A001147 -id:A001147 - cp's OEIS frontend

A035342 The convolution matrix of the double factorial of odd numbers (A001147).

1, 3, 1, 15, 9, 1, 105, 87, 18, 1, 945, 975, 285, 30, 1, 10395, 12645, 4680, 705, 45, 1, 135135, 187425, 82845, 15960, 1470, 63, 1, 2027025, 3133935, 1595790, 370125, 43890, 2730, 84, 1, 34459425, 58437855, 33453945, 8998290

Offset: 1

Wolfdieter Lang

Previous name was: A triangle of numbers related to the triangle A035324; generalization of Stirling numbers of second kind A008277 and Lah numbers A008297.
If one replaces in the recurrence the '2' by '0', resp. '1', one obtains the Lah-number, resp. Stirling-number of 2nd kind, triangle A008297, resp. A008277.
The product of two lower triangular Jabotinsky matrices (see A039692 for the Knuth 1992 reference) is again such a Jabotinsky matrix: J(n,m) = Sum_{j=m..n} J1(n,j)*J2(j,m). The e.g.f.s of the first columns of these triangular matrices are composed in the reversed order: f(x)=f2(f1(x)). With f1(x)=-(log(1-2*x))/2 for J1(n,m)=|A039683(n,m)| and f2(x)=exp(x)-1 for J2(n,m)=A008277(n,m) one has therefore f2(f1(x))=1/sqrt(1-2*x) - 1 = f(x) for J(n,m)=a(n,m). This proves the matrix product given below. The m-th column of a Jabotinsky matrix J(n,m) has e.g.f. (f(x)^m)/m!, m>=1.
a(n,m) gives the number of forests with m rooted ordered trees with n non-root vertices labeled in an organic way. Organic labeling means that the vertex labels along the (unique) path from the root with label 0 to any leaf (non-root vertex of degree 1) is increasing. Proof: first for m=1 then for m>=2 using the recurrence relation for a(n,m) given below. - Wolfdieter Lang, Aug 07 2007
Also the Bell transform of A001147(n+1) (adding 1,0,0,... as column 0). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 19 2016

			Matrix begins:
    1;
    3,   1;
   15,   9,   1;
  105,  87,  18,   1;
  945, 975, 285,  30,   1;
  ...
Combinatoric meaning of a(3,2)=9: The nine increasing path sequences for the three rooted ordered trees with leaves labeled with 1,2,3 and the root labels 0 are: {(0,3),[(0,1),(0,2)]}; {(0,3),[(0,2),(0,1)]}; {(0,3),(0,1,2)}; {(0,1),[(0,3),(0,2)]}; [(0,1),[(0,2),(0,3)]]; [(0,2),[(0,1),(0,3)]]; {(0,2),[(0,3),(0,1)]}; {(0,1),(0,2,3)}; {(0,2),(0,1,3)}.

Reinhard Zumkeller, Rows n = 1..125 of triangle, flattened
Peter Bala, Generalized Dobinski formulas
J. Fernando Barbero G., Jesús Salas, and Eduardo J. S. Villaseñor, Bivariate Generating Functions for a Class of Linear Recurrences. I. General Structure, arXiv:1307.2010 [math.CO], 2013-2014.
Pawel Blasiak, Karol A. Penson, and Allan I. Solomon, The Boson Normal Ordering Problem and Generalized Bell Numbers, arXiv:quant-ph/0212072, 2002.
Pawel Blasiak, Karol A. Penson, and Allan I. Solomon, The general boson normal ordering problem, arXiv:quant-ph/0402027, 2004.
Richell O. Celeste, Roberto B. Corcino and Ken Joffaniel M. Gonzales. Two Approaches to Normal Order Coefficients. Journal of Integer Sequences, Vol. 20 (2017), Article 17.3.5.
Tom Copeland, A Class of Differential Operators and the Stirling Numbers, 2015.
Tom Copeland, Addendum to Mathemagical Forests, 2010.
Tom Copeland, Mathemagical Forests, 2008.
Askar Dzhumadildaev and Damir Yeliussizov, Path decompositions of digraphs and their applications to Weyl algebra, arXiv preprint arXiv:1408.6764v1 [math.CO], 2014. [Version 1 contained many references to the OEIS, which were removed in Version 2. - _N. J. A. Sloane_, Mar 28 2015]
Askar Dzhumadil'daev and Damir Yeliussizov, Walks, partitions, and normal ordering, Electronic Journal of Combinatorics, 22(4) (2015), #P4.10.
Milan Janjic, Some classes of numbers and derivatives, JIS 12 (2009) #09.8.3.
Donald E. Knuth, Convolution polynomials, arXiv:math/9207221 [math.CA], 1992; Mathematica J. 2.1 (1992), no. 4, 67-78.
Wolfdieter Lang, On generalizations of Stirling number triangles, J. Integer Seqs., Vol. 3 (2000), #00.2.4.
Wolfdieter Lang, First 10 rows.
Shi-Mei Ma, Some combinatorial sequences associated with context-free grammars, arXiv:1208.3104v2 [math.CO], 2012.
Toufik Mansour, Matthias Schork and Mark Shattuck, The Generalized Stirling and Bell Numbers Revisited, Journal of Integer Sequences, Vol. 15 (2012), #12.8.3.
Erich Neuwirth, Recursively defined combinatorial functions: Extending Galton's board, Discrete Math. (2001) Vol. 239, 33-51.
Mathias Pétréolle and Alan D. Sokal, Lattice paths and branched continued fractions. II. Multivariate Lah polynomials and Lah symmetric functions, arXiv:1907.02645 [math.CO], 2019.
Umesh Shankar, Log-concavity of rows of triangular arrays satisfying a certain super-recurrence, arXiv:2508.12467 [math.CO], 2025. See p. 4.

The column sequences are A001147, A035101, A035119, ...
Row sums: A049118(n), n >= 1.
Cf. A000108, A035324, A008277, A008297, A094638.

a035342 n k = a035342_tabl !! (n-1) !! (k-1)
a035342_row n = a035342_tabl !! (n-1)
a035342_tabl = map fst $ iterate (\(xs, i) -> (zipWith (+)
   ([0] ++ xs) $ zipWith (*) [i..] (xs ++ [0]), i + 2)) ([1], 3)
-- Reinhard Zumkeller, Mar 12 2014

T := (n,k) -> 2^(k-n)*hypergeom([k-n,k+1],[k-2*n+1],2)*GAMMA(2*n-k)/
(GAMMA(k)*GAMMA(n-k+1)); for n from 1 to 9 do seq(simplify(T(n,k)),k=1..n) od; # Peter Luschny, Mar 31 2015
T := (n, k) -> local j; 2^n*add((-1)^(k-j)*binomial(k, j)*pochhammer(j/2, n), j = 1..k)/k!: for n from 1 to 6 do seq(T(n, k), k=1..n) od;  # Peter Luschny, Mar 04 2024

a[n_, k_] := 2^(n+k)*n!/(4^n*n*k!)*Sum[(j+k)*2^(j)*Binomial[j + k - 1, k-1]*Binomial[2*n - j - k - 1, n-1], {j, 0, n-k}]; Flatten[Table[a[n,k], {n, 1, 9}, {k, 1, n}] ] [[1 ;; 40]] (* Jean-François Alcover, Jun 01 2011, after Vladimir Kruchinin *)

a(n,k):=2^(n+k)*n!/(4^n*n*k!)*sum((j+k)*2^(j)*binomial(j+k-1,k-1)*binomial(2*n-j-k-1,n-1),j,0,n-k); /* Vladimir Kruchinin, Mar 30 2011 */

# uses[bell_matrix from A264428]
# Adds a column 1,0,0,0, ... at the left side of the triangle.
print(bell_matrix(lambda n: A001147(n+1), 9)) # Peter Luschny, Jan 19 2016

a(n, m) = Sum_{j=m..n} |A039683(n, j)|*S2(j, m) (matrix product), with S2(j, m) := A008277(j, m) (Stirling2 triangle). Priv. comm. to Wolfdieter Lang by E. Neuwirth, Feb 15 2001; see also the 2001 Neuwirth reference. See the comment on products of Jabotinsky matrices.

a(n, m) = n!*A035324(n, m)/(m!*2^(n-m)), n >= m >= 1; a(n+1, m)= (2*n+m)*a(n, m)+a(n, m-1); a(n, m) := 0, n

E.g.f. of m-th column: ((x*c(x/2)/sqrt(1-2*x))^m)/m!, where c(x) = g.f. for Catalan numbers A000108.

From Vladimir Kruchinin, Mar 30 2011: (Start)

G.f. (1/sqrt(1-2*x) - 1)^k = Sum_{n>=k} (k!/n!)*a(n,k)*x^n.

a(n,k) = 2^(n+k) * n! / (4^n*n*k!) * Sum_{j=0..n-k} (j+k) * 2^(j) * binomial(j+k-1,k-1) * binomial(2*n-j-k-1,n-1). (End)

From Peter Bala, Nov 25 2011: (Start)

E.g.f.: G(x,t) = exp(t*A(x)) = 1 + t*x + (3*t + t^2)*x^2/2! + (15*t + 9*t^2 + t^3)*x^3/3! + ..., where A(x) = -1 + 1/sqrt(1-2*x) satisfies the autonomous differential equation A'(x) = (1+A(x))^3.

The generating function G(x,t) satisfies the partial differential equation t*(dG/dt+G) = (1-2*x)*dG/dx, from which follows the recurrence given above.

The row polynomials are given by D^n(exp(x*t)) evaluated at x = 0, where D is the operator (1+x)^3*d/dx. Cf. A008277 (D = (1+x)*d/dx), A105278 (D = (1+x)^2*d/dx), A035469 (D = (1+x)^4*d/dx) and A049029 (D = (1+x)^5*d/dx). (End)

The n-th row polynomial R(n,x) is given by the Dobinski-type formula R(n,x) = exp(-x)*Sum_{k>=1} k*(k+2)*...*(k+2*n-2)*x^k/k!. - Peter Bala, Jun 22 2014

T(n,k) = 2^(k-n)*hypergeom([k-n,k+1],[k-2*n+1],2)*Gamma(2*n-k)/(Gamma(k)*Gamma(n-k+1)). - Peter Luschny, Mar 31 2015

T(n,k) = 2^n*Sum_{j=1..k} ((-1)^(k-j)*binomial(k, j)*Pochhammer(j/2, n)) / k!. - Peter Luschny, Mar 04 2024

Simpler name from Peter Luschny, Mar 31 2015

A112934 a(0) = 1; a(n+1) = Sum_{k=0..n} a(k)*A001147(n-k), where A001147 = double factorial numbers.

Original entry on oeis.org

1, 1, 2, 6, 26, 158, 1282, 13158, 163354, 2374078, 39456386, 737125446, 15279024026, 347786765150, 8621313613954, 231139787526822, 6663177374810266, 205503866668090750, 6751565903597571842

Offset: 0

Philippe Deléham and Paul D. Hanna, Oct 09 2005

nonn

			A(x) = 1 + x + 2*x^2 + 6*x^3 + 26*x^4 + 158*x^5 + 1282*x^6 + ...
1/A(x) = 1 - x - x^2 - 3*x^3 - 15*x^4 - 105*x^5 - ... - A001147(n)*x^(n+1) - ...
a(4) = a(3+1) = Sum_{k=0..3} a(k)*A001147(3-k) = a(0)*5!! + a(1)*3!! + a(2)*1 + a(3)*1 = 1*15 + 1*3 + 2*1 + 6*1 = 26. - _Michael B. Porter_, Jul 22 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A001147, A112935 (log derivative); A112936, A112937, A112938, A112939, A112940, A112941, A112942, A112943.

a_list := proc(len) local A, n; A[0] := 1; A[1] := 1;
for n from 2 to len-1 do A[n] := (2*n-1)*A[n-1] - add(A[j]*A[n-j], j=1..n-1) od;
convert(A, list) end: a_list(19); # Peter Luschny, May 22 2017
# Alternative:
T := proc(n, k) option remember; if k = 0 then 1 else if k = n then T(n, k-1)
else (n - k) * T(n, k - 1) + T(n - 1, k) fi fi end:
a := n -> T(n, n): seq(a(n), n = 0..18);  # Peter Luschny, Oct 02 2023

a[0] = 1; a[n_] := a[n] = Sum[a[k]*(2n - 2k - 3)!!, {k, 0, n - 1}]; Table[ a[n], {n, 0, 19}] (* Robert G. Wilson v, Oct 12 2005 *)

{a(n)=local(F=1+x+x*O(x^n));for(i=1,n,F=1+x+2*x^2*deriv(F)/F); return(polcoeff(F,n,x))}

{a(n) = local(A); if( n<1, n==0, A = vector(n); A[1] = 1; for( k=2, n, A[k] = (2*k - 1) * A[k-1] - sum( j=1, k-1, A[j] * A[k-j])); A[n])} /* Michael Somos, Jul 23 2011 */

INVERT transform of double factorials (A001147), shifted right one place, where g.f. A(x) satisfies: A(x) = 1 + x*[d/dx x*A(x)^2]/A(x)^2.
G.f. A(x) satisfies: A(x) = 1+x + 2*x^2*[d/dx A(x)]/A(x) (log derivative).
G.f.: A(x) = 1+x +2*x^2/(1-3*x -2*2*1*x^2/(1-7*x -2*3*3*x^2/(1-11*x -2*4*5*x^2/(1-15*x - ... -2*n*(2*n-3)*x^2/(1-(4*n-1)*x - ...)))) (continued fraction).
G.f.: A(x) = 1/(1-x/(1 -1*x/(1-2*x/(1 -3*x/(1-4*x(1 - ...))))))) (continued fraction).
From Paul Barry, Dec 04 2009: (Start)
The g.f. of a(n+1) is 1/(1-2x/(1-x/(1-4x/(1-3x/(1-6x/(1-5x/(1-.... (continued fraction).
The Hankel transform of a(n+1) is A137592. (End)
a(n) = Sum_{k=0..n} A111106(n,k). - Philippe Deléham, Jun 20 2006
From Gary W. Adamson, Jul 08 2011: (Start)
a(n) is the upper left term in M^n, M = the production matrix:
  1, 1;
  1, 1, 2;
  1, 1, 2, 3;
  1, 1, 2, 3, 4;
  1, 1, 2, 3, 4, 5;
  ... (End)
From Gary W. Adamson, Jul 21 2016: (Start)
Another production matrix Q is:
  1, 1, 0, 0, 0, ...
  1, 0, 3, 0, 0, ...
  1, 0, 0, 5, 0, ...
  1, 0, 0, 0, 7, ...
  ...
The sequence is generated by extracting the upper left term of powers of Q. By extracting the top row of Q^n, we obtain a triangle with the sequence in the left column and row sums = (1, 2, 6, 26, 158, ...): (1), (1, 1), (2, 1, 3), (6, 2, 3, 15), (26, 6, 6, 15, 105), ... (End)
a(n) = (2*n - 1) * a(n-1) - Sum_{k=1..n-1} a(k) * a(n-k) if n>1. - Michael Somos, Jul 23 2011
G.f.: 1 / (1 - b(0)*x / (1 - b(1)*x / ...)) where b = A028310. - Michael Somos, Mar 31 2012
From Sergei N. Gladkovskii, Aug 11 2012,  Aug 12 2012, Dec 26 2012, Mar 20 2013, Jun 02 2013, Aug 14 2013, Oct 22 2013: (Start) Continued fractions:
G.f. 1/(G(0)-x) where G(k) = 1 - x*(k+1)/G(k+1).
G.f. 1 + x/(G(0)-x) where G(k) = 1 - x*(k+1)/G(k+1).
G.f.: A(x) = 1 + x/(G(0) - x) where G(k) = 1 + (2*k+1)*x - x*(2*k+2)/G(k+1).
G.f.: Q(0) where Q(k) = 1 - x*(2*k-1)/(1 - x*(2*k+2)/Q(k+1)).
G.f.: 2/G(0) where G(k) = 1 + 1/(1 - x/(x + 1/(2*k-1)/G(k+1))).
G.f.: 3*x - G(0) where G(k) = 3*x - 2*x*k - 1 - x*(2*k-1)/G(k+1).
G.f.: 1 + x*Q(0) where Q(k) = 1 - x*(2*k+2)/(x*(2*k+2) - 1/(1 - x*(2*k+1)/(x*(2*k+1) - 1/Q(k+1)))). (End)
a(n) ~ n^(n-1) * 2^(n-1/2) / exp(n). - Vaclav Kotesovec, Feb 22 2014

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

Original entry on oeis.org

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

A132382 Lower triangular array T(n,k) generator for group of arrays related to A001147 and A102625.

Original entry on oeis.org

1, -1, 1, -1, -2, 1, -3, -3, -3, 1, -15, -12, -6, -4, 1, -105, -75, -30, -10, -5, 1, -945, -630, -225, -60, -15, -6, 1, -10395, -6615, -2205, -525, -105, -21, -7, 1, -135135, -83160, -26460, -5880, -1050, -168, -28, -8, 1, -2027025, -1216215, -374220, -79380, -13230, -1890, -252, -36, -9, 1

Offset: 0

Tom Copeland, Nov 11 2007, Nov 12 2007, Nov 19 2007, Dec 04 2007, Dec 06 2007

Let b(n) = LPT[ A001147 ] = -A001147(n-1) for n > 0 and 1 for n=0, where LPT represents the action of the list partition transform described in A133314.
Then T(n,k) = binomial(n,k) * b(n-k) .
Form the matrix of polynomials TB(n,k,t) = T(n,k) * t^(n-k) = binomial(n,k) * b(n-k) * t^(n-k) = binomial(n,k) * Pb(n-k,t),
beginning as
     1;
    -1,       1;
    -1*t,    -2,       1;
    -3*t^2,  -3*t,    -3,       1;
   -15*t^3, -12*t^2,  -6*t,    -4,    1;
  -105*t^4, -75*t^3, -30*t^2, -10*t, -5, 1;
Let Pc(n,t) = LPT(Pb(.,t)).
Then [TB(t)]^(-1) = TC(t) = [ binomial(n,k) * Pc(n-k,t) ] = LPT(TB),
whose first column is
Pc(0,t) = 1
Pc(1,t) = 1
Pc(2,t) = 2 + t
Pc(3,t) = 6 + 6*t + 3*t^2
Pc(4,t) = 24 + 36*t + 30*t^2 + 15*t^3
Pc(5,t) = 120 + 240*t + 270*t^2 + 210*t^3 + 105*t^4.
The coefficients of these polynomials are given by the reverse of A102625 with the highest order coefficients given by A001147 with an additional leading 1.
Note this is not the complete matrix TC. The complete matrix is formed by multiplying along the diagonal of the lower triangular Pascal matrix by these polynomials, embedding trees of coefficients in the matrix.
exp[Pb(.,t)*x] = 1 + [(1-2t*x)^(1/2) - 1] / (t-0) = [1 + a finite diff. of [(1-2t*x)^(1/2)] with step t] = e.g.f. of the first column of TB.
exp[Pc(.,t)*x] = 1 / { 1 + [(1-2t*x)^(1/2) - 1] / t } = 1 / exp[Pb(.,t)*x) = e.g.f. of the first column of TC.
TB(t) and TC(t), being inverse to each other, are the generators of an Abelian group.
TB(0) and TC(0) are generators for a subgroup representing the iterated Laguerre operator described in A132013 and A132014.
Let sb(t,m) and sc(t,m) be the associated sequences under the LPT to TB(t)^m = B(t,m) and TC(t)^m = C(t,m).
Let Esb(t,m) and Esc(t,m) be e.g.f.'s for sb(t,m) and sc(t,m), rB(t,m) and rC(t,m) be the row sums of B(t,m) and C(t,m) and aB(t,m) and aC(t,m) be the alternating row sums.
Then B(t,m) is the inverse of C(t,m), Esb(t,m) is the reciprocal of Esc(t,m) and sb(t,m) and sc(t,m) form a reciprocal pair under the LPT. Similar relations hold among the row sums and the alternating sign row sums and associated quantities.
All the group members have the form B(t,m) * C(u,p) = TB(t)^m * TC(u)^p = [ binomial(n,k) * s(n-k) ]
with associated e.g.f. Es(x) = exp[m * Pb(.,t) * x] * exp[p * Pc(.,u) * x] for the first column of the matrix, with terms s(n), so group multiplication is isomorphic to matrix multiplication and to multiplication of the e.g.f.'s for the associated sequences (see examples).
These results can be extended to other groups of integer-valued arrays by replacing the 2 by any natural number in the expression for exp[Pb(.,t)*x].
More generally,
[ G.f. for M = Product_{i=0..j} B[s(i),m(i)] * C[t(i),n(i)] ]
= exp(u*x) * Product_{i=0..j} { exp[m(i) * Pb(.,s(i)) * x] * exp[n(i) * Pc(.,t(i)) * x] }
= exp(u*x) * Product_{i=0..j} { 1 + [ (1 - 2*s(i)*x)^(1/2) - 1 ] / s(i) }^m(i) / { 1 + [ (1 - 2*t(i)*x)^(1/2) - 1 ] / t(i) }^n(i)
= exp(u*x) * H(x)
[ E.g.f. for M ] = I_o[2*(u*x)^(1/2)] * H(x).
M is an integer-valued matrix for m(i) and n(i) positive integers and s(i) and t(i) integers. To invert M, change B to C in Product for M.
H(x) is the e.g.f. for the first column of M and diagonally multiplying the Pascal matrix by the terms of this column generates M. See examples.
The G.f. for M, i.e., the e.g.f. for the row polynomials of M, implies that the row polynomials form an Appell sequence (see Wikipedia and Mathworld). - Tom Copeland, Dec 03 2013

			Some group members and associated arrays are
(t,m) :: Array :: Asc. Matrix :: Asc. Sequence :: E.g.f. for sequence
..............................................................................
(0,1).::.B..::..A132013.::.(1,-1,0,0,0,0,...).....::.s(x).=.1-x
(0,1).::.C..::..A094587.::.(0!,1!,2!,3!,...)......::.1./.s(x)
(0,1).::.rB.::.~A055137.::.(1,0,-1,-2,-3,-4,...)..::.exp(x).*.s(x)
(0,1).::.rC.::....-.....::..A000522...............::.exp(x)./.s(x)
(0,1).::.aB.::....-.....::.(1,-2,3,-4,5,-6,...)...::.exp(-x).*.s(x)
(0,1).::.aC.::..A008290.::..A000166...............::.exp(-x)./.s(x)
..............................................................................
(0,2).::.B..::..A132014.::.(1,-2,2,0,0,0,0...)....::.s(x).=.(1-x)^2
(0,2).::.C..::..A132159.::.(1!,2!,3!,4!,...)......::..1./.s(x).
(0,2).::.rB.::...-......::.(1,-1,-1,1,5,11,19,29,)::.exp(x).*.s(x).
(0,2).::.rC.::...-......::..A001339...............::.exp(x)./.s(x).
(0,2).::.aB.::...-......::.(-1)^n.A002061(n+1)....::.exp(-x).*.s(x).
(0,2).::.aC.::...-......::..A000255...............::.exp(-x)./.s(x).
..............................................................................
(1,1).::.B..::..T.......::.(1,-A001147(n-1))......::.s(x).=.(1-2x)^(1/2)
(1,1).::.C..::.~A113278.::..A001147...............::.1./.s(x)...
(1,1).::.rB.::...-......::..A055142...............::.exp(x).*.s(x).
(1,1).::.rC.::...-......::..A084262...............::.exp(x)./.s(x).
(1,1).::.aB.::...-......::.(1,-2,2,-4,-4,-56,...).::.exp(-x).*.s(x).
(1,1).::.aC.::...-......::..A053871...............::.exp(-x)./.s(x).
..............................................................................
(2,1).::.B..::...-......::.(1,-A001813)...........::.s=[1+(1-4x)^(1/2)]/2....
(2,1).::.C..::...-......::..A001761...............::.1./.s(x)..
(2,1).::.rB.::...-......::.(1,0,-3,-20,-183,...)..::.exp(x).*.s(x)..
(2,1).::.rC.::...-......::.(1,2,7,46,485,...).....::.exp(x)./.s(x).
(2,1).::.aB.::...-......::.(1,-2,1,-10,-79,...)...::.exp(-x).*.s(x).
(2,1).::.aC.::...-......::.(1,0,3,20,237,...).....::.exp(-x)./.s(x)
..............................................................................
(1,2).::.B..::.~A134082.::.(1,-2,0,0,0,0,...).....::.s(x).=.1.-.2x
(1,2).::.C..::....-.....::..A000165...............::.1./.s(x)..
(1,2).::.rB.::....-.....::.(1,-1,-3,-5,-7,-9,...).::.exp(x).*.s(x).
(1,2).::.rC.::....-.....::..A010844...............::.exp(x)./.s(x)..
(1,2).::.aB.::....-.....::.(1,-3,5,-7,9,-11,...)..::.exp(-x).*.s(x).
(1,2).::.aC.::....-.....::..A000354...............::.exp(-x)./.s(x).
..............................................................................
(The tilde indicates the match is not exact--specifically, there are differences in signs from the true matrices.)
Note the row sums correspond to binomial transforms of s(x) and the alternating row sums, to inverse binomial transforms, or, finite differences.
Some additional examples:
C(1,2)*B(0,1) = B(1,-2)*C(0,-1) = [ binomial(n,k)*A002866(n-k) ] with asc. e.g.f. (1-x) / (1-2x).
B(1,2)*C(0,1) = C(1,-2)*B(0,-1) = 2I - A094587 with asc. e.g.f. (1-2x) / (1-x).

[G.f. for TB(n,k,t)] = GTB(u,x,t) = exp(u*x) * { 1 + [ (1 - 2t*x)^(1/2) - 1 ] / t } = exp[(u+Pb(.,t))*x] where TB(n,k,t) = (D_x)^n (D_u)^k /k! GTB(u,x,t) eval. at u=x=0.
[G.f. for TC(n,k,t)] = GTC(u,x,t) = exp(u*x) / { 1 + [ (1 - 2t*x)^(1/2) - 1 ] / t } = exp[(u+Pc(.,t))*x] where TC(n,k,t) = (D_x)^n (D_u)^k /k! GTC(u,x,t) eval. at u=x=0.
[E.g.f. for TB(n,k,t)] = I_o[2*(u*x)^(1/2)] * { 1 + [ (1 - 2t*x)^(1/2) - 1 ] / t } and
[E.g.f. for TC(n,k,t)] = I_o[2*(u*x)^(1/2)] / { 1 + [ (1 - 2t*x)^(1/2) - 1 ] / t }
where I_o is the zeroth modified Bessel function of the first kind, i.e.,
I_o[2*(u*x)^(1/2)] = Sum_{j>=0} (u^j/j!) * (x^j/j!).
So [e.g.f. for TB(n,k)] = I_o[2*(u*x)^(1/2)] * (1 - 2x)^(1/2).

More terms from Tom Copeland, Dec 05 2007

A034430 Convolution of A001147 (double factorial numbers) with itself.

Original entry on oeis.org

1, 2, 7, 36, 249, 2190, 23535, 299880, 4426065, 74294010, 1397669175, 29123671500, 665718201225, 16560190196550, 445300709428575, 12869793995058000, 397815487883438625, 13095523164781307250, 457362512442763302375, 16890682269050394304500

Offset: 0

Jim FitzSimons (cherry(AT)neta.com)

nonn

Old name was "Expand arctan(sqrt(x)*sqrt(x+2))/(sqrt(x)*sqrt(x+2)) and multiply n-th term by 1.3.5...(2n+1)".

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A000111, A001147, A233481.

A034430 := proc(n) option remember; if n=0 then 1 elif n=1 then 2 else
(3*n-1)*A034430(n-1)-(1+2*n^2-3*n)*A034430(n-2) fi end: seq(A034430(n),n=0..19); # Peter Luschny, Dec 14 2013

Range[0, 19]! * CoefficientList[Series[1/(1 - x)/Sqrt[1 - 2*x], {x, 0, 19}], x] (* David Scambler, May 24 2012 *)

E.g.f.: 1/(1-x)/sqrt(1-2*x). - Vladeta Jovovic, May 11 2003
a(n) = Integral_{x=-infinity..infinity} x^(2*n+1)*exp(-x^2)*erfi(x/sqrt(2)), with erfi the imaginary error function. - Groux Roland, Mar 26 2011
E.g.f.: d/dx(F(x)^(-1)) where (-1) denotes the compositional inverse and F(x) = sin(x)/(1+sin(x)) = x - 2*x^2/2! + 5*x^3/3! - 16*x^4/4! + .... See A000111. - Peter Bala, Jun 24 2012
E.g.f.: E(x) = 1/sqrt(1-2*x)/(1-x) = (1 + x/(U(0)-x))/(1-x), where U(k) = (2*k+1)*x + (k+1) - (k+1)*(2*k+3)*x/U(k+1); (continued fraction Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Jun 27 2012
G.f.: hypergeom([1,1/2],[],2*x)^2. - Mark van Hoeij, May 16 2013
a(n-1)*n = A233481(n) for n >= 1. - Peter Luschny, Dec 14 2013
D-finite with recurrence: a(n) = (3*n-1)*a(n-1)-(2*n-1)*(n-1)*a(n-2) for n >= 2. - Peter Luschny, Dec 14 2013
a(n) ~ 2^(n+3/2) * n^n / exp(n). - Vaclav Kotesovec, Dec 20 2013
a(n) = 2*Pochhammer(1/2, n+1)*hyper2F1([1/2, -n], [3/2], -1). - Peter Luschny, Aug 02 2014
a(n) = -(2*n+1)!! * 2^(-n-1) * Im(Beta(2, n+1, 1/2)). - Vladimir Reshetnikov, Apr 23 2016
Expansion of square of continued fraction 1/(1 - x/(1 - 2*x/(1 - 3*x/(1 - 4*x/(1 - 5*x/(1 - ...)))))). - Ilya Gutkovskiy, Apr 19 2017

Better name from Philippe Deléham, Mar 21 2005

A132101 a(n) = (A001147(n) + A047974(n))/2.

Original entry on oeis.org

1, 1, 3, 11, 65, 513, 5363, 68219, 1016481, 17243105, 327431363, 6874989963, 158118876449, 3952936627361, 106729080101235, 3095142009014843, 95949394016339393, 3166329948046914369, 110821547820208233731, 4100397266856761733515

Offset: 0

Keith F. Lynch, Oct 31 2007

Also, number of distinct Tsuro tiles which are digonal in shape and have n points per side. Turning over is not allowed. See A132100 for definition and comments.
See the Burns et al. papers for another interpretation.
From Ross Drewe, Mar 16 2008: (Start)
This is also the number of arrangements of n pairs which are equivalent under the joint operation of sequence reversal and permutations of labels. Assume that the elements of n distinct pairs are labeled to show the pair of origin, e.g., [1 1], [2 2]. The number of distinguishable ways of arranging these elements falls as the conditions are made more general:
a(n) = A000680: element order is significant and the labels are distinguishable;
b(n) = A001147: element order is significant but labels are not distinguishable, i.e., all label permutations of a given sequence are equivalent;
c(n) = A132101: element order is weakened (reversal allowed) and all label permutations are equivalent;
d(n) = A047974: reversal allowed, all label permutations are equivalent and equivalence class maps to itself under joint operation.
Those classes that do not map to themselves form reciprocal pairs of classes under the joint operation and their number is r(n). Then c = b - r/2 = b - (b - d)/2 = (b+d)/2. A formula for r(n) is not available, but formulas are available for b(n) = A001147 and d(n) = A047974, allowing an explicit formula for this sequence.
c(n) is useful in extracting structure information without regard to pair ordering (see example). c(n) terms also appear in formulas related to binary operators, e.g., the number of binary operators in a k-valued logic that are invertible in 1 operation.
a(n) = (b(n) + c(n))/2, where b(n) = (2n)!/(2^n * n!) = A001147(n), c(n) = Sum_{k=0..floor(n/2)} n!/((n-2*k)! * k!) = A047974(n).
For 3 pairs, the arrangement A = [112323] is the same as B = [212133] under the permutation of the labels [123] -> [312] plus reversal of the elements, or vice versa. The unique structure common to A and B is {1 intact pair + 2 interleaved pairs}, where the order is not significant (contrast A001147). (End)

			a(2)=3 counts the arrangements [1122], [1212] and [1221]. - _R. J. Mathar_, Oct 18 2019

G. C. Greubel, Table of n, a(n) for n = 0..400
Jonathan Burns, Assembly Graph Words - Single Transverse Component (Counts).
Jonathan Burns, Egor Dolzhenko, Natasa Jonoska, Tilahun Muche, and Masahico Saito, Four-Regular Graphs with Rigid Vertices Associated to DNA Recombination, Discrete Applied Mathematics, Volume 161, Issues 10-11, July 2013, Pages 1378-1394.
R. J. Mathar, Feynman diagrams of the QED vacuum polarization, vixra:1901.0148 (2019), Section IV.

Cf. A000680, A001147, A047974, A007769, A054499.

Cf. A132100, A132101, A132102, A132103, A132104, A132105.

R:=PowerSeriesRing(Rationals(), 30);
Coefficients(R!(Laplace( (Exp(x+x^2) + 1/Sqrt(1-2*x))/2 ))); // G. C. Greubel, Jul 12 2024

A132101 := proc(n)
    (A001147(n)+A047974(n))/2 ;
end proc:
seq(A132101(n),n=0..30) ; # R. J. Mathar, Dec 20 2020

Table[((2n-1)!!+I^(-n)*HermiteH[n,I/2])/2,{n,0,30}] (* Jonathan Burns, Apr 05 2016 *)

[(factorial(n)*binomial(2*n,n) + (-2*i)^n*hermite(n,i/2))/2^(n+1) for n in range(31)] # G. C. Greubel, Jul 12 2024

D-finite with recurrence -(n-3)*a(n) +2*(n^2-3*n+1)*a(n-1) -(n-1)*a(n-2) -2*(2*n-5)*(n-1)*(n-2)*a(n-3) = 0. - R. J. Mathar, Dec 20 2020

E.g.f.: (1/2)*( exp(x+x^2) + 1/sqrt(1-2*x) ). - G. C. Greubel, Jul 12 2024

Entry revised by N. J. A. Sloane, Nov 04 2011

A076729 a(n) = A001147(n+1) * Integral_{x=0..1} (1 + x^2)^n dx.

Original entry on oeis.org

1, 4, 28, 288, 3984, 70080, 1506240, 38384640, 1133072640, 38038533120, 1431213235200, 59645279232000, 2726781752217600, 135661078090137600, 7295806823277772800, 421717409630060544000, 26071235813929033728000, 1716456412254215503872000, 119894838461795743137792000

Offset: 0

Al Hakanson (hawku(AT)hotmail.com), Oct 28 2002

nonn

Numerator of the integral where denominator is equal to (2n+1)!! = A001147(n+1).
Also numerator of the integral (1-x^2)^-(n+1/2) for x from 0 to sqrt(1/2). Here the sequence starts at n=1; at n=2 the function is 4.
a(n) = Integral_{x=0..log(1+sqrt(2))} cosh(x)^(2*n-1) dx where the denominators are b(n) = (2*n)!/(n!*2^n). E.g., a(3)=28 and b(3)=15; both offsets are 1. - Al Hakanson (hawkuu(AT)excite.com), Mar 02 2004
Self-convolution of A001813. - Vladimir Reshetnikov, Oct 11 2016

			For n = 3, (2n+1)!! = 105 and the integral is 96/35 = 288/105, so a(3) = 288.

Cf. A001147, A077595, A077745, A086891, A034430, A001813.

seq((doublefactorial(2*n+1))*add((binomial(n, i))/(2*i+1), i=0..n), n=0..20) ; # John M. Campbell, Feb 06 2016
A076729 := n -> 2^n*n!*JacobiP(n, 1/2, -1 - n, 3):
seq(simplify(A076729(n)), n = 0..18);  # Peter Luschny, Jan 22 2025

a[n_] := (2n + 1)!!*Integrate[(1 + x^2)^n, {x, 0, 1}]; Table[ a[n], {n, 0, 16}] (* Robert G. Wilson v, Feb 27 2004 *)
Round@Table[-(2 n + 1)!! Im[Beta[2, n + 1, 1/2]]/2, {n, 0, 20}] (* Round is equivalent to FullSimplify here, but is much faster - Vladimir Reshetnikov, Oct 08 2016 *)
nxt[{n_,a_}]:={n+1,2a(n+1)+(2(n+1))!/(n+1)!}; NestList[nxt,{0,1},20][[All,2]] (* Harvey P. Dale, Feb 04 2023 *)

a(n)=if(n<0,0,subst(intformal((1+x^2)^n),x,1)*(2*n+1)!/2^n/n!)

a(n) = 2*n*a(n-1) + (2*n)!/n!.
a(n) = 2^n*Sum_{k=0..n} A001147(k)*A001147(n-k).
a(n) = (2*n+1)!*Sum_{k=0..n} k!*(-2)^k/((2*k+1)!*(n-k)!).
a(n) = (2*n+1)!!*hypergeom([1/2, -n], [3/2], -1). - Vladeta Jovovic, Dec 05 2002
E.g.f.: 1/((1-2*x)*sqrt(1-4*x)). - Vladeta Jovovic, May 11 2003
G.f.: hypergeom([1,1/2],[],4*x)^2  - Mark van Hoeij, May 16 2013
a(n) ~ 2^(2*n+3/2)*n^n/exp(n). - Vaclav Kotesovec, Oct 05 2013
a(n) = (2n+1)!!*Sum_{i=0..n} binomial(n,i)/(2i+1). - John M. Campbell, Feb 06 2016
From Vladimir Reshetnikov, Oct 08 2016: (Start)
a(n) = 2^n*A034430(n) = -(2*n+1)!! * Im(Beta(2, n+1, 1/2))/2.
Recurrence: 2*(3*n+2)*a(n) = a(n+1) + 4*n*(2*n+1)*a(n-1). (End)
Expansion of square of continued fraction 1/(1 - 2*x/(1 - 4*x/(1 - 6*x/(1 - 8*x/(1 - 10*x/(1 - ...)))))). - Ilya Gutkovskiy, Apr 19 2017
a(n) = 2^n*n!*JacobiP(n, 1/2, -1 - n, 3). - Peter Luschny, Jan 22 2025

A051579 a(n) = (2*n+5)!!/5!!, related to A001147 (odd double factorials).

Original entry on oeis.org

1, 7, 63, 693, 9009, 135135, 2297295, 43648605, 916620705, 21082276215, 527056905375, 14230536445125, 412685556908625, 12793252264167375, 422177324717523375, 14776206365113318125, 546719635509192770625

Offset: 0

Wolfdieter Lang

Row m=5 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 (rows m=0..4).

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

List([0..20], n-> Product([0..n-1], j-> 2*j+7) ); # G. C. Greubel, Nov 12 2019

[1] cat [(&*[2*j+7: j in [0..n-1]]): n in [1..20]]; // G. C. Greubel, Nov 12 2019

df:=doublefactorial; seq(df(2*n+5)/df(5), n = 0..20); # G. C. Greubel, Nov 12 2019

Table[2^n*Pochhammer[7/2, n], {n,0,20}] (* G. C. Greubel, Nov 12 2019 *)

vector(20, n, prod(j=1,n-1, 2*j+5) ) \\ G. C. Greubel, Nov 12 2019

[product( (2*j+7) for j in (0..n-1)) for n in (0..20)] # G. C. Greubel, Nov 12 2019

a(n) = (2*n+5)!!/4!!.
E.g.f.: 1/(1-2*x)^(7/2).
a(n) ~ 8/15*sqrt(2)*n^3*2^n*e^-n*n^n*(1 + 107/24*n^-1 + ...). - Joe Keane (jgk(AT)jgk.org), Nov 23 2001
G.f.: G(0)/(10*x) -1/(5*x), where G(k)= 1 + 1/(1 - x*(2*k+5)/(x*(2*k+5) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 13 2013
From Peter Bala, May 26 2017: (Start)
a(n+1) = (2*n + 7)*a(n) with a(0) = 1.
O.g.f. satisfies the Riccati differential equation 2*x^2*A(x)' = (1 - 7*x)*A(x) - 1 with A(0) = 1.
G.f. as an S-fraction: A(x) = 1/(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 - ...))))))))) (by Stokes 1982).
Reciprocal as an S-fraction: 1/A(x) = 1/(1 + 7*x/(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 - ...)))))))))). (End)
From Amiram Eldar, Dec 11 2022: (Start)
Sum_{n>=0} 1/a(n) = 15 * sqrt(e*Pi/2) * erf(1/sqrt(2)) - 20, where erf is the error function.
Sum_{n>=0} (-1)^n/a(n) = 15 * sqrt(Pi/(2*e)) * erfi(1/sqrt(2)) - 10, where erfi is the imaginary error function. (End)

A112935 Logarithmic derivative of A112934 such that a(n)=(1/2)*A112934(n+1) for n>0, where A112934 equals the INVERT transform of double factorials A001147.

Original entry on oeis.org

1, 3, 13, 79, 641, 6579, 81677, 1187039, 19728193, 368562723, 7639512013, 173893382575, 4310656806977, 115569893763411, 3331588687405133, 102751933334045375, 3375782951798785921, 117693183724386637635

Offset: 1

Paul D. Hanna, Oct 09 2005

nonn

			log(1+x + 2*x*[x + 3*x^2 + 13*x^3 + 79*x^4 + 641*x^5 +...])
= x + 3/2*x^2 + 13/3*x^3 + 79/4*x^4 + 641/5*x^5 +...

Cf. A001147, A112934; A112936, A112937, A112938, A112939, A112940, A112941, A112942, A112943.

{a(n)=local(F=1+x+x*O(x^n));for(i=1,n,F=1+x+2*x^2*deriv(F)/F); return(n*polcoeff(log(F),n,x))}

G.f.: log(1 + x + 2*x*[Sum_{n>=1} a(n)*x^n]) = Sum_{k>=1} a(n)/n*x^n.
G.f.: (1 - 1/Q(0))/x where Q(k) = 1 - x*(2*k-1)/(1 - x*(2*k+4)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 19 2013
G.f.: 1/(x*G(0)) - 1/(2*x), where G(k)= 1 + 1/(1 - 2*x*(2*k+2)/(2*x*(2*k+2) - 1 + 2*x*(2*k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 31 2013

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)

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cp's OEIS Frontend

A035342 The convolution matrix of the double factorial of odd numbers (A001147).

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A112934 a(0) = 1; a(n+1) = Sum_{k=0..n} a(k)*A001147(n-k), where A001147 = double factorial numbers.

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

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A132382 Lower triangular array T(n,k) generator for group of arrays related to A001147 and A102625.

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A034430 Convolution of A001147 (double factorial numbers) with itself.

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A132101 a(n) = (A001147(n) + A047974(n))/2.

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