A088617 -id:A088617 - cp's OEIS frontend

A190909 Triangle read by rows: T(n,k) = binomial(n+k,n-k) * k! / floor(k/2)!^2.

1, 1, 1, 1, 3, 2, 1, 6, 10, 6, 1, 10, 30, 42, 6, 1, 15, 70, 168, 54, 30, 1, 21, 140, 504, 270, 330, 20, 1, 28, 252, 1260, 990, 1980, 260, 140, 1, 36, 420, 2772, 2970, 8580, 1820, 2100, 70, 1, 45, 660, 5544, 7722, 30030, 9100, 16800, 1190, 630

Offset: 0

Peter Luschny, May 24 2011

The triangle may be regarded as a generalization of the triangle A063007.
A063007(n,k) = binomial(n+k, n-k)*(2*k)$;
T(n,k) = binomial(n+k, n-k)*(k)$.
Here n$ denotes the swinging factorial A056040(n). As A063007 is a decomposition of the central Delannoy numbers A001850, a combinatorial interpretation of T(n,k) in terms of lattice paths can be expected.
T(n,n) = A056040(n) which can be seen as extended central binomial numbers.

			[0]  1
[1]  1,  1
[2]  1,  3,   2
[3]  1,  6,  10,    6
[4]  1, 10,  30,   42,   6
[5]  1, 15,  70,  168,  54,   30
[6]  1, 21, 140,  504, 270,  330,  20
[7]  1, 28, 252, 1260, 990, 1980, 260, 140

Peter Luschny, The lost Catalan numbers
R. A. Sulanke, Objects counted by the central Delannoy numbers, J. Integer Seq. 6 (2003), no. 1, Article 03.1.5.

Cf. Row sums: A190910; A056040, A063007, A085478, A088617.

A190909 := (n,k) -> binomial(n+k,n-k)*k!/iquo(k,2)!^2:
seq(print(seq(A190909(n,k),k=0..n)),n=0..7);

Flatten[Table[Binomial[n+k,n-k] k!/(Floor[k/2]!)^2,{n,0,10},{k,0,n}]] (* Harvey P. Dale, Mar 25 2012 *)

T(n,1) = A000217(n). T(n,2) = 2*binomial(n+2,4) (Cf. A034827).

A258820 Reversed rows of A178252 presented as diagonals of an irregular triangle.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 2, 1, 1, 5, 2, 1, 1, 3, 10, 1, 1, 7, 5, 5, 1, 1, 4, 7, 5, 1, 1, 9, 28, 35, 3, 1, 1, 5, 12, 14, 7, 1, 1, 11, 15, 21, 14, 7, 1, 1, 6, 55, 30, 126, 28, 1, 1, 13, 22, 165, 42, 21, 4, 1

Offset: 0

Tom Copeland, Jun 18 2015

The diagonals of T are the reversed rows of A178252. E.g., the fifth diagonal of T is (1,2,2,1,1) from the example below, which is the fifth reversed row of A178252.
Factoring out the greatest common divisor (gcd) of the coefficients of the sub-polynomials in the indeterminate q of the polynomials in s presented on p. 12 of the Alexeev et al. link and then evaluating the sub-polynomials at q=1 gives the first nine rows of T given in the example below. E.g., for k=6 (the seventh row), q*s^6 + (6*q + 9*q^2) s^4 + (15*q + 15*q^2) s^2 + 5  = q*s^6 + 3*(2*q + 3*q^2)*s^4 + 15*(q + q^2)*s^2 + 5 generates (1,2+3,1+1,1)=(1,5,2,1).
The row length sequence of this irregular triangle is A008619(n) = 1 + floor(n/2). - Wolfdieter Lang, Aug 25 2015

			The irregular triangle T(n,k) starts
n\k  0 1  2  3 4 5 ...
0:   1
1:   1
2:   1 1
3:   1 1
4:   1 3  1
5:   1 2  1
6:   1 5  2  1
7:   1 3 10  1
8:   1 7  5  5 1
9:   1 4  7  5 1
10:  1 9 28 35 3 1
... reformatted. - _Wolfdieter Lang_, Aug 25 2015

N. Alexeev, J. Andersen, R. Penner, P. Zograf, Enumeration of chord diagrams on many intervals and their non-orientable analogs, arXiv:1307.0967 [math.CO], 2013-2014.

Cf. A050169, A161642, A055151, A088617, A178252, A107711, A165661, A003989, A008619.

max = 15; coes = Table[ PadRight[ CoefficientList[ BernoulliB[n, x], x], max], {n, 0, max-1}]; inv = Inverse[coes] // Numerator; t[n_, k_] := inv[[n, k]]; t[n_, k_] /; k == n+1 = 1; Table[t[n-k+1, k], {n, 2, max+1}, {k, 2, Floor[n/2]+1}] // Flatten (* Jean-François Alcover, Jul 22 2015 *)

T(n,k) = A178252(n-k,n-2k) = A055151(n,k) / A161642(n,k) = A007318(n,2k) * A000108(k) / A161642(n,k) = n! / [(n-2k)! k! (k+1)! A161642(n,k)] = A003989(n-k+1,k+1) * (n-k)! / [ (n-2k)! (k+1)! ], where A003989(j,k) = gcd(j,k).

A190907 Triangle read by rows: T(n,k) = binomial(n+k, n-k) k! / (floor(k/2)! * floor((k+2)/2)!).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 6, 5, 3, 1, 10, 15, 21, 2, 1, 15, 35, 84, 18, 10, 1, 21, 70, 252, 90, 110, 5, 1, 28, 126, 630, 330, 660, 65, 35, 1, 36, 210, 1386, 990, 2860, 455, 525, 14, 1, 45, 330, 2772, 2574, 10010, 2275, 4200, 238, 126

Offset: 0

Peter Luschny, May 24 2011

The triangle may be regarded as a generalization of the triangle A088617.
A088617(n,k) = binomial(n+k,n-k)*(2*k)$/(k+1);
T(n,k) = binomial(n+k,n-k)*(k)$ /(floor(k/2)+1).
Here n$ denotes the swinging factorial A056040(n). As A088617 is a decomposition of the large Schroeder numbers A006318, a combinatorial interpretation of T(n,k) in terms of lattice paths can be expected.
T(n,n) = A057977(n) which can be seen as extended Catalan numbers.

			[0]  1
[1]  1,  1
[2]  1,  3,   1
[3]  1,  6,   5,   3
[4]  1, 10,  15,  21,   2
[5]  1, 15,  35,  84,  18,  10
[6]  1, 21,  70, 252,  90, 110,  5
[7]  1, 28, 126, 630, 330, 660, 65, 35

Peter Luschny, The lost Catalan numbers.

Cf. Row sums: A190908; A056040, A085478, A088617, A060693.

A190907 := (n,k) -> binomial(n+k,n-k)*k!/(floor(k/2)!*floor((k+2)/2)!);
seq(print(seq(A190907(n,k), k=0..n)), n=0..7);

Flatten[Table[Binomial[n+k,n-k] k!/(Floor[k/2]!Floor[(k+2)/2]!),{n,0,10},{k,0,n}]] (* Harvey P. Dale, May 05 2012 *)

T(n,1) = A000217(n). T(n,2) = (n-1)*n*(n+1)*(n+2)/24 (Cf. A000332).

A350499 Unsigned coefficients of free moment partition polynomials determining the free cumulants from the free moments of free probability theory. Irregular triangle with row lengths given by A000041, n >= 1.

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 4, 2, 10, 5, 1, 5, 5, 15, 15, 35, 14, 1, 6, 6, 3, 21, 42, 7, 56, 84, 126, 42, 1, 7, 7, 7, 28, 56, 28, 28, 84, 252, 84, 210, 420, 462, 132, 1, 8, 8, 8, 4, 36, 72, 72, 36, 36, 120, 360, 180, 360, 30, 330, 1320, 660, 792, 1980, 1716, 429

Offset: 1

Tom Copeland, Jan 01 2022

Coefficients are listed in Abramowitz and Stegun order (A036036).
Irregular triangular matrix of the unsigned coefficients of the free moment partition polynomials of free probability theory, for a single variable, that give the free formal cumulants given the free formal moments. This set of partition polynomials together with those of A134264 are the counterparts to the exp-log relations for the classical formal moments and cumulants associated with A036040 and A127671.
Associations with a compositional inverse pair of Laurent series, Kac-Schwarz operators of 2-D quantum theory, Virasoro / Witt / Heisenberg group actions, and KP and KdV integrable hierarchies are noted in references supplied in the MathOverflow link as well as a geometric combinatorial model based upon noncrossing partitions.

			Triangle begins:
  1;
  1, 1;
  1, 3, 2;
  1, 4, 2, 10,  5;
  1, 5, 5, 15, 15, 35, 14;
  ...
___________
The first few free cumulants in terms of the free moments are
  c_1 = m_1
  c_2 = m_2 - m_1^2
  c_3 = m_3 - 3 m_2 m_1 + 2 m_1^3
  c_4 = m_4 - 2 m_2^2 - 4m_3 m_1 + 10 m_2 m_1^2 - 5 m_1^4
  c_5 = m_5 - 5 m_2  m_3 - 5  m_4 m_1 + 15  m_2^2 m_1 + 15 m_3 m_1^2 - 35 m_2 m_1^3 + 14 m_1^5
___________
Conversely, from A134264, these free moments in terms of the free cumulants are
  m_1 = c_1
  m_2 = c_2 + c_1^2
  m_3 = c_3 + 3 c_2 c_1 + c_1^3
  m_4 = c_4 + + 2 c_2^2 + 4 c_3 c_1 + 6 c_2 c_1^2 + c_1^4
  m_5 = c_5 + 5 c_2 c_3 + 5 c_4 c_1 + 10 c_2^2 c_1 + 10 c_3 c_1^2  + 10 c_2 c_1^3 + c_1^5
___________

Tom Copeland, Ruling the inverse universe, the inviscid Hopf-Burgers evolution equation: Compositional inversion, free probability, associahedra, diff ids, integrable hierarchies, and translation, 2022
MathOverflow, Combinatorics for the action of Virasoro / Kac-Schwarz operators: partition polynomials of free probability theory, a MO question posed by Tom Copeland, 2021.
J. Zhou, Quantum deformation theory of the Airy curve and the mirror symmetry of a point, arXiv preprint arXiv:1405.5296 [math.AG], 2014.

Cf. A000041, A000108, A001764, A002293, A006318, A036040, A060693, A086810, A088617, A111785, A127671, A133437, A133932, A134685, A134264, A276850.

mv(n)={eval(Str("'m",n))}
Trm(m,v)={my(S=Set(v)); for(i=1, #S, my(x=S[i]); m=polcoef(m, #select(y->y==x, v), mv(x))); m}
Q(n)={polcoef(-x/serreverse(x*(1 + sum(k=1, n, -x^k*mv(k), O(x*x^n)))), n)}
row(n)={my(q=Q(n)); [Trm(q,Vec(v)) | v<-partitions(n)]}
{ for(n=1, 7, print(row(n))) } \\ Andrew Howroyd, Feb 01 2022

C(v)={my(n=vecsum(v), S=Set(v)); (n+#v-2)!/(n-1)!/prod(i=1, #S, my(x=S[i]); (#select(y->y==x, v))!)}
row(n)=[C(Vec(p)) | p<-partitions(n)]
{ for(n=1, 7, print(row(n))) } \\ Andrew Howroyd, Feb 01 2022

O.g.f.: C(x) = 1 + c_1 x + c_2 x^2 + ... = x / (x + m_1 x^2 + m_2 x^3 + m_3 x^4 + ...)^(-1) = x / M^(-1)(x), the shifted reciprocal of the compositional inverse of a power series M(x) = x + m_1 x^2 + m_2 x^3 + ..., the o.g.f. of the free moments m_n in free probability theory.
Row sums: big Schroeder numbers A006318.
Refinement of A060693 and A088617, i.e., by letting m_n = -t and removing all resulting signs, the elements of these two lower triangular matrices are generated.
The coefficients of the highest order terms in m_1^n of the free moment partition polynomials are the signed Catalan numbers A000108.
Taking the derivative with respect to the indeterminate m_1 generates the Lagrange inversion partition polynomials, with shifted indices, of A133437 and A111785 with an overall scale factor. These Lagrange inversion polynomials are the refined Euler characteristic polynomials of the associahedra. E.g.,
  D_{m_1} c_5 = 5 (- m_4 + 3 m_2^2 + 6 m_3 m_1 - 21 m_2 m_1^2 + 14 m_1^4). An analogous differential formula that applies to the classical formal cumulants in relation to the permutahedra is stated in my 2012 comment in A127671.
The o.g.f. satisfies the partial differential equations D_{m_1} (x / C(x)) = -(1/3) D_x (x / C(x))^3 and D_{m_1} (C(x) / x) = D_x (x / C(x)), where D_{m_1} and D_x are the partial derivatives with respect to m_1 and x.
More generally, D_{m_n} (x / C(x)) = -(1/(n+2)) D_x (x / C(x))^{n+2), equivalent to  D_{m_n} M^(-1)(x) = -(1/(n+2)) D_x (M^(-1)(x))^{n+2). Equations of this type are found in Zhou (see eqn. 44 on p. 11), characterizing the KdV hierarchy. These differential equations can be transformed into the inviscid Burgers-Hopf partial differential equation (see, e.g., A133437, A086810, A001764, A002293, A133932, A134685, and A276850).
The formal free cumulants when identified as the indeterminates of the noncrossing Lagrange inversion partition polynomials NCP_n(c_1,c_2,...,c_n) = m_n of A134264 (as in the example section) satisfy the partial differential equations D_{m_k} NCP_n(c_1, ..., c_n) = d(m_n)/dm_k = delta_{n-k}, where delta_{n} is the Kronecker delta which is zero for all integers n other than n = 0, for which it evaluates as unity. This provides a recursion method for determining the partial derivatives dc_n/dm_k from the partial derivatives dc_p/dm_k and cumulants c_p with k <= p < n. For example, dc_k/dm_j = 0 for j > k and dc_k/dm_k = 1, so dm_3/dm_2 = 0 = D_{m_2} (c_3 + 3 c_2 c_1 + c_1^3) = dc_3/dm_2 + 3 c_1  dc_2/dm_2 = dc_3/dm_2 + 3 c_1 , implying dc_3/dm_2 = -3 c_1 = -3 m_1.
T(n,k) = (n+j-2)!/((n-1)!*Product_{i>=1} s_i!), where (1*s_1 + 2*s_2 + ... = n) is the k-th partition of n and j = s_1 + s_2 + ... is the number of parts. - Andrew Howroyd, Feb 01 2022
Conjecture: free cumulants in terms of the free moments are R(n,1) for n > 0 where R(n,k) = R(n-1,k+1) - Sum_{j=1..n-1} R(j,k)*R(n-j,1) for n > 1, k > 0 with R(1,k) = m_k for k > 0. - Mikhail Kurkov, Mar 30 2025

Terms a(19) and beyond from Andrew Howroyd, Feb 01 2022

A213380 a(n) = 132*binomial(n,12).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 132, 1716, 12012, 60060, 240240, 816816, 2450448, 6651216, 16628040, 38798760, 85357272, 178474296, 356948592, 686439600, 1274816400, 2294669520, 4015671660, 6850263420, 11417105700, 18627909300, 29804654880, 46835886240, 72382733280, 110147637600, 165221456400, 244527755472, 357386719536, 516225261552, 737464659360

Offset: 0

N. J. A. Sloane, Jun 10 2012. This sequence was in the 1973 "Handbook", but was later dropped from the database. I have now reinstated it. - N. J. A. Sloane, Jun 10 2012

Charles Jordan, Calculus of Finite Differences, Chelsea 1965, p. 450.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

Index entries for linear recurrences with constant coefficients, signature (13,-78,286,-715,1287,-1716,1716,-1287,715,-286,78,-13,1).

Cf. A010965, A000217, A034827, A000910, A088625, A088626. A column of A088617.

[132*Binomial(n,12):n in [0..40]]; // Vincenzo Librandi, Feb 17 2020

132*Binomial[Range[0,40],12] (* Harvey P. Dale, Feb 16 2020 *)

G.f.: -132*x^12 / (x-1)^13. - Colin Barker, Jul 22 2013

a(n) = 132*A010965(n). - R. J. Mathar, Jul 15 2015

A253284 Triangle read by rows, T(n,k) = (k+1)(n+1)!(n+k)!/((k+1)!^2*(n-k)!) with n >= 0 and 0 <= k <= n.

Original entry on oeis.org

1, 2, 2, 6, 18, 12, 24, 144, 240, 120, 120, 1200, 3600, 4200, 1680, 720, 10800, 50400, 100800, 90720, 30240, 5040, 105840, 705600, 2116800, 3175200, 2328480, 665280, 40320, 1128960, 10160640, 42336000, 93139200, 111767040, 69189120, 17297280

Offset: 0

Peter Luschny, Mar 23 2015

G_n(x) = - Sum_{k=0..n} T(n,k)/(x-1)^(n+k+1) are generating functions, for n=0 of A000012, for n=1 of A002378, for n=2 of A083374 (with offset 0) and for n=3 for A253285. In general G_n(x) is the generating function of the sequence k -> ((n+k)!/k!)*C(n+k-1,k-1). These sequences are associated with the rows of the square array of unsigned Lah numbers (compare A253283 for the columns).

			Triangle begins:
1;
2, 2;
6, 18, 12;
24, 144, 240, 120;
120, 1200, 3600, 4200, 1680;
720, 10800, 50400, 100800, 90720, 30240;
5040, 105840, 705600, 2116800, 3175200, 2328480, 665280.

Cf. A000142, A001804, A001813, A002378, A006318, A083374, A088617, A253283, A253285.

/* As triangle: */ [[(k + 1)*Factorial(n + 1)*Factorial(n + k)/(Factorial(k + 1)^2*Factorial(n - k)): k in [0..n]]: n in [0..10]]; // Bruno Berselli, Mar 23 2015

T := (n,k) -> ((k+1)*(n+1)!*(n+k)!)/((k+1)!^2*(n-k)!);
for n from 0 to 6 do seq(T(n,k), k=0..n) od;

f[n_] := Rest@ Flatten@ Reap@ Block[{i, k, t}, For[i = 0, i <= n, i++, For[k = 0, k <= i, k++, Sow[(i + 1)!*Binomial[i + k, i]*Binomial[i, k]/(k + 1)]]]]; f@ 7 (* Michael De Vlieger, Mar 23 2015 *)

tabl(nn) = {for (n=0, nn, for (k=0, n, print1((n+1)!*binomial(n+k,n)*binomial(n,k)/(k+1), ", ");); print(););} \\ Michel Marcus, Mar 23 2015

T(n,k) = (n+1)!*binomial(n+k,n)*binomial(n,k)/(k+1).
T(n,k) = (n+1)!*A088617(n,k).
T(n,0) = n! = A000142(n).
T(n,1) = A001804(n+1) for n>0.
T(n,n) = (2*n)!/n! = A001813(n).
Sum_{k=0..n} T(n,k) = (n+1)!*hypergeom([-n, n+1], [2], -1) = (n+1)!*A006318(n).

A351385 Triangle read by rows: T(n,k) = Sum_{j=k..n} binomial(n + j, n)*binomial(n, j)/(j + 1).

Original entry on oeis.org

1, 2, 1, 6, 5, 2, 22, 21, 15, 5, 90, 89, 79, 49, 14, 394, 393, 378, 308, 168, 42, 1806, 1805, 1784, 1644, 1224, 594, 132, 8558, 8557, 8529, 8277, 7227, 4917, 2145, 429, 41586, 41585, 41549, 41129, 38819, 31889, 19877, 7865, 1430, 206098, 206097, 206052, 205392, 200772, 182754, 140712, 80652, 29172, 4862

Offset: 0

David Callan, Feb 09 2022

T(n,k) is the number of central Delannoy paths of steps E = (1,0), N = (0,1), D = (1,1) from the origin to (n,n) with k E steps above the diagonal line y=x. For example, T(3,1) = 5 counts ENNE, NEEN, NED, NDE, DNE. That the titular sum counts these paths is a consequence of the following equidistribution result: among the central Delannoy n-paths with j E steps, the statistic "number of E steps above y=x" is uniformly distributed over {0,1,...,j}. So, for k <= j <= n, there are binomial(n + j, n) binomial(n, j)/(j + 1) central Delannoy n-paths with j E steps, k of which are above y = x.

			Triangle begins:
   n
  [0]  1;
  [1]  2,  1;
  [2]  6,  5,  2;
  [3] 22, 21, 15,  5;
  [4] 90, 89, 79, 49, 14;
      ...

Andrew Howroyd, Table of n, a(n) for n = 0..1325 (rows 0..50)

Columns k=0..1 give: A006318, A035011.
Main diagonal gives A000108.
Row sums give A001850.
Cf. A001003, A002695, A080243, A088617 gives summands in title.

Flatten[Table[
  Sum[Binomial[n + j, n] Binomial[n, j]/(j + 1), {j, k, n}], {n, 0,
   10}, {k, 0, n}]]

T(n,k)={sum(j=k, n, binomial(n+j, n)*binomial(n,j)/(j+1))} \\ Andrew Howroyd, Feb 09 2022

G.f.: 2/(sqrt(1 - 6*x + x^2) + sqrt(1 - 2*x + x^2 - 4*x*y)).
From Alois P. Heinz, Feb 09 2022: (Start)
Sum_{k=0..n} k * T(n,k) = A002695(n).
Sum_{k=0..n} (-1)^k * T(n,k) = A001003(n).
Sum_{k=0..n} (-1)^k * T(n,n-k) = A080243(n). (End)

cp's OEIS Frontend

A190909 Triangle read by rows: T(n,k) = binomial(n+k,n-k) * k! / floor(k/2)!^2.

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A258820 Reversed rows of A178252 presented as diagonals of an irregular triangle.

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A190907 Triangle read by rows: T(n,k) = binomial(n+k, n-k) k! / (floor(k/2)! * floor((k+2)/2)!).

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A350499 Unsigned coefficients of free moment partition polynomials determining the free cumulants from the free moments of free probability theory. Irregular triangle with row lengths given by A000041, n >= 1.

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A213380 a(n) = 132*binomial(n,12).

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A253284 Triangle read by rows, T(n,k) = (k+1)*(n+1)!*(n+k)!/((k+1)!^2*(n-k)!) with n >= 0 and 0 <= k <= n.

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A351385 Triangle read by rows: T(n,k) = Sum_{j=k..n} binomial(n + j, n)*binomial(n, j)/(j + 1).

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A253284 Triangle read by rows, T(n,k) = (k+1)(n+1)!(n+k)!/((k+1)!^2*(n-k)!) with n >= 0 and 0 <= k <= n.