,k)*C(x+k,2n) = x*((x+1)*(x+2)*...*(x+n-1))^2*(x+n);
sum {k = n..2n} T(2n+1,k)*C(x+k,2n+1) = x*((x+1)*(x+2)*...*(x+n))^2; and for the odd numbered rows read in reverse order, sum {k = n..2n} T(2n+1,3n-k)*C(x+k,2n+1) = (x*(x+1)*(x+2)*...*(x+n-1))^2*(x+n).

A136718 Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} having k entries divisible by 3 that are followed by a smaller entry (n>=1, k>=0).

Original entry on oeis.org

1, 2, 2, 4, 12, 12, 72, 48, 72, 456, 192, 960, 3120, 960, 10800, 23760, 5760, 10800, 133920, 183600, 34560, 241920, 1572480, 1572480, 241920, 4233600, 18869760, 14878080, 1935360, 4233600, 84309120, 233331840, 141644160, 15482880
Offset: 1

Views

Author

Peter Bala, Jan 23 2008

Keywords

Comments

Define a permutation (p(1),p(2),...,p(n)) in the symmetric group S_n to have a multiplicative 3-descent at position i, 1 <= i <= n-1, if p(i) > p(i+1) and 3|p(i). The (n,k)-th entry in this array gives the number of permutations in S_n with k multiplicative 3-descents.
Compare with A008292, the triangle of Eulerian numbers, which enumerates permutations by the usual descent number and A134434, which enumerates permutations by multiplicative 2-descents. This multiplicative 3-descent statistic is equidistributed on the symmetric group S_n with a multiplicative 3-excedance statistic - see A136717 for details.

Examples

			T(3,1) = 4; the four permutations in the group S_3 with a single multiplicative 3-descent are (1,3,2), (3,1,2), (2,3,1) and (3,2,1). The remaining two permutations in S_3, namely (1,2,3) and (2,1,3), have no multiplicative 3-descents.
Table starts
n\k|..0....1....2
-----------------
1..|..1
2..|..2
3..|..2....4
4..|.12...12
5..|.72...48
6..|.72..456..192
		

Crossrefs

Cf. A000142 (row sums), A008292, A134434, A136717.

Programs

  • Mathematica
    T[n_?Positive, k_] /; 0 <= k <= Floor[n/3] := T[n, k] = Switch[Mod[n, 3], 1|2, (k+1)T[n-1, k+1] + (n-k)T[n-1, k], 0, (k+1)T[n-1, k] + (n-k)T[n-1, k-1]];
    T[1, 0] = 1; T[, ] = 0;
    Table[T[n, k], {n, 1, 12}, {k, 0, Floor[n/3]}] // Flatten (* Jean-François Alcover, Nov 12 2019 *)

Formula

Recurrence relations: T(3n+1,k) = (k+1)*T(3n,k+1) + (3n+1-k)*T(3n,k); T(3n+2,k) = (k+1)*T(3n+1,k+1) + (3n+2-k)*T(3n+1,k); T(3n+3,k) = (k+1)*T(3n+2,k) + (3n+3-k)*T(3n+2,k-1); with boundary conditions T(n,-1) = 0 for all n, T(1,0) = 1, T(1,k) = 0 for k > 0.
The row polynomials A(n,x) := sum {k = 0..floor(n/3)} T(n,k)*x^k satisfy the recursion relations: A(3n+1,x) = (1-x)*d/dx(A(3n,x)) + (3n+1)*A(3n,x); A(3n+2,x) = (1-x)*d/dx(A(3n+1,x)) + (3n+2)*A(3n+1,x); A(3n+3,x) = (x-x^2)*d/dx(A(3n+2,x)) + ((3n+2)x+1)*A(3n+2,x).

Extensions

Keyword corrected by Peter Bala, Oct 30 2008
Showing 1-5 of 5 results.