A051579 -id:A051579 - cp's OEIS frontend

A051577 a(n) = (2*n + 3)!!/3 = A001147(n+2)/3.

1, 5, 35, 315, 3465, 45045, 675675, 11486475, 218243025, 4583103525, 105411381075, 2635284526875, 71152682225625, 2063427784543125, 63966261320836875, 2110886623587616875, 73881031825566590625, 2733598177545963853125, 106610328924292590271875

Offset: 0

Wolfdieter Lang

Row m = 3 of the array A(3; m,n) := (2*n+m)!!/m!!, m >= 0, n >= 0.

G. C. Greubel, Table of n, a(n) for n = 0..250
H. W. Gould, Harris Kwong, and Jocelyn Quaintance, On Certain Sums of Stirling Numbers with Binomial Coefficients, J. Integer Sequences, 18 (2015), Article 15.9.6.
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.
Michael Wallner, A bijection of plane increasing trees with relaxed binary trees of right height at most one, arXiv:1706.07163 [math.CO], 2017, p. 10.
Eric Weisstein's World of Mathematics, Double Factorial.
Wikipedia, Double factorial.

Cf. A000165, A001147, A002866(n+1) (m=0..2 rows of A(3; m,n)).

Cf. A051578, A051579, A051580, A051581, A051582, A051583, A178647.

F:=Factorial;; List([0..25], n-> F(2*n+4)/(12*2^n*F(n+2)) ); # G. C. Greubel, Nov 12 2019

F:=Factorial; [F(2*n+4)/(12*2^n*F(n+2)): n in [0..25]]; // G. C. Greubel, Nov 12 2019

seq( doublefactorial(2*n+3)/3,n=0..10) ; # R. J. Mathar, Sep 29 2013

Table[(2*n + 3)!!/3!!, {n, 0, 25}] (* G. C. Greubel, Jan 22 2017 *)
a[n_] := Sum[(-1)^k*Binomial[2*n + 1, n + k]*StirlingS1[n + k + 1 ,k], {k , 1, n + 1}]; Flatten[Table[a[n], {n, 0, 18}]] (* Detlef Meya, Jan 17 2024 *)

vector(26, n, (2*n+2)!/(6*2^n*(n+1)!) ) \\ G. C. Greubel, Nov 12 2019

f=factorial; [f(2*n+4)/(12*2^n*f(n+2)) for n in (0..25)] # G. C. Greubel, Nov 12 2019

a(n) = (2*n + 3)!!/3!!.
E.g.f.: 1/(1 - 2*x)^(5/2).
a(n) ~ (4/3) * sqrt(2) * n^2 * 2^n * e^(-n) * n^n *{1 + (47/24)*n^(-1) + ...}. - Joe Keane (jgk(AT)jgk.org), Nov 23 2001
Ramanujan polynomials -psi_n(n, x) evaluated at 0. - Ralf Stephan, Apr 16 2004
a(n) = 2^(2 + n) * Gamma(n + 5/2)/(3 * sqrt(Pi)). - Gerson Washiski Barbosa, May 05 2010
From Peter Bala, May 26 2017: (Start)
D-finite with recurrence a(n+1) = (2*n + 5)*a(n) with a(0) = 1.
O.g.f. A(x) satisfies the Riccati differential equation 2*x^2*A(x)' = (1 - 5*x)*A(x) - 1 with A(0) = 1.
G.f. as an S-fraction: A(x) = 1/(1 - 5*x/(1 - 2*x/(1 - 7*x/(1 - 4*x/(1 - 9*x/(1 - 6*x/(1 - ... - (2*n+3)*x/(1 - 2*n*x/(1 - ...))))))))) (by Stokes, 1982).
Reciprocal as an S-fraction: 1/A(x) = 1/(1 + 5*x/(1 - 7*x/(1 - 2*x/(1 - 9*x/(1 - 4*x/(1 - 11*x/(1 - 6*x/(1 - ... - (2*n + 5)*x/(1 - 2*n*x/(1 - ...)))))))))). (End)
From Amiram Eldar, Dec 11 2022: (Start)
Sum_{n>=0} 1/a(n) = 3*(sqrt(e*Pi/2) * erf(1/sqrt(2)) - 1), where erf is the error function.
Sum_{n>=0} (-1)^n/a(n) = 3*(1 - sqrt(Pi/(2*e)) * erfi(1/sqrt(2))), where erfi is the imaginary error function. (End)
a(n) = Sum_{k=1..n+1} (-1)^k*binomial(2*n + 1, n + k)*Stirling1(n + k + 1, k). - Detlef Meya, Jan 17 2024

A051578 a(n) = (2*n+4)!!/4!!, related to A000165 (even double factorials).

Original entry on oeis.org

1, 6, 48, 480, 5760, 80640, 1290240, 23224320, 464486400, 10218700800, 245248819200, 6376469299200, 178541140377600, 5356234211328000, 171399494762496000, 5827582821924864000, 209792981589295104000, 7972133300393213952000, 318885332015728558080000

Offset: 0

Wolfdieter Lang

Row m=4 of the array A(3; m,n) := (2*n+m)!!/m!!, m >= 0, n >= 0.

G. C. Greubel, Table of n, a(n) for n = 0..400
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 521.
A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.

Cf. A000165, A001147(n+1), A002866(n+1), A051577 (rows m=0..3), A051579, A051580, A051581, A051582, A051583.

Cf. A052587 (essentially the same).

List([0..20], n-> 2^(n-1)*Factorial(n+2) ); # G. C. Greubel, Nov 11 2019

[2^(n-1)*Factorial(n+2): n in [0..20]]; // G. C. Greubel, Nov 11 2019

a:= proc(n) option remember; `if`(n=0, 1, 2*(n+2)*a(n-1)) end:
seq(a(n), n=0..20);  # Alois P. Heinz, Apr 29 2019
seq(2^(n-1)*(n+2)!, n=0..20); # G. C. Greubel, Nov 11 2019

Table[2^(n-1)(n+2)!, {n,0,20}] (* Jean-François Alcover, Oct 05 2019 *)
Table[(2n+4)!!/8,{n,0,20}] (* Harvey P. Dale, Apr 06 2023 *)

vector(21, n, 2^(n-2)*(n+1)! ) \\ G. C. Greubel, Nov 11 2019

apply( {A051578(n)=(n+2)!<<(n-1)}, [0..18]) \\ M. F. Hasler, Nov 10 2024

[2^(n-1)*factorial(n+2) for n in (0..20)] # G. C. Greubel, Nov 11 2019

a(n) = (2*n+4)!!/4!!.
E.g.f.: 1/(1-2*x)^3.
a(n) ~ 2^(-1/2)*Pi^(1/2)*n^(5/2)*2^n*e^-n*n^n*{1 + 37/12*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Nov 23 2001
a(n) = (n+2)!*2^(n-1). - Zerinvary Lajos, Sep 23 2006. [corrected by Gary Detlefs, Apr 29 2019]
a(n) = 2^n*A001710(n+2). - R. J. Mathar, Feb 22 2008
From Peter Bala, May 26 2017: (Start)
a(n+1) = (2*n + 6)*a(n) with a(0) = 1.
O.g.f. satisfies the Riccati differential equation 2*x^2*A(x)' = (1 - 6*x)*A(x) - 1 with A(0) = 1.
G.f. as an S-fraction: A(x) = 1/(1 - 6*x/(1 - 2*x/(1 - 8*x/(1 - 4*x/(1 - 10*x/(1 - 6*x/(1 - ... - (2*n + 4)*x/(1 - 2*n*x/(1 - ...))))))))) (by Stokes 1982).
Reciprocal as an S-fraction: 1/A(x) = 1/(1 + 6*x/(1 - 8*x/(1 - 2*x/(1 - 10*x/(1 - 4*x/(1 - 12*x/(1 - 6*x/(1 - ... - (2*n + 6)*x/(1 - 2*n*x/(1 - ...)))))))))). (End)
From Amiram Eldar, Dec 11 2022: (Start)
Sum_{n>=0} 1/a(n) = 8*sqrt(e) - 12.
Sum_{n>=0} (-1)^n/a(n) = 8/sqrt(e) - 4. (End)
a(n) = A052587(n+2) for n > 0. - M. F. Hasler, Nov 10 2024

A051580 a(n) = (2*n+6)!!/6!!, related to A000165 (even double factorials).

Original entry on oeis.org

1, 8, 80, 960, 13440, 215040, 3870720, 77414400, 1703116800, 40874803200, 1062744883200, 29756856729600, 892705701888000, 28566582460416000, 971263803654144000, 34965496931549184000, 1328688883398868992000

Offset: 0

Wolfdieter Lang

Row m=6 of the array A(3; m,n) := (2*n+m)!!/m!!, m >= 0, n >= 0.

G. C. Greubel, Table of n, a(n) for n = 0..400
A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.

Cf. A000165, A001147(n+1), A002866(n+1), A051577, A051578, A051579 (rows m=0..5), A051581, A051582, A051583.

List([0..20], n-> Product([1..n], j-> 2*j+6) ); # G. C. Greubel, Nov 11 2019

[1] cat [(&*[2*j+6: j in [1..n]]): n in [1..20]]; // G. C. Greubel, Nov 11 2019

seq( mul(2*j+6, j=1..n), n=0..20); # G. C. Greubel, Nov 11 2019

Table[2^n*Pochhammer[4, n], {n,0,20}] (* G. C. Greubel, Nov 11 2019 *)
Table[(2n+6)!!/6!!,{n,0,20}] (* Harvey P. Dale, Mar 03 2022 *)

vector(20, n, prod(j=1,n-1, 2*j+6) ) \\ G. C. Greubel, Nov 11 2019

[product( (2*j+6) for j in (1..n)) for n in (0..20)] # G. C. Greubel, Nov 11 2019

a(n) = (2*n+6)!!/6!!.
E.g.f.: 1/(1-2*x)^4.
a(n) = n!*2^(n-4)/3, n>=3. - Zerinvary Lajos, Sep 23 2006
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - x/(x + 1/(2*k+8)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 02 2013
From Peter Bala, May 26 2017: (Start)
a(n+1) = (2*n + 8)*a(n) with a(0) = 1.
O.g.f. satisfies the Riccati differential equation 2*x^2*A(x)' = (1 - 8*x)*A(x) - 1 with A(0) = 1.
G.f. as an S-fraction: A(x) = 1/(1 - 8*x/(1 - 2*x/(1 - 10*x/(1 - 4*x/(1 - 12*x/(1 - 6*x/(1 - ... - (2*n + 6)*x/(1 - 2*n*x/(1 - ...))))))))) (by Stokes 1982).
Reciprocal as an S-fraction: 1/A(x) = 1/(1 + 8*x/(1 - 10*x/(1 - 2*x/(1 - 12*x/(1 - 4*x/(1 - 14*x/(1 - 6*x/(1 - ... - (2*n + 8)*x/(1 - 2*n*x/(1 - ...)))))))))). (End)
From Amiram Eldar, Dec 11 2022: (Start)
Sum_{n>=0} 1/a(n) = 48*sqrt(e) - 78.
Sum_{n>=0} (-1)^n/a(n) = 30 - 48/sqrt(e). (End)

A051582 a(n) = (2*n+8)!!/8!!, related to A000165 (even double factorials).

Original entry on oeis.org

1, 10, 120, 1680, 26880, 483840, 9676800, 212889600, 5109350400, 132843110400, 3719607091200, 111588212736000, 3570822807552000, 121407975456768000, 4370687116443648000, 166086110424858624000, 6643444416994344960000

Offset: 0

Wolfdieter Lang

Row m=8 of the array A(3; m,n) := (2*n+m)!!/m!!, m >= 0, n >= 0.

G. C. Greubel, Table of n, a(n) for n = 0..399
A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.

Cf. A000165, A001147(n+1), A002866(n+1).

Cf. A051577, A051578, A051579, A051580, A051581 (rows m=0..7), A051583.

F:=Factorial;; List([0..20], n-> 2^n*F(n+4)/F(4) ); # G. C. Greubel, Nov 12 2019

F:=Factorial; [2^n*F(n+4)/F(4): n in [0..20]]; // G. C. Greubel, Nov 12 2019

seq(2^n*pochhammer(5, n), n=0..20); # G. C. Greubel, Nov 12 2019

(2Range[0,20]+8)!!/8!! (* Harvey P. Dale, Feb 03 2013 *)
Table[2^n*Pochhammer[5, n], {n,0,20}] (* G. C. Greubel, Nov 12 2019 *)

vector(20, n, n--; (n+4)!*2^(n-1)/12) \\ Michel Marcus, Feb 09 2015

f=factorial; [2^n*f(n+4)/f(4) for n in (0..20)] # G. C. Greubel, Nov 12 2019

a(n) = (2*n+8)!!/8!!.
E.g.f.: 1/(1-2*x)^5.
a(n) = (n+4)!*2^(n-1)/12. - Zerinvary Lajos, Sep 23 2006
From Peter Bala, May 26 2017: (Start)
a(n+1) = (2*n + 10)*a(n) with a(0) = 1.
O.g.f. satisfies the Riccati differential equation 2*x^2*A(x)' = (1 - 10*x)*A(x) - 1 with A(0) = 1.
G.f. as an S-fraction: A(x) = 1/(1 - 10*x/(1 - 2*x/(1 - 12*x/(1 - 4*x/(1 - 14*x/(1 - 6*x/(1 - ... - (2*n + 8)*x/(1 - 2*n*x/(1 - ...))))))))) (by Stokes 1982).
Reciprocal as an S-fraction: 1/A(x) = 1/(1 + 10*x/(1 - 12*x/(1 - 2*x/(1 - 14*x/(1 - 4*x/(1 - 16*x/(1 - 6*x/(1 - ... - (2*n + 10)*x/(1 - 2*n*x/(1 - ...)))))))))). (End)
From Amiram Eldar, Dec 11 2022: (Start)
Sum_{n>=0} 1/a(n) = 384*sqrt(e) - 632.
Sum_{n>=0} (-1)^n/a(n) = 384/sqrt(e) - 232. (End)

A051581 a(n) = (2*n+7)!!/7!!, related to A001147 (odd double factorials).

Original entry on oeis.org

1, 9, 99, 1287, 19305, 328185, 6235515, 130945815, 3011753745, 75293843625, 2032933777875, 58955079558375, 1827607466309625, 60311046388217625, 2110886623587616875, 78102805072741824375, 3046009397836931150625

Offset: 0

Wolfdieter Lang

Row m=7 of the array A(3; m,n) := (2*n+m)!!/m!!, m >= 0, n >= 0.

G. C. Greubel, Table of n, a(n) for n = 0..400
A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.

Cf. A000165, A001147(n+1), A002866(n+1).

Cf. A051577, A051578, A051579, A051580 (rows m=0..6), A051582, A051583.

List([0..20], n-> Product([0..n-1], j-> 2*j+9) ); # G. C. Greubel, Nov 12 2019

[1] cat [(&*[2*j+9: j in [0..n-1]]): n in [1..20]]; // G. C. Greubel, Nov 12 2019

df:=doublefactorial; seq(df(2*n+7)/df(7), n = 0..20); # G. C. Greubel, Nov 12 2019

Table[2^n*Pochhammer[9/2, n], {n,0,20}] (* G. C. Greubel, Nov 12 2019 *)

vector(20, n, prod(j=1,n-1, 2*j+7) ) \\ G. C. Greubel, Nov 12 2019

[product( (2*j+9) for j in (0..n-1)) for n in (0..20)] # G. C. Greubel, Nov 12 2019

a(n) = (2*n+7)!!/7!!.
E.g.f.: 1/(1-2*x)^(9/2).
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - x/(x + 1/(2*k+9)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 02 2013
From Peter Bala, May 26 2017: (Start)
a(n+1) = (2*n + 9)*a(n) with a(0) = 1.
O.g.f. satisfies the Riccati differential equation 2*x^2*A(x)' = (1 - 9*x)*A(x) - 1 with A(0) = 1.
G.f. as an S-fraction: A(x) = 1/(1 - 9*x/(1 - 2*x/(1 - 11*x/(1 - 4*x/(1 - 13*x/(1 - 6*x/(1 - ... - (2*n + 7)*x/(1 - 2*n*x/(1 - ...))))))))) (by Stokes 1982).
Reciprocal as an S-fraction: 1/A(x) = 1/(1 + 9*x/(1 - 11*x/(1 - 2*x/(1 - 13*x/(1 - 4*x/(1 - 15*x/(1 - 6*x/(1 - ... - (2*n + 9)*x/(1 - 2*n*x/(1 - ...)))))))))). (End)
From Amiram Eldar, Dec 11 2022: (Start)
Sum_{n>=0} 1/a(n) = 105 * sqrt(e*Pi/2) * erf(1/sqrt(2)) - 147, where erf is the error function.
Sum_{n>=0} (-1)^n/a(n) = 77 - 105 * sqrt(Pi/(2*e)) * erfi(1/sqrt(2)), where erfi is the imaginary error function. (End)

A051583 a(n) = (2*n+9)!!/9!!, related to A001147 (odd double factorials).

Original entry on oeis.org

1, 11, 143, 2145, 36465, 692835, 14549535, 334639305, 8365982625, 225881530875, 6550564395375, 203067496256625, 6701227376468625, 234542958176401875, 8678089452526869375, 338445488648547905625, 13876265034590464130625

Offset: 0

Wolfdieter Lang

Row m=9 of the array A(3; m,n) := (2*n+m)!!/m!!, m >= 0, n >= 0.

G. C. Greubel, Table of n, a(n) for n = 0..398
A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.

Cf. A000165, A001147(n+1), A002866(n+1), A178647.

Cf. A051577, A051578, A051579, A051580, A051581, A051582 (rows m=0..8).

List([0..20], n-> Product([0..n-1], j-> 2*j+11) ); # G. C. Greubel, Nov 12 2019

[1] cat [(&*[2*j+11: j in [0..n-1]]): n in [1..20]]; // G. C. Greubel, Nov 12 2019

seq(2^n*pochhammer(11/2,n), n = 0..20); # G. C. Greubel, Nov 12 2019

(2*Range[0,20]+9)!!/945 (* Harvey P. Dale, Apr 10 2019 *)
Table[2^n*Pochhammer[11/2, n], {n,0,20}] (* G. C. Greubel, Nov 12 2019 *)

vector(20, n, prod(j=0,n-2, 2*j+11) ) \\ G. C. Greubel, Nov 12 2019

[product( (2*j+11) for j in (0..n-1)) for n in (0..20)] # G. C. Greubel, Nov 12 2019

a(n) = (2*n+9)!!/9!!.
E.g.f.: 1/(1-2*x)^(11/2).
From Peter Bala, May 26 2017: (Start)
a(n+1) = (2*n + 11)*a(n) with a(0) = 1.
O.g.f. satisfies the Riccati differential equation 2*x^2*A(x)' = (1 - 11*x)*A(x) - 1 with A(0) = 1.
G.f. as an S-fraction: A(x) = 1/(1 - 11*x/(1 - 2*x/(1 - 13*x/(1 - 4*x/(1 - 15*x/(1 - 6*x/(1 - ... - (2*n + 9)*x/(1 - 2*n*x/(1 - ...))))))))) (by Stokes 1982).
Reciprocal as an S-fraction: 1/A(x) = 1/(1 + 11*x/(1 - 13*x/(1 - 2*x/(1 - 15*x/(1 - 4*x/(1 - 17*x/(1 - 6*x/(1 - ... - (2*n + 11)*x/(1 - 2*n*x/(1 - ...)))))))))). (End)
From Amiram Eldar, Dec 11 2022: (Start)
Sum_{n>=0} 1/a(n) = 945 * sqrt(e*Pi/2) * erf(1/sqrt(2)) - 1332, where erf is the error function.
Sum_{n>=0} (-1)^n/a(n) = 945 * sqrt(Pi/(2*e)) * erfi(1/sqrt(2)) - 684, where erfi is the imaginary error function. (End)

A112292 An invertible triangle of ratios of double factorials.

Original entry on oeis.org

1, 1, 1, 3, 3, 1, 15, 15, 5, 1, 105, 105, 35, 7, 1, 945, 945, 315, 63, 9, 1, 10395, 10395, 3465, 693, 99, 11, 1, 135135, 135135, 45045, 9009, 1287, 143, 13, 1, 2027025, 2027025, 675675, 135135, 19305, 2145, 195, 15, 1, 34459425, 34459425, 11486475, 2297295, 328185, 36465, 3315, 255, 17, 1

Offset: 0

Paul Barry, Sep 01 2005

As a square array read by antidiagonals, column k has e.g.f. (1/(1-2x)^(1/2))*(1/(1-2x))^k. - Paul Barry, Sep 04 2005

Let G(m, k, p) = (-p)^k*Product_{j=0..k-1}(j - m - 1/p) and T(n, k, p) = G(n-1, n-k, p) then T(n, k, 1) = A094587(n, k), T(n, k, 2) is this sequence and T(n, k, 3) = A136214. - Peter Luschny, Jun 01 2009, revised Jun 18 2019

			Triangle begins
      1;
      1,     1;
      3,     3,    1;
     15,    15,    5,  1;
    105,   105,   35,  7,  1;
    945,   945,  315, 63,  9,  1;
  10395, 10395, 3465,693, 99, 11, 1;
Inverse is A112295, which begins
   1;
  -1,  1;
   0, -3,  1;
   0,  0, -5,  1;
   0,  0,  0, -7,  1;
   0,  0,  0,  0, -9,  1;
Similar results arise for higher factorials.

Columns include A001147, A051577, A051579.
Row sums are A112293.
Diagonal sums are A112294.
Cf. A094587 (p=1), this sequence (p=2), A136214 (p=3).

T[n_, k_] := If[k <= n, (2n-1)!!/(2k-1)!!, 0];
Table[T[n, k], {n, 0, 9}, {k, 0, n}] (* Jean-François Alcover, Jun 13 2019 *)

T(n, k)=if(k<=n, (2n-1)!!/(2k-1)!!, 0);
T(n, k)=if(k<=n, n!*C(2n, n)2^(k-n)/(k!*C(2k, k)), 0);
T(n, k)=if(k<=n, 2^(n-k)(n-1/2)!/(k-1/2)!, 0);
T(n, k)=if(k<=n, (n+1)!*C(n)2^(k-n)/((k+1)!*C(k)), 0).

A370419 A(n, k) = 2^n*Pochhammer(k/2, n). Square array read by ascending antidiagonals.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 3, 2, 1, 0, 15, 8, 3, 1, 0, 105, 48, 15, 4, 1, 0, 945, 384, 105, 24, 5, 1, 0, 10395, 3840, 945, 192, 35, 6, 1, 0, 135135, 46080, 10395, 1920, 315, 48, 7, 1, 0, 2027025, 645120, 135135, 23040, 3465, 480, 63, 8, 1

Offset: 0

Peter Luschny, Mar 04 2024

			The array starts:
[0] 1,   1,    1,     1,     1,     1,     1,      1,      1, ...
[1] 0,   1,    2,     3,     4,     5,     6,      7,      8, ...
[2] 0,   3,    8,    15,    24,    35,    48,     63,     80, ...
[3] 0,  15,   48,   105,   192,   315,   480,    693,    960, ...
[4] 0, 105,  384,   945,  1920,  3465,  5760,   9009,  13440, ...
[5] 0, 945, 3840, 10395, 23040, 45045, 80640, 135135, 215040, ...
.
Seen as the triangle T(n, k) = A(n - k, k):
[0] 1;
[1] 0,   1;
[2] 0,   1,   1;
[3] 0,   3,   2,   1;
[4] 0,  15,   8,   3,  1;
[5] 0, 105,  48,  15,  4, 1;
[6] 0, 945, 384, 105, 24, 5, 1;
.
From _Werner Schulte_, Mar 07 2024: (Start)
Illustrating the LU decomposition of A:
    / 1                \   / 1 1 1 1 1 ... \   / 1   1   1   1    1 ... \
    | 0   1            |   |   1 2 3 4 ... |   | 0   1   2   3    4 ... |
    | 0   3   2        | * |     1 3 6 ... | = | 0   3   8  15   24 ... |
    | 0  15  18   6    |   |       1 4 ... |   | 0  15  48 105  192 ... |
    | 0 105 174 108 24 |   |         1 ... |   | 0 105 384 945 1920 ... |
    | . . .            |   | . . .         |   | . . .                  |. (End)

Paolo Xausa, Table of n, a(n) for n = 0..11324 (first 150 antidiagonals, flattened).

Columns: A000007, A001147, A000165, A001147 (shifted), A002866, A051577, A051578, A051579, A051580.
Rows: A000012, A001477, A005563, A370912, A190577.
Cf. A035342, A265649, A370890, A370982 (row sums of the triangle), A370915, A371025, A371077.

A := (n, k) -> 2^n*pochhammer(k/2, n):
for n from 0 to 5 do seq(A(n, k), k = 0..9) od;
T := (n, k) -> A(n - k, k): seq(seq(T(n, k), k = 0..n), n = 0..9);
# Using the exponential generating functions of the columns:
EGFcol := proc(k, len) local egf, ser, n; egf := (1 - 2*x)^(-k/2);
ser := series(egf, x, len+2): seq(n!*coeff(ser, x, n), n = 0..len) end:
seq(lprint(EGFcol(n, 9)), n = 0..8);
# Using the generating polynomials for the rows:
P := (n, x) -> local k; add(Stirling1(n, k)*(-2)^(n - k)*x^k, k=0..n):
seq(lprint([n], seq(P(n, k), k = 0..8)), n = 0..5);
# Implementing the comment of Werner Schulte about the LU decomposition of A:
with(LinearAlgebra):
L := Matrix(7, 7, (n, k) -> A371025(n - 1,  k - 1)):
U := Matrix(7, 7, (n, k) -> binomial(n - 1, k - 1)):
MatrixMatrixMultiply(L, Transpose(U));  #  Peter Luschny, Mar 08 2024

A370419[n_, k_] := 2^n*Pochhammer[k/2, n];
Table[A370419[n-k, k], {n, 0, 10}, {k, 0, n}] (* Paolo Xausa, Mar 06 2024 *)

def A(n, k): return 2**n * rising_factorial(k/2, n)
for n in range(6): print([A(n, k) for k in range(9)])

The polynomials P(n, x) = Sum_{k=0..n} Stirling1(n, k)*(-2)^(n-k)*x^k are ordinary generating functions for row n, i.e., A(n, k) = P(n, k).
From Werner Schulte, Mar 07 2024: (Start)
A(n, k) = Product_{i=1..n} (2*i - 2 + k).
E.g.f. of column k: Sum_{n>=0} A(n, k) * t^n / (n!) = (1/sqrt(1 - 2*t))^k.
A(n, k) = A(n+1, k-2) / (k - 2) for k > 2.
A(n, k) = Sum_{i=0..k-1} i! * A265649(n, i) * binomial(k-1, i) for k > 0.
E.g.f. of row n > 0: Sum_{k>=1} A(n, k) * x^k / (k!) = (Sum_{k=1..n} A035342(n, k) * x^k) * exp(x).
Sum_{n>=0, k>=0} A(n, k) * x^k * t^n / (k! * n!) = exp(x/sqrt(1 - 2*t)).
Sum_{n>=0, k>=0} A(n, k) * x^k * t^n / (n!) = 1 / (1 - x/sqrt(1 - 2*t)).
The LU decomposition of this array is given by the upper triangular matrix U which is the transpose of A007318 and the lower triangular matrix L, where L is defined L(n, k) = A035342(n, k) * k! for 1 <= k <= n and L(n, 0) = 0^n. Note that L(n, k) + L(n, k+1) = A265649(n, k) * k! for 0 <= k <= n. (End)

A153272 Triangle T(n, k) = Product_{j=0..k} (j*n + prime(m)), with T(n, 0) = prime(m) and m = 4, read by rows.

Original entry on oeis.org

7, 7, 56, 7, 63, 693, 7, 70, 910, 14560, 7, 77, 1155, 21945, 504735, 7, 84, 1428, 31416, 848232, 27143424, 7, 91, 1729, 43225, 1339975, 49579075, 2131900225, 7, 98, 2058, 57624, 2016840, 84707280, 4150656720, 232436776320, 7, 105, 2415, 74865, 2919735, 137227545, 7547514975, 475493443425, 33760034483175

Offset: 0

Roger L. Bagula, Dec 22 2008

Row sums are {7, 63, 763, 15547, 527919, 28024591, 2182864327, 236674216947, 34243215666247, 6391699984166119, 1497639790982770659, ...}.

			Triangle begins as:
  7;
  7, 56;
  7, 63,  693;
  7, 70,  910, 14560;
  7, 77, 1155, 21945,  504735;
  7, 84, 1428, 31416,  848232, 27143424;
  7, 91, 1729, 43225, 1339975, 49579075, 2131900225;

G. C. Greubel, Rows n = 0..100 of triangle, flattened

Cf. A153270 (m=2), A153271 (m=3), this sequence (m=4).

Cf. A001730, A051579, A051604.

m:=4;
function T(n,k)
  if k eq 0 then return NthPrime(m);
  else return (&*[j*n + NthPrime(m): j in [0..k]]);
  end if; return T; end function;
[T(n,k): k in [0..n], n in [0..10]]; // G. C. Greubel, Dec 03 2019

m:=4; seq(seq(`if`(k=0, ithprime(m), mul(j*n + ithprime(m), j=0..k)), k=0..n), n=0..10); # G. C. Greubel, Dec 03 2019

T[n_, k_, m_]:= If[k==0, Prime[m], Product[j*n + Prime[m], {j,0,k}]];
Table[T[n,k,4], {n,0,10}, {k,0,n}]//Flatten

T(n,k) = my(m=4); if(k==0, prime(m), prod(j=0,k, j*n + prime(m)) ); \\ G. C. Greubel, Dec 03 2019

def T(n, k):
    m=4
    if (k==0): return nth_prime(m)
    else: return product(j*n + nth_prime(m) for j in (0..k))
[[T(n, k) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Dec 03 2019

T(n, k) = Product_{j=0..k} (j*n + prime(m)), with T(n, 0) = prime(m) and m = 4.

Edited by G. C. Greubel, Dec 03 2019

cp's OEIS Frontend

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