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 1-6 of 6 results.

A286725 Third column of triangle A286724: Lah[2,1](n+2, 2), n >= 0.

Original entry on oeis.org

1, 18, 288, 4800, 86400, 1693440, 36126720, 836075520, 20901888000, 562028544000, 16186422067200, 497364605337600, 16247243774361600, 562404592189440000, 20567939371499520000, 792551263781781504000, 32098326183162150912000, 1363234794367239585792000
Offset: 0

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Author

Wolfdieter Lang, Jun 17 2017

Keywords

Comments

See A286725 for the generalized Lah numbers Lah[2,1].

Crossrefs

Programs

  • Mathematica
    a[n_] := 2^(n - 1)*(n + 2)! * Binomial[n + 2, 2]; Array[a, 20, 0] (* Amiram Eldar, Dec 11 2022 *)

Formula

E.g.f.: d^2/dx^2 (x^2/(2*(1-2*x)^3)) = (1 + 8*x + 4*x^2)/(1 - 2*x)^5.
a(n) = 2^(n-1)*(n+2)!*binomial(n+2,2), n >= 0.
From Amiram Eldar, Dec 11 2022: (Start)
Sum_{n>=0} 1/a(n) = 32 - 16*sqrt(e) + 8*gamma - 8*Ei(1/2) - 8*log(2), where Ei(x) is the exponential integral and gamma is Euler's constant (A001620).
Sum_{n>=0} (-1)^n/a(n) = 24*gamma - 16/sqrt(e) - 24*Ei(-1/2) - 24*log(2). (End)

A048854 Triangle read by rows. A generalization of unsigned Lah numbers, called L[4,1].

Original entry on oeis.org

1, 2, 1, 12, 12, 1, 120, 180, 30, 1, 1680, 3360, 840, 56, 1, 30240, 75600, 25200, 2520, 90, 1, 665280, 1995840, 831600, 110880, 5940, 132, 1, 17297280, 60540480, 30270240, 5045040, 360360, 12012, 182, 1, 518918400, 2075673600, 1210809600, 242161920, 21621600, 960960, 21840, 240, 1, 17643225600, 79394515200, 52929676800, 12350257920, 1323241920, 73513440, 2227680, 36720, 306, 1
Offset: 0

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Author

Keywords

Comments

s(n,x) := Sum_{m=0..n} T(n,m)*x^m are monic polynomials satisfying s(n,x+y) = Sum_{k=0..n} binomial(n,k)*s(k,x)*p(n-k,y), with polynomials p(n,x) = Sum_{m=1..n} A048786(n,m)*x^m (row polynomials of triangle A048786) and p(0,x)=1.
In the umbral calculus (see the Roman reference, p. 21) the s(n,x) are called Sheffer polynomials for(1/sqrt(1+4*t),t/(1+4*t)). Here the Sheffer notation differs. See the W. Lang link under A006232.
For the general L[d,a] triangles see A286724, also for references.
This is the generalized signless Lah number triangle L[4,1], the Sheffer triangle ((1 - 4*t)^(-1/2), t/(1 - 4*t)). It is defined as transition matrix
risefac[4,1](x, n) = Sum_{m=0..n} L[4,1](n, m)*fallfac[4,1](x, m), where risefac[4,1](x, n) := Product_{0..n-1} (x + (1 + 4*j)) for n >= 1 and risefac[4,1](x, 0) := 1, and fallfac[4,1](x, n) := Product_{0..n-1} (x - (1 + 4*j)) for n >= 1 and fallfac[4,1](x, 0) := 1.
In matrix notation: L[4,1] = S1phat[4,1]*S2hat[4,1] with the unsigned scaled Stirling1 and the scaled Stirling2 generalizations A290319 and A111578 (but here with offsets 0), respectively.
The a- and z-sequences for this Sheffer matrix have e.g.f.s Ea(t) = 1 + 4*t and Ez(t) = (1 + 4*t)*(1 - (1 + 4*t)^(-1/2))/t, respectively. That is, a = {1, 4, repeat(0)} and z(n) = 2*A292220(n). See the W. Lang link on a- and z-sequences there.
The inverse matrix T^(-1) = L^(-1)[4,1] is Sheffer ((1 + 4*t)^(-1/2), t/(1 + 4*t)). This means that T^(-1)(n, m) = (-1)^(n-m)*T(n, m).
fallfac[4,1](x, n) = Sum_{m=0..n} (-1)^(n-m)*T(n, m)*risefac[4,1](x, m), n >= 0.
Diagonal sequences have o.g.f. G(d, x) = A001813(d)*Sum_{m=0..d} A091042(d, m)*x^m/(1 - x)^{2*d + 1}, for d >= 0 (d=0 main diagonal). G(d, x) generates {A001813(d)*binomial(2*(n + d),2*d)}{n >= 0}. See the second W. Lang link on how to compute o.g.f.s of diagonal sequences of general Sheffer triangles. - _Wolfdieter Lang, Oct 12 2017

Examples

			The triangle T(n, m) begins:
n\m         0          1          2         3        4      5     6   7 8  ...
0:          1
1:          2          1
2:         12         12          1
3:        120        180         30         1
4:       1680       3360        840        56        1
5:      30240      75600      25200      2520       90      1
6:     665280    1995840     831600    110880     5940    132     1
7:   17297280   60540480   30270240   5045040   360360  12012   182   1
8:  518918400 2075673600 1210809600 242161920 21621600 960960 21840 240 1
...
n = 9: 17643225600 79394515200 52929676800 12350257920 1323241920 73513440 2227680 36720 306 1,
n = 10: 670442572800 3352212864000 2514159648000 670442572800 83805321600 5587021440 211629600 4651200 58140 380 1.
...
Recurrence from a-sequence: T(4, 2) = 2*T(3, 1) + 4*4*T(3, 2) = 2*180 + 16*30 = 840.
Recurrence from z-sequence: T(4, 0) = 4*(z(0)*T(3, 0) + z(1)*T(3, 1) + z(2)*T(3, 2)+ z(3)*T(3, 3)) = 4*(2*120 + 2*180 - 8*30 + 60*1) = 1680.
Four term recurrence: T(4, 2) = T(3, 1) + 2*13*T(3, 2) - 8*3*5*T(2, 2) =  180 + 26*30 - 120*1 = 840.
Meixner type identity for n = 2: (D_x - 4*(D_x)^2)*(12 + 12*x + 1*x^2) = (12 + 2*x) - 4*2 = 2*(2 + x).
Sheffer recurrence for R(3, x): [(2 + x) + 8*(1 + x)*D_x + 16*x*(D_x)^2] (12 + 12*x + 1*x^2) = (2 + x)*(12 + 12*x + x^2) + 8*(1 + x)*(12 + 2*x) + 16*2*x = 120 + 180*x + 30*x^2 + x^3 = R(3, x).
Boas-Buck recurrence for column m = 2 with n = 4: T(4, 2) = (4!*10/2)*(1*30/3! + 4*1/2!) = 840.
Diagonal sequence d = 2: {12, 180, 840 ...} has o.g.f. 12*(1 + 10*x + 5*x^2)/(1 - x)^5  (see A001813(2) and row n=2 of A091042) generating
{12*binomial(2*(n + 2), 4)}_{n >= 0}. - _Wolfdieter Lang_, Oct 12 2017
		

References

  • S. Roman, The Umbral Calculus, Academic Press, New York, 1984.

Crossrefs

Related to triangle A046521. Cf. A048786. a(n, 0) = A001813.
A111578, A271703 L[1,0], A286724 L[2,1], A290319, A290596 L[3,1], A290597 L[3,2], A292220.
The diagonal sequences are: A000012, 2*A000384(n+1), 12*A053134, 120*A053135, 1680*A053137, ... - Wolfdieter Lang, Oct 12 2017

Programs

  • Maple
    A290604_row := proc(n) exp(x*t/(1-4*t))/sqrt(1-4*t): series(%, t, n+2): seq(n!*coeff(coeff(%,t,n),x,j), j=0..n) end: seq(A290604_row(n), n=0..9); # Peter Luschny, Sep 23 2017
  • Mathematica
    T[n_, m_] := n!/m! * Binomial[2*n, n] * Binomial[n, m] / Binomial[2*m, m]; Table[a[n, m], {n, 0, 8}, {m, 0, n}] // Flatten (* Jean-François Alcover, Jul 05 2013 *)
    T[0, 0] = 1; T[-1, ] = T[, -1] = 0; T[n_, m_] /; n < m = 0; T[n_, m_] := T[n, m] = T[n-1, m-1] + 2*(4*n-3)*T[n-1, m] - 8*(n-1)*(2*n-3)*T[n-2, m]; Table[T[n, m], {n, 0, 9}, {m, 0, n}] // Flatten (* Jean-François Alcover, Sep 23 2017 *)

Formula

T(n, m) = (n!/m!)*A046521(n, m) = (n!/m!)* binomial(2*n, n)*binomial(n, m)/binomial(2*m, m), n >= m >= 0, a(n, m) := 0, n < m.
Sum_{n>=0, k>=0} T(n, k)*x^n*y^k/(2*n)! = exp(x)*cosh(sqrt(x*y)). - Vladeta Jovovic, Feb 21 2003
T(n, m) = L[4,1](n,m) = Sum_{k=m..n} A290319(n, k)*A111578(k+1, m+1), 0 <= m <= n.
E.g.f. of row polynomials R(n, x) := Sum_{m=0..n} T(n, m)*x^m:
(1 - 4*t)^(-1/2)*exp(x*t/(1 - 4*t)) (this is the e.g.f. for the triangle).
E.g.f. of column m: (1 - 4*t)^(-1/2)*(t/(1 - 4*t))^m/m!, m >= 0.
Three term recurrence for column entries m >= 1: T(n, m) = (n/m)*T(n-1, m-1) + 4*n*T(n-1, m) with T(n, m) = 0 for n < m, and for the column m = 0: T(n, 0) = n*Sum_{j=0..n-1} z(j)*T(n-1, j), n >= 1, T(0, 0) = 0, from the a-sequence {1, 4 repeat(0)} and the z(j) = 2*A292220(j) (see above).
Four term recurrence: T(n, m) = T(n-1, m-1) + 2*(4*n - 3)*T(n-1, m) - 8*(n-1)*(2*n - 3)*T(n-2, m), n >= m >= 0, with T(0, 0) =1, T(-1, m) = 0, T(n, -1) = 0 and T(n, m) = 0 if n < m.
Meixner type identity for (monic) row polynomials: (D_x/(1 + 4*D_x)) * R(n, x) = n * R(n-1, x), n >= 1, with R(0, x) = 1 and D_x = d/dx. That is, Sum_{k=0..n-1} (-4)^k*(D_x)^(k+1)*R(n, x) = n*R(n-1, x), n >= 1.
General recurrence for Sheffer row polynomials (see the Roman reference, p. 50, Corollary 3.7.2, rewritten for the present Sheffer notation):
R(n, x) = [(2 + x)*1 + 8*(1 + x)*D_x + 16*x*(D_x)^2]*R(n-1, x), n >= 1, with R(0, x) = 1.
Boas-Buck recurrence for column m (see a comment in A286724 with references): T(n, m) = (n!/(n-m))*(2 + 4*m)*Sum_{p=0..n-1-m} 4^p*T(n-1-p, m)/(n-1-p)!, for n > m >= 0, with input T(m, m) = 1.
Explicit form (from the o.g.f.s of diagonal sequences): ((2*(n-m))!/(n-m)!)*binomial(2*n,2*(n-m)), n >= m >= 0, and vanishing for n < m. - Wolfdieter Lang, Oct 12 2017

Extensions

Name changed, after merging my newer duplicate, from Wolfdieter Lang, Oct 10 2017

A290596 Triangle read by rows. A generalization of unsigned Lah numbers, called L[3,1].

Original entry on oeis.org

1, 2, 1, 10, 10, 1, 80, 120, 24, 1, 880, 1760, 528, 44, 1, 12320, 30800, 12320, 1540, 70, 1, 209440, 628320, 314160, 52360, 3570, 102, 1, 4188800, 14660800, 8796480, 1832600, 166600, 7140, 140, 1, 96342400, 385369600, 269758720, 67439680, 7663600, 437920, 12880, 184, 1, 2504902400, 11272060800, 9017648640, 2630147520, 358656480, 25618320, 1004640, 21528, 234, 1, 72642169600, 363210848000, 326889763200, 108963254400, 17335063200, 1485862560, 72836400, 2081040, 33930, 290, 1
Offset: 0

Views

Author

Wolfdieter Lang, Sep 13 2017

Keywords

Comments

For the general L[d,a] triangles see A286724, also for references.
This is the generalized signless Lah number triangle L[3,1], the Sheffer triangle ((1 - 3*t)^(-2/3), t/(1 - 3*t)). It is defined as transition matrix
risefac[3,1](x, n) = Sum_{m=0..n} L[3,1](n, m)*fallfac[3,1](x, m), where risefac[3,1](x, n):= Product_{0..n-1} (x + (1 + 3*j)) for n >= 1 and risefac[3,1](x, 0) := 1, and fallfac[3,1](x, n):= Product_{0..n-1} (x - (1 + 3*j)) for n >= 1 and fallfac[3,1](x, 0) := 1.
In matrix notation: L[3,1] = S1phat[3,1]*S2hat[3,1] with the unsigned scaled Stirling1 and the scaled Stirling2 generalizations A286718 and A111577 (but here with offsets 0), respectively.
The a- and z-sequences for this Sheffer matrix has e.g.f.s Ea(t) = 1 + 3*t and (Ez(t) = (1 + 3*t)*(1 - (1 + 3*t)^(-2/3))/t, respectively. That is, a = {1, 3, repeat(0)} and z(n) = A290597(n)/A038500(n+1). For the proof see the second W. Lang link. See also a W. Lang link under A006232 for Sheffer a- and z-sequences with references (in the Riordan case).
The inverse matrix T^(-1) = L^(-1)[3,1] is Sheffer ((1 + 3*t)^(-2/3), t/(1 + 3*t)). This means that T^(-1)(n, m) = (-1)^(n-m)*T(n, m).
fallfac[3,1](x, n) = Sum_{m=0..n} (-1)^(n-m)*T(n, m)*risefac[3,1](x, m), n >= 0.

Examples

			The triangle T(n, m) begins:
n\m         0         1         2        3       4      5     6   7 8  ...
0:          1
1:          2         1
2:         10        10         1
3:         80       120        24        1
4:        880      1760       528       44       1
5:      12320     30800     12320     1540      70      1
6:     209440    628320    314160    52360    3570    102     1
7:    4188800  14660800   8796480  1832600  166600   7140   140   1
8:   96342400 385369600 269758720 67439680 7663600 437920 12880 184 1
...
n = 9: 2504902400 11272060800 9017648640 2630147520 358656480 25618320 1004640 21528 234 1,
n = 10: 72642169600 363210848000 326889763200 108963254400 17335063200 1485862560 72836400 2081040 33930 290 1.
...
Recurrence from a-sequence:  T(4, 2) = 2*T(3, 1) + 3*4*T(3, 2) = 2*120 + 12*24 = 528.
Recurrence from z-sequence: T(4, 0) = 4*(z(0)*T(3, 0) + z(1)*T(3, 1) + z(2)*T(3, 2) + z(3)*T(3, 3)) = 4*(2*80 + 1*120 - (10/3)*24 + 20*1) = 880.
Four term recurrence: T(4, 2) = T(3, 1) + 2*10*T(3, 2) - 3*3*8*T(2, 2) =  120 + 20*24 - 72*1 = 528.
Meixner type identity for n = 2: (D_x - 3*(D_x)^2)*(10 + 10*x + x^2 ) = (10 + 2*x) - 3*2 = 2*(2 + x).
Sheffer recurrence for R(3, x): [(2 + x) + 6*(1 + x)*D_x + 9*x*(D_x)^2] (10 + 10*x + x^2) = (2 + x)*(10 + 10*x + x^2) + 6*(1 + x)*(10 +2*x) + 9*2*x = 80 + 120*x + 24*x^2 + x^3 = R(3, x).
Boas-Buck recurrence for column m = 2 with n = 4: T(4, 2) = (4!*8/2)*(1*24/3! + 3*1/2!) = 528.
		

References

  • Steven Roman, The Umbral Calculus, Academic press, Orlando, London, 1984, p. 50.

Crossrefs

Cf. A008544 (column m=0), A038500, A111577, A271703 L[1,0], A286718, A286724 L[2,1], A290597, A290598 L[3,2].

Formula

T(n, m) = L[3,1](n,m) = Sum_{k=m..n} A286718(n, k)*A111577(k+1, m+1), 0 <= m <= n.
E.g.f. of row polynomials R(n, x) := Sum_{m=0..n} T(n, m)*x^m:
(1 - 3*t)^(-2/3)*exp(x*t/(1 - 3*t)) (this is the e.g.f. for the triangle).
E.g.f. of column m: (1 - 3*t)^(-2/3)*(t/(1 - 3*t))^m/m!, m >= 0.
Three term recurrence for column entries m >= 1: T(n, m) = (n/m)*T(n-1, m-1) + 3*n*T(n-1, m) with T(n, m) = 0 for n < m, and for the column m = 0: T(n, 0) = n*Sum_{j=0}^(n-1) z(j)*T(n-1, j), from the a-sequence {1, 3 repeat(0)} and the z-sequence given above.
Four term recurrence: T(n, m) = T(n-1, m-1) + 2*(3*n - 2)*T(n-1, m) - 3*(n-1)*(3*n - 4)*T(n-2, m), n >= m >= 0, with T(0, 0) = 1, T(-1, m) = 0, T(n, -1) = 0 and T(n, m) = 0 if n < m.
Meixner type identity for (monic) row polynomials: (D_x/(1 + 3*D_x)) * R(n, x) = n*R(n-1, x), n >= 1, with R(0, x) = 1 and D_x = d/dx. That is, Sum_{k=0..n-1} (-3)^k*(D_x)^(k+1)*R(n, x) = n*R(n-1, x), n >= 1.
General recurrence for Sheffer row polynomials (see the Roman reference, p. 50, Corollary 3.7.2, rewritten for the present Sheffer notation):
R(n, x) = [(2 + x)*1 + 6*(1 + x)*D_x + 3^2*x*(D_x)^2]*R(n-1, x), n >= 1, with R(0, x) = 1.
Boas-Buck recurrence for column m (see a comment in A286724 with references): T(n, m) = (n!/(n-m))*(2 + 3*m)*Sum_{p=0..n-1-m} 3^p*T(n-1-p, m)/(n-1-p)!, for n > m >= 0, with input T(m, m) = 1.

A292219 Triangle read by rows. A generalization of unsigned Lah numbers, called L[4,3].

Original entry on oeis.org

1, 6, 1, 60, 20, 1, 840, 420, 42, 1, 15120, 10080, 1512, 72, 1, 332640, 277200, 55440, 3960, 110, 1, 8648640, 8648640, 2162160, 205920, 8580, 156, 1, 259459200, 302702400, 90810720, 10810800, 600600, 16380, 210, 1, 8821612800, 11762150400, 4116752640, 588107520, 40840800, 1485120, 28560, 272, 1
Offset: 0

Views

Author

Wolfdieter Lang, Sep 23 2017

Keywords

Comments

For the general L[d,a] triangles see A286724, also for references.
This is the generalized signless Lah number triangle L[4,3], the Sheffer triangle ((1 - 4*t)^(-3/2), t/(1 - 4*t)). It is defined as transition matrix
risefac[4,3](x, n) = Sum_{m=0..n} L[4,3](n, m)*fallfac[4,3](x, m), where risefac[4,3](x, n) := Product_{0..n-1} (x + (3 + 4*j)) for n >= 1 and risefac[4,3](x, 0) := 1, and fallfac[4,3](x, n):= Product_{0..n-1} (x - (3 + 4*j)) for n >= 1 and fallfac[4,3](x, 0) := 1.
In matrix notation: L[4,3] = S1phat[4,3]*S2hat[4,3] with the unsigned scaled Stirling1 and the scaled Stirling2 generalizations A225471 and A225469, respectively.
The a- and z-sequences for this Sheffer matrix have e.g.f.s Ea(t) = 1 + 4*t and Ez(t) = (1 + 4*t)*(1 - (1 + 4*t)^(-3/2))/t, respectively. That is, a = {1, 4, repeat(0)} and z(n) = 2*A292221(n). See the W. Lang link on a- and z-sequences there.
The inverse matrix T^(-1) = L^(-1)[4,3] is Sheffer ((1 + 4*t)^(-3/2), t/(1 + 4*t)). This means that T^(-1)(n, m) = (-1)^(n-m)*T(n, m).
fallfac[4,3](x, n) = Sum_{m=0..n} (-1)^(n-m)*T(n, m)*risefac[4,3](x, m), n >= 0.
Diagonal sequences have o.g.f. G(d, x) = A001813(d)*Sum_{m=0..d} A103327(d, m)*x^m/(1 - x)^(2*d + 1), for d >= 0 (d=0 main diagonal). G(d, x) generates {A001813(d)*binomial(2*(m + d) + 1, 2*d)}{m >= 0}. See the second W. Lang link on how to compute o.g.f.s of diagonal sequences of general Sheffer triangles. - _Wolfdieter Lang, Oct 12 2017

Examples

			The triangle T(n, m) begins:
  n\m          0           1          2         3        4       5     6   7  8
  0:           1
  1:           6           1
  2:          60          20          1
  3:         840         420         42         1
  4:       15120       10080       1512        72        1
  5:      332640      277200      55440      3960      110       1
  6:     8648640     8648640    2162160    205920     8580     156     1
  7:   259459200   302702400   90810720  10810800   600600   16380   210   1
  8:  8821612800 11762150400 4116752640 588107520 40840800 1485120 28560 272  1
  ...
Recurrence from a-sequence: T(4, 2) = (4/2)*T(3, 1) + 4*4*T(3, 2) = 2*420 + 16*42 = 1512.
Recurrence from z-sequence: T(4, 0) = 4*(z(0)*T(3, 0) + z(1)*T(3, 1) + z(2)*T(3, 2)+ z(3)*T(3, 3)) = 4*(6*840 - 6*420 + 40*42 -420*1) = 15120.
Meixner type identity for n = 2: (D_x - 4*(D_x)^2)*(60 + 20*x + 1*x^2 ) = (20 + 2*x) - 4*2 = 2*(6 + x).
Sheffer recurrence for R(3, x): [(6 + x) + 8*(3 + x)*D_x + 16*x*(D_x)^2] (60 + 20*x + 1*x^2) = (6 + x)*(60 + 20*x + x^2) + 8*(3 + x)*(20 + 2*x) + 16*2*x = 840 + 420*x + 42*x^2 + x^3 = R(3, x).
Boas-Buck recurrence for column m = 2 with n = 4: T(4, 2) = (2*4!/2)*(3 + 2*2)*(1*42/3! + 4*1/2!) = 1512.
Diagonal sequence d = 2: {60, 420, 1512, ...} has o.g.f. 12*(5 + 10*x + x^2)/(1 - x)^5 (see A001813(2) and row n=2 of A103327) generating {12*binomial(2*(m + 2) + 1, 4)}_{m >= 0}. - _Wolfdieter Lang_, Oct 12 2017
		

References

  • Steven Roman, The Umbral Calculus, Academic press, Orlando, London, 1984, p. 50.

Crossrefs

Cf. A225469, A225471, A271703 L[1,0], A286724 L[2,1], A290596 L[3,1], A290597 L[3,2], A048854 L[4,1], A292221, A103327,
Diagonal sequences: A000012, 2*A014105(m+1), 12*A053126(m+4), 120*A053128(m+6), A053130(n+8), ... - Wolfdieter Lang, Oct 12 2017

Formula

T(n, m) = L[4,3](n,m) = Sum_{k=m..n} A225471(n, k)*A225469(k, m), 0 <= m <= n.
E.g.f. of row polynomials R(n, x) := Sum_{m=0..n} T(n, m)*x^m:
(1 - 4*t)^(-3/2)*exp(x*t/(1 - 4*t)) (this is the e.g.f. for the triangle).
E.g.f. of column m: (1 - 4*t)^(-3/2)*(t/(1 - 4*t))^m/m!, m >= 0.
Three term recurrence for column entries k >= 1: T(n, m) = (n/m)*T(n-1, m-1) + 4*n*T(n-1, m) with T(n, m) = 0 for n < m, and for the column m = 0: T(n, 0) = n*Sum_{j=0}^(n-1) z(j)*T(n-1, j), n >= 1, T(0, 0) = 0, from the a-sequence {1, 4 repeat(0)} and z(j) = 2*A292221(j) (see above).
Four term recurrence: T(n, m) = T(n-1, m-1) + 2*(4*n - 1)*T(n-1, m) - 8*(n-1)*(2*n - 1)*T(n-2, m), n >= m >= 0, with T(0, 0) =1, T(-1, m) = 0, T(n, -1) = 0 and T(n, m) = 0 if n < m.
Meixner type identity for (monic) row polynomials: (D_x/(1 + 4*D_x)) * R(n, x) = n * R(n-1, x), n >= 1, with R(0, x) = 1 and D_x = d/dx. That is, Sum_{k=0..n-1} (-4)^k*{D_x)^(k+1)*R(n, x) = n*R(n-1, x), n >= 1.
General recurrence for Sheffer row polynomials (see the Roman reference, p. 50, Corollary 3.7.2, rewritten for the present Sheffer notation):
R(n, x) = [(6 + x)*1 + 8*(3 + x)*D_x + 16*x*(D_x)^2]*R(n-1, x), n >= 1, with R(0, x) = 1.
Boas-Buck recurrence for column m (see a comment in A286724 with references): T(n, m) = (2*n!/(n-m))*(3 + 2*m)*Sum_{p=0..n-1-m} 4^p*T(n-1-p, m)/(n-1-p)!, for n > m >= 0, with input T(m, m) = 1.
Explicit form (from the o.g.f.s of diagonal sequences): ((2*(n-m))!/(n-m)!)*binomial(2*n + 1, 2*(n-m)), n >= m >= 0, and vanishing for n < m. - Wolfdieter Lang, Oct 12 2017

A290598 Triangle read by rows. A generalization of unsigned Lah numbers, called L[3,2].

Original entry on oeis.org

1, 4, 1, 28, 14, 1, 280, 210, 30, 1, 3640, 3640, 780, 52, 1, 58240, 72800, 20800, 2080, 80, 1, 1106560, 1659840, 592800, 79040, 4560, 114, 1, 24344320, 42602560, 18258240, 3043040, 234080, 8778, 154, 1, 608608000, 1217216000, 608608000, 121721600, 11704000, 585200, 15400, 200, 1, 17041024000, 38342304000, 21909888000, 5112307200, 589881600, 36867600, 1293600, 25200, 252, 1, 528271744000, 1320679360000, 849008160000, 226402176000, 30477216000, 2285791200, 100254000, 2604000, 39060, 310, 1
Offset: 0

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Author

Wolfdieter Lang, Sep 13 2017

Keywords

Comments

For the general L[d,a] triangles see A286724, also for references.
This is the generalized signless Lah number triangle L[3,2], the Sheffer triangle ((1 - 3*t)^(-4/3), t/(1 - 3*t)). It is defined as transition matrix risefac[3,2](x, n) = Sum_{m=0..n} L[3,2](n, m)*fallfac[3,2](x, m), where risefac[3,2](x, n):= Product_{0..n-1} (x + (2 + 3*j)) for n >= 1 and risefac[3,2](x, 0) := 1, and fallfac[3,2](x, n):= Product_{0..n-1} (x - (2 + 3*j)) for n >= 1 and fallfac[3,2](x, 0) := 1.
In matrix notation: L[3,2] = S1phat[3,2]*S2hat[3,2] with the unsigned scaled Stirling1 and the scaled Stirling2 generalizations A225470 and A225468, respectively.
The a- and z-sequences for this Sheffer matrix have e.g.f.s 1 + 3*t and (1 + 3*t)*(1 - (1 + 3*t)^(-4/3))/t, respectively. That is, a = {1, 3, repeat(0)} and z(n) = A290603(n)/A038500(n+1). See a W. Lang link under A006232 for these types of sequences with a reference, and also the present link, eq. (142).
The inverse matrix T^(-1) = L^(-1)[3,2] is Sheffer ((1 + 3*t)^(-4/3), t/(1 + 3*t)). This means that T^(-1)(n, m) = (-1)^(n-m)*T(n, m).
fallfac[3,2](x, n) = Sum_{m=0..n} (-1)^(n-m)*T(n, m)*risefac[3,2](x, m), n >= 0.

Examples

			The triangle T(n, m) begins:
n\m          0          1         2         3        4      5     6   7 8  ...
0:           1
1:           4          1
2:          28         14         1
3:         280        210        30         1
4:        3640       3640       780        52        1
5:       58240      72800     20800      2080       80      1
6:     1106560    1659840    592800     79040     4560    114     1
7:    24344320   42602560  18258240   3043040   234080   8778   154   1
8:   608608000 1217216000 608608000 121721600 11704000 585200 15400 200 1
...
n = 9: 17041024000 38342304000 21909888000 5112307200 589881600 36867600 1293600 25200 252 1,
n = 10: 528271744000 1320679360000 849008160000 226402176000 30477216000 2285791200 100254000 2604000 39060 310 1.
...
Recurrence from a-sequence:  T(4, 2) = (4/2)*T(3, 1) + 3*4*T(3, 2) = 2*210 + 12*30 = 780.
Recurrence from z-sequence: T(4, 0) = 4*(z(0)*T(3, 0) + z(1)*T(3, 1) + z(2)*T(3, 2) + z(3)*T(3, 3)) = 4*(4* 280  - 2*210 + (28/3)*30 - 70*1) = 3640.
Four term recurrence: T(4, 2) = T(3, 1) + 2*11*T(3, 2) - 3*3*10*T(2, 2) =  210 + 22*30  - 90*1 = 780.
Meixner type identity for n = 2: (D_x - 3*(D_x)^2)*(28 + 14*x + x^2) = (14 + 2*x) - 3*2 = 2*(4 + x).
Sheffer recurrence for R(3, x): [(4 + x) + 6*(2 + x)*D_x + 9*x*(D_x)^2] (28 + 14*x + x^2) = (4 + x)*(28 + 14*x + x^2) + 6*(2 + x)*(14 + 2*x) + 9*2*x= 280 + 210*x + 30*x^2 + x^3 =  R(3, x).
Boas-Buck recurrence for column m = 2 with n = 4: T(4, 2) = (4!*(4 + 3*2)/2)*(1*30/3! + 3*1/2!) = 780.
		

References

  • Steven Roman, The Umbral Calculus, Academic press, Orlando, London, 1984, p. 50.

Crossrefs

Cf. A007559(n+1) (column m=0), A225468, A225470, A271703 L[1,0], A286724 L[2,1], A290596, L[3,1], A290603.

Formula

T(n, m) = L[3,2](n,m) = Sum_{k=m..n} A225470(n, k) * A225468(k, m), 0 <= m <= n.
E.g.f. of row polynomials R(n, x) := Sum_{m=0..n} T(n, m)*x^m:
(1 - 3*t)^(-4/3)*exp(x*t/(1 - 3*t)) (this is the e.g.f. for the triangle).
E.g.f. of column m: (1 - 3*t)^(-4/3)*(t/(1 - 3*t))^m/m!, m >= 0.
Three term recurrence for column entries m >= 1: T(n, m) = (n/m)*T(n-1, m-1) + 3*n*T(n-1, m) with T(n, m) = 0 for n < m, and for the column m = 0: T(n, 0) = n*Sum_{j=0}^(n-1) z(j)*T(n-1, j), from the a-sequence {1, 3 repeat(0)} and the z-sequence given above.
Four term recurrence: T(n, m) = T(n-1, m-1) + 2*(3*n - 1)*T(n-1, m) - 3*(n-1)*(3*n - 2)*T(n-2, m), n >= m >= 0, with T(0, 0) =1, T(-1, m) = 0, T(n, -1) = 0 and T(n, m) = 0 if n < m.
Meixner type identity for (monic) row polynomials: (D_x/(1 + 3*D_x)) * R(n, x) = n*R(n-1, x), n >= 1, with R(0, x) = 1 and D_x = d/dx. That is, Sum_{k=0..n-1} (-3)^k*{D_x)^(k+1)*R(n, x) = n*R(n-1, x), n >= 1.
General recurrence for Sheffer row polynomials (see the Roman reference, p. 50, Corollary 3.7.2, rewritten for the present Sheffer notation):
R(n, x) = [(4 + x)*1 + 6*(2 + x)*D_x + 3^2*x*(D_x)^2]*R(n-1, x), n >= 1, with R(0, x) = 1.
Boas-Buck recurrence for column m (see a comment in A286724 with references): T(n, m) = (n!/(n-m))*(4 + 3*m)*Sum_{p=0..n-1-m} 3^p*T(n-1-p, m)/(n-1-p)!, for n > m >= 0, with input T(m, m) = 1.

A368235 Triangle read by rows: n-th row polynomial equals the numerator of the rational function (-1)^n*f(x) * (d/dx)^n (1/f(x)), where f(x) = sqrt(x + x^2).

Original entry on oeis.org

1, 1, 2, 3, 8, 8, 15, 54, 72, 48, 105, 480, 864, 768, 384, 945, 5250, 12000, 14400, 9600, 3840, 10395, 68040, 189000, 288000, 259200, 138240, 46080, 135135, 1018710, 3333960, 6174000, 7056000, 5080320, 2257920, 645120, 2027025, 17297280, 65197440, 142248960, 197568000, 180633600, 108380160, 41287680, 10321920
Offset: 0

Views

Author

Peter Bala, Dec 18 2023

Keywords

Comments

Unsigned row reverse of A123516.
The row polynomials also occur on repeated integration of 1/sqrt(x + x^2). See the example section.

Examples

			Triangle begins
 n\k |     0       1        2        3        4        5       6
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
  0  |     1
  1  |     1       2
  2  |     3       8        8
  3  |    15      54       72       48
  4  |   105     480      864      768      384
  5  |   945    5250    12000    14400     9600     3840
  6  | 10395   68040   189000   288000   259200   138240   46080
 ...
Repeated integration of 1/f(x), where f(x) = sqrt(x + x^2):
Let I denote the integral operator h(x) -> Integral_{t = 0..x} h(t) dt.
Let g(x) = I(1/f(x)) = log(2*x + 1 + 2*f(x)). Then
(2^1 * 1!^2) * I^(2)(1/f(x)) = (2*x + 1)*g(x) - 2*f(x).
(2^2 * 2!^2) * I^(3)(1/f(x)) = (8*x^2 + 8*x + 3)*g(x) - 6*(2*x + 1)*f(x).
(2^3 * 3!^2) * I^(4)(1/f(x)) = (48*x^3 + 72*x^2 + 54*x + 15)*g(x) - 2*(44*x^2 + 44*x + 15)*f(x).
(2^4 * 4!^2) * I^(5)(1/f(x)) = (384*x^4 + 768*x^3 + 864*x^2 + 480*x + 105)*g(x) - 10*(2*x + 1)*(40*x^2 + 40*x + 21)*f(x).
		

Crossrefs

Cf. A001147 (column 1), 2*A161120 (column 2), A000165 (main diagonal) A014479 (first subdiagonal), 3*A286725 (second subdiagonal).

Programs

  • Maple
    # sequence in triangular form
    T := (n, k) -> n! * 2^(2*k-n) * binomial(n, k)*binomial(2*n-2*k, n-k):
    for n from 0 to 8 do seq(T(n, k), k = 0..n) od;

Formula

T(n, k) = n! * 2^(2*k-n) * binomial(n, k) * binomial(2*n-2*k, n-k).
k*T(n, k) = (2*n^2)*T(n-1, k-1) for k >= 1 with T(n, 0) = (2*n - 1)!! = A001147(n).
T(n, 1) = 2*A161120(n).
T(n, n) = 2^n * n! = A000165(n); T(n+1, n) = 2^n * n! * (n+1)^2 = A014479(n);
T(n+2, n) = 3 * 2^(n-1)*(n+2)!*binomial(n+2, 2) = 3 * A286725(n).
More generally, T(n+r, n) = (2*r - 1)!! * A286724(n+r, r).
E.g.f.: Sum_{k >= 0} (1/2^k)*binomial(2*k, k)*t^k/(1 - 2*t*x)^(k+1) = 1 + (1 + 2*x)*t + (3 + 8*x + 8*x^2)*t^2/2! + (15 + 54*x + 72*x^2 + 48*x^3)*t^3/3! + ....
n-th row polynomial R(n, x) = (-2)^n*(x + x^2)^(n+1/2)*(d/dx)^n (1/sqrt(x + x^2)).
Recurrence for row polynomials:
R(n+1, x) = (2*x + 1)*(2*n + 1)*R(n, x) - 4*x*(x + 1)*n^2*R(n-1, x), with R(0, x) = 1.
R'(n, x) = 2*n^2 * R(n-1, x) for n >= 1.
Functional equation: R(n, -1 - x) = (-1)^n * R(n, x).
Conjecture: the zeros of the polynomial R(n, -x) lie on the vertical line Re(x) = 1/2 in the complex plane.
(-1)^n * x^n * R(n, (- 1 - x)/x) equals the n-th row polynomial of A123516.
(1 - x)^n * R(n, x/(1 - x)) equals the n-th row polynomial of A059366.
Let D denote the operator (1/x)*d/dx. Then D^(n+1)( arcsinh(x) ) = (-1)^n*R(n, x^2)/(x*sqrt(1 + x^2))^(2*n+1).
R(n, 1/2) = A331817(n); R(n, -1/2) = A177145(n+1);
(2^n) * R(n, 1/4) = A098461(n).
Alternating row sums R(n, -1) = (-1)^n * A001147(n).
Showing 1-6 of 6 results.