A112934 -id:A112934 - cp's OEIS frontend

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.

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

A355721 Square table, read by antidiagonals: the g.f. for row n is given recursively by (2n-1)xR(n,x) = 1 + (2n-3)x - 1/R(n-1,x) for n >= 1 with the initial value R(0,x) = Sum_{k >= 0} A112934(k+1)x^k.

Original entry on oeis.org

1, 1, 2, 1, 2, 6, 1, 2, 10, 26, 1, 2, 14, 74, 158, 1, 2, 18, 138, 706, 1282, 1, 2, 22, 218, 1686, 8162, 13158, 1, 2, 26, 314, 3194, 24162, 110410, 163354, 1, 2, 30, 426, 5326, 53890, 394254, 1708394, 2374078, 1, 2, 34, 554, 8178, 102722, 1019250, 7191018, 29752066, 39456386

Offset: 0

Peter Bala, Jul 15 2022

Compare with A111528, which has a similar definition.

			Square array begins
1, 2,  6,  26,   158,    1282,   13158,    163354,    2374078,     39456386, ...
1, 2, 10,  74,   706,    8162,  110410,   1708394,   29752066,    576037442, ...
1, 2, 14, 138,  1686,   24162,  394254,   7191018,  144786006,   3188449602, ...
1, 2, 18, 218,  3194,   53890, 1019250,  21256090,  483426010,  11895873410, ...
1, 2, 22, 314,  5326,  102722, 2197558,  51355514, 1297759918,  35208930050, ...
1, 2, 26, 426,  8178,  176802, 4206618, 108577674, 3011332338,  89141101506, ...
1, 2, 30, 554, 11846,  283042, 7396830, 208569034, 6288011206, 201404591042, ...
...

A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.

Cf. A112934 (row 0), A000698 (row 1), A355722 (row 2), A355723 (row 3), A355724 (row 4), A355725 (row 5). Cf. A001147, A111528.

T := (n,k) -> coeff(series(hypergeom([n+1/2, 1], [], 2*x)/ hypergeom([n-1/2, 1], [], 2*x), x, 21), x, k):
# display as a sequence
seq(seq(T(n-k,k), k = 0..n), n = 0..10);
# display as a square array
seq(print(seq(T(n,k), k = 0..10)), n = 0..10);

Let d(n) = Product_{k = 1..n} 2*k-1 = A001147(n) denote the double factorial of odd numbers.
O.g.f. for row n: R(n,x) = ( Sum_{k >= 0} d(n+k)/d(n)*x^k )/( Sum_{k >= 0} d(n-1+k)/d(n-1)*x^k ).
R(n,x)/(1 - (2*n-1)*x*R(n,x)) = Sum_{k >= 0} d(n+k)/d(n)*x^k.
R(n,x) = 1/(1 + (2*n-1)*x - (2*n+1)*x/(1 + (2*n+1)*x - (2*n+3)*x/(1 + (2*n+3)*x - (2*n+5)*x/(1 + (2*n+5)*x - ... )))).
R(n,x) satisfies the Riccati differential equation 2*x^2*d/dx(R(n,x)) + (2*n-1)*x*R(n,x)^2 - (1 + (2*n-3)*x)*R(n,x) + 1 = 0 with R(n,0) = 1.
Applying Stokes 1982 gives A(x) = 1/(1 - 2*x/(1 - (2*n+1)*x/(1 - 4*x/(1 - (2*n+3)*x/(1 - 6*x/(1 - (2*n+5)*x/(1 - ... - 2*m*x/(1 - (2*n+2*m-1)*x/(1 - ... ))))))))), a continued fraction of Stieltjes type.
Row 0: A112934(n+1); Row 1; A000698(n+1).

A028310 Expansion of (1 - x + x^2) / (1 - x)^2 in powers of x.

Original entry on oeis.org

1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71

Offset: 0

N. J. A. Sloane

1 followed by the natural numbers.
Molien series for ring of Hamming weight enumerators of self-dual codes (with respect to Euclidean inner product) of length n over GF(4).
Engel expansion of e (see A006784 for definition) [when offset by 1]. - Henry Bottomley, Dec 18 2000
Also the denominators of the series expansion of log(1+x). Numerators are A062157. - Robert G. Wilson v, Aug 14 2015
The right-shifted sequence (with a(0)=0) is an autosequence (of the first kind - see definition in links). - Jean-François Alcover, Mar 14 2017

			G.f. = 1 + x + 2*x^2 + 3*x^3 + 4*x^4 + 5*x^5 + 6*x^6 + 7*x^7 + 8*x^8 + 9*x^9  + ...

G. C. Greubel, Table of n, a(n) for n = 0..10000
Andrei Asinowski, Cyril Banderier, and Valerie Roitner, Generating functions for lattice paths with several forbidden patterns, (2019).
Daniel Birmajer, Juan B. Gil, Jordan O. Tirrell, and Michael D. Weiner, Pattern-avoiding stabilized-interval-free permutations, arXiv:2306.03155 [math.CO], 2023.
Olivia Nabawanda and Fanja Rakotondrajao, The sets of flattened partitions with forbidden patterns, arXiv:2011.07304 [math.CO], 2020.
G. Nebe, E. M. Rains and N. J. A. Sloane, Self-Dual Codes and Invariant Theory, Springer, Berlin, 2006.
Oeis Wiki, Autosequence
E. M. Rains and N. J. A. Sloane, Self-dual codes, pp. 177-294 of Handbook of Coding Theory, Elsevier, 1998 (Abstract, pdf, ps).
Michael Somos, Rational Function Multiplicative Coefficients
Index entries for linear recurrences with constant coefficients, signature (2,-1).
Index entries for Molien series
Index entries for sequences related to Engel expansions

Cf. A000007, A000027, A000660 (boustrophedon transform).

Cf. A001477, A004001, A005229, A112934, A123110, A212393, A204503.

a028310 n = 0 ^ n + n
a028310_list = 1 : [1..]  -- Reinhard Zumkeller, Nov 06 2012

[n eq 0 select 1 else n: n in [0..75]]; // G. C. Greubel, Jan 05 2024

a:= n-> `if`(n=0, 1, n):
seq(a(n), n=0..60);

Denominator@ CoefficientList[Series[Log[1+x], {x,0,75}], x] (* or *)
CoefficientList[ Series[(1 -x +x^2)/(1-x)^2, {x,0,75}], x] (* Robert G. Wilson v, Aug 14 2015 *)
Join[{1}, Range[75]] (* G. C. Greubel, Jan 05 2024 *)
LinearRecurrence[{2,-1},{1,1,2},80] (* Harvey P. Dale, Jan 29 2025 *)

{a(n) = (n==0) + max(n, 0)} /* Michael Somos, Feb 02 2004 */

A028310(n)=n+!n  \\ M. F. Hasler, Jan 16 2012

def A028310(n): return n|bool(n)^1 # Chai Wah Wu, Jul 13 2023

[n + int(n==0) for n in range(76)] # G. C. Greubel, Jan 05 2024

Binomial transform is A005183. - Paul Barry, Jul 21 2003
G.f.: (1 - x + x^2) / (1 - x)^2 = (1 - x^6) /((1 - x) * (1 - x^2) * (1 - x^3)) = (1 + x^3) / ((1 - x) * (1 - x^2)). a(0) = 1, a(n) = n if n>0.
Euler transform of length 6 sequence [ 1, 1, 1, 0, 0, -1]. - Michael Somos Jul 30 2006
G.f.: 1 / (1 - x / (1 - x / (1 + x / (1 - x)))). - Michael Somos, Apr 05 2012
G.f. of A112934(x) = 1 / (1 - a(0)*x / (1 - a(1)*x / ...)). - Michael Somos, Apr 05 2012
a(n) = A000027(n) unless n=0.
a(n) = Sum_{k=0..n} A123110(n,k). - Philippe Deléham, Oct 06 2009
E.g.f: 1+x*exp(x). - Wolfdieter Lang, May 03 2010
a(n) = sqrt(floor[A204503(n+3)/9]). - M. F. Hasler, Jan 16 2012
E.g.f.: 1-x + x*E(0), where E(k) = 2 + x/(2*k+1 - x/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Dec 24 2013
a(n) = A001477(n) + A000007(n). - Miko Labalan, Dec 12 2015 (See the first comment.)

A098694 Double-superfactorials: a(n) = Product_{k=1..n} (2k)!.

Original entry on oeis.org

1, 2, 48, 34560, 1393459200, 5056584744960000, 2422112183371431936000000, 211155601241022491077587763200000000, 4417964278440225627098723475313498521600000000000

Offset: 0

Ralf Stephan, Sep 22 2004

nonn

Hankel transform of double factorial numbers A001147. - Paul Barry, Jan 28 2008

Hankel transform of A112934(n+1). - Paul Barry, Dec 04 2009

Seiichi Manyama, Table of n, a(n) for n = 0..28
C. Radoux, Déterminants de Hankel et théorème de Sylvester, Séminaire Lotharingien de Combinatoire, B28b (1992), 9 pp.

Cf. A000178, A010050, A074962, A168467, A268504, A268505, A268506, A271946, A271947.

[&*[ Factorial(2*k): k in [0..n] ]: n in [0..10]]; // Vincenzo Librandi, Dec 11 2016

Table[Product[(2k)!,{k,1,n}],{n,0,10}] (* Vaclav Kotesovec, Nov 13 2014 *)

a(n) = prod(k=1, n, (2*k)!); \\ Michel Marcus, Dec 11 2016

from math import prod
def A098694(n): return prod(((k+1)*((k<<1)+1)<<1)**(n-k) for k in range(1,n+1))<Chai Wah Wu, Nov 26 2023

a(n) = Product_{k=0..n} (2*(k+1)*(2*k+1))^(n-k). - Paul Barry, Jan 28 2008
a(n) = A000178(n)*A057863(n)*A006125(n+1) = A121835(n)*A006125(n+1). - Paul Barry, Jan 28 2008
G.f.: G(0)/(2*x)-1/x, where G(k)= 1  + 1/(1 - 1/(1 + 1/(2*k+2)!/x/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 14 2013
a(n) ~ 2^(n^2+2*n+17/24) * n^(n^2+3*n/2+11/24) * Pi^((n+1)/2) / (A^(1/2) * exp(3*n^2/2+3*n/2-1/24)), where A = 1.2824271291... is the Glaisher-Kinkelin constant (see A074962). - Vaclav Kotesovec, Nov 13 2014
a(n) = A^(3/2)*2^(n^2+n-1/24)*Pi^(-n/2-1/4)*G(n+3/2)*G(n+2)/exp(1/8), where G(n) is the Barnes G-function and A is the Glaisher-Kinkelin constant. - Ilya Gutkovskiy, Dec 11 2016
a(n) = A000178(2*n + 1) / A168467(n). - Vaclav Kotesovec, Oct 28 2017
For n > 0, a(n) = 2^((n+1)/2) * n * sqrt(BarnesG(2*n)*Gamma(n)) * Gamma(2*n). - Vaclav Kotesovec, Nov 27 2024

A112936 INVERT transform (with offset) of triple factorials (A008544), where g.f. satisfies: A(x) = 1 + x[d/dx xA(x)^3]/A(x)^3.

Original entry on oeis.org

1, 1, 3, 15, 111, 1131, 14943, 243915, 4742391, 106912131, 2739347103, 78569371275, 2492748594471, 86650852740531, 3274367635513263, 133625238021647835, 5856377114106629751, 274320168321004350531

Offset: 0

Paul D. Hanna, Oct 09 2005

nonn

			A(x) = 1 + x + 3*x^2 + 15*x^3 + 111*x^4 + 1131*x^5 + 14943*x^6 +...
1/A(x) = 1 - x - 2*x^2 - 10*x^3 - 80*x^4 - 880*x^5 -...-A008544(n)*x^(n+1)-...

Cf. A008544, A112937 (log derivative); A112934, A112935, A112938, A112939, A112940, A112941, A112942, A112943.

a = ConstantArray[0,20]; a[[1]]=1; Do[a[[n]] = (3*n-2)*a[[n-1]] - Sum[a[[k]]*a[[n-k]],{k,1,n-1}],{n,2,20}]; Flatten[{1,a}] (* Vaclav Kotesovec after Michael Somos, Feb 22 2014 *)
CoefficientList[Series[1/(1+(1/3*ExpIntegralE[2/3,-1/(3*x)])/E^(1/(3*x))), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 22 2014 *)

{a(n)=local(F=1+x+x*O(x^n));for(i=1,n,F=1+x+3*x^2*deriv(F)/F); return(polcoeff(F,n,x))}

{a(n) = my(A); if( n<1, n==0, A = vector(n, k, 1); for(k=2, n, A[k] = (3*k - 2)*A[k-1] - sum(j=1, k-1, A[j] * A[k-j])); A[n])}; /* Michael Somos, Jul 23 2011 */

{a(n) = if( n<1, n==0, polcoeff( 1 / sum(k=0, n, x^k * prod(i=1, k, 3*i - 4), x * O(x^n)), n))}; /* Michael Somos, Oct 17 2016 */

{a(n) = my(A); if( n<0, 0, A = O(x); for(k=0, n, A = (x + sqrt(x^2 + 4*x^5*A')) / 2); polcoeff(A, 3*n + 1))}; /* Michael Somos, Oct 17 2016 */

{a(n) = my(A); if( n<1, n==0, A = x; for(k=1, n, A = truncate(A) + O(x^(3*k + 4)); A += A + x^4*A' - A^2/x); polcoeff(A, 3*n + 1))}; /* Michael Somos, Oct 17 2016 */

G.f. satisfies: A(x) = 1+x + 3*x^2*[d/dx A(x)]/A(x) (log derivative).
G.f.: A(x) = 1+x +3*x^2/(1-5*x -3*2*2*x^2/(1-11*x -3*3*5*x^2/(1-17*x -3*4*8*x^2/(1-23*x -... -3*n*(3*n-4)*x^2/(1-(6*n-1)*x -...)))) (continued fraction).
G.f.: A(x) = 1/(1-x/(1 -2*x/(1-3*x/(1 -5*x/(1-6*x/(1 -8*x/(1-9*x/(1 -...)))))))) (continued fraction).
a(n) = (3*n - 2) * a(n-1) - Sum_{k=1..n-1} a(k) * a(n-k) if n>1. - Michael Somos, Jul 23 2011
G.f.: Q(0) where Q(k) = 1 - x*(3*k-1)/(1 - x*(3*k+3)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 20 2013
G.f.: 2/G(0)+4*x, where G(k)= 1 + 1/(1 - x*(3*k+3)/(x*(3*k+5) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 25 2013
G.f.: 2/G(0), where G(k)= 1 + 1/(1 - x*(3*k-1)/(x*(3*k-1) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 06 2013
G.f.: -4/Sum_{n>=0} [Product_{k=0..n} (3*k-4)]*x^n. - Sergei N. Gladkovskii, Jun 06 2013
a(n) ~ n! * 3^(n-1) / (GAMMA(2/3) * n^(4/3)) . - Vaclav Kotesovec, Feb 22 2014
Given g.f. A(x), then y = x * A(x^3) satisfies y^2 = x*y + x^5*y'. - Michael Somos, Oct 17 2016

A111146 Triangle T(n,k), read by rows, given by [0, 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, ...] DELTA [1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 0, 2, 0, 0, 1, 4, 0, 0, 2, 5, 8, 0, 0, 6, 15, 17, 16, 0, 0, 24, 62, 68, 49, 32, 0, 0, 120, 322, 359, 243, 129, 64, 0, 0, 720, 2004, 2308, 1553, 756, 321, 128, 0, 0, 5040, 14508, 17332, 11903, 5622, 2151, 769, 256, 0, 0, 40320, 119664

Offset: 0

Philippe Deléham, Oct 19 2005

Let R(m,n,k), 0<=k<=n, the Riordan array (1, x*g(x)) where g(x) is g.f. of the m-fold factorials . Then Sum_{k, 0<=k<=n} = R(m,n,k) = Sum_{k, 0<=k<=n} T(n,k)*m^(n-k).
For m = -1, R(-1,n,k) is A026729(n,k).
For m = 0, R(0,n,k) is A097805(n,k).
For m = 1, R(1,n,k) is A084938(n,k).
For m = 2, R(2,n,k) is A111106(n,k).

			Triangle begins:
.1;
.0, 1;
.0, 0, 2;
.0, 0, 1, 4;
.0, 0, 2, 5, 8;
.0, 0, 6, 15, 17, 16;
.0, 0, 24, 62, 68, 49, 32;
.0, 0, 120, 322, 359, 243, 129, 64;
.0, 0, 720, 2004, 2308, 1553, 756, 321, 128;
.0, 0, 5040, 14508, 17332, 11903, 5622, 2151, 769, 256;
.0, 0, 40320, 119664, 148232, 105048, 49840, 18066, 5756, 1793, 512;
....................................................................
At y=2: Sum_{k=0..n} 2^k*T(n,k) = A113327(n) where (1 + 2*x + 8*x^2 + 36*x^3 +...+ A113327(n)*x^n +..) = 1/(1 - 2/1!*x*(1! + 2!*x + 3!*x^2 + 4!*x^3 +..) ).
At y=3: Sum_{k=0..n} 3^k*T(n,k) = A113328(n) where (1 + 3*x + 18*x^2 + 117*x^3 +...+ A113328(n)*x^n +..) = 1/(1 - 3/2!*x*(2! + 3!*x + 4!*x^2 + 5!*x^3 +..) ).
At y=4: Sum_{k=0..n} 4^k*T(n,k) = A113329(n) where (1 + 4*x + 32*x^2 + 272*x^3 +...+ A113329(n)*x^n +..) = 1/(1 - 4/3!*x*(3! + 4!*x + 5!*x^2 + 6!*x^3 +..) ).

Alois P. Heinz, Rows n = 0..140, flattened

Cf. A000045, A026729, A051295, A084938, A097805, A111106, A112934.
Cf. m-fold factorials : A000142, A001147, A007559, A007696, A008548, A008542.
Cf. A113326, A113327 (y=2), A113328 (y=3), A113329 (y=4), A113330 (y=5), A113331 (y=6).

T[n_, k_] := Module[{x = X + X*O[X]^n, y = Y + Y*O[Y]^k}, A = 1/(1 - x*y*Sum[x^j*Product[y + i, {i, 0, j - 1}], {j, 0, n}]); Coefficient[ Coefficient[A, X, n], Y, k]];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 26 2019, from PARI *)

{T(n,k)=local(x=X+X*O(X^n),y=Y+Y*O(Y^k)); A=1/(1-x*y*sum(j=0,n,x^j*prod(i=0,j-1,y+i))); return(polcoeff(polcoeff(A,n,X),k,Y))} (Hanna)

Sum_{k, 0<=k<=n} (-1)^(n-k)*T(n, k) = A000045(n+1), Fibonacci numbers.
Sum_{k, 0<=k<=n} T(n, k) = A051295(n).
Sum_{k, 0<=k<=n} 2^(n-k)*T(n, k) = A112934(n).
T(0, 0) = 1, T(n, n) = 2^(n-1).
G.f.: A(x, y) = 1/(1 - x*y*Sum_{j>=0} (y-1+j)!/(y-1)!*x^j ). - Paul D. Hanna, Oct 26 2005

A168467 a(n) = Product_{k=0..n} ((2k+2)(2*k+3))^(n-k).

Original entry on oeis.org

1, 6, 720, 3628800, 1316818944000, 52563198423859200000, 327312129899898454671360000000, 428017682605583614976547335700480000000000, 152240508705590071980086429193304853792686080000000000000

Offset: 0

Paul Barry, Nov 26 2009

Hankel transform of A000698(n+1).
The sequence 1,1,6,720,... with general term Product_{k=0..n, ((2k+1)(2k+0^k))^(n-k)} is the Hankel transform of A112934. - Paul Barry, Dec 04 2009
a(n) is also the determinant of the n X n matrix M(i,j) = i^(2*j)*sinh(2*j*arccsch(i))/(2*sqrt(i^2+1)), with i and j from 1 to n, which is the same matrix generated by sequences of length n by the linear recurrences with kernel { 2*(k^2 + z), -k^4 }, and initial conditions { 1, 2*(k^2 + z) }, with k from 1 to n, and z = 2. Regardless of the value of z, for every n, the determinant of the n X n matrix of polynomials generated gives always a(n) as result. - Federico Provvedi, Feb 01 2021

			From _Federico Provvedi_, Apr 01 2021: (Start)
From both formulas in the comment above and in particular with z=2 from the linear recurrences, the determinant of the 5 X 5 matrix: ( (1,6,35,204,1189), (1,12,128,1344,14080),(1,22,403,7084,123205), (1,36,1040,28224,749824), (1,54,2291,89964,3426181) ) = 1316818944000 = a(5).
For a generic z, the determinant doesn't change as shown in this example, where the determinant of the 3 X 3 square matrix:
( ( 1, 2*(z+1), (2*z + 1)*(2*z+3)  ),
  ( 1, 2*(z+4), 4*(z+6)*(z+2)      ),
  ( 1, 2*(z+9), (2*z + 9)(2*z + 27)) ) = 720 = a(3). (End)

Seiichi Manyama, Table of n, a(n) for n = 0..28

Cf. A000178, A000698, A098694, A112934, A306635, A001109, A099157, A246645, A202768, A000142.

Table[2^(n^2 + 2*n + 23/24) Glaisher^(3/2) Pi^(-n/2 - 3/4) BarnesG[n + 2] BarnesG[n + 5/2]/E^(1/8), {n, 0, 10}] (* Vladimir Reshetnikov, Sep 06 2016 *)
Table[Product[((2k+2)(2k+3))^(n-k),{k,0,n}],{n,0,10}] (* Harvey P. Dale, Dec 26 2019 *)
Table[Det@Table[LinearRecurrence[{2*k^2,-k^4},{1, 2*k^2},n], {k, 1, n}], {n,1,20}] (* Federico Provvedi, Feb 01 2021 *)
Det@Expand@Array[(#1^(2 #2))/(4 Sqrt[1 + #1^2])((Sqrt[1+1/#1^2]+1/#1)^(2 #2)-(Sqrt[1+1/#1^2]-1/#1)^(2 #2))&,{#,#}]&/@Range[20] (* Federico Provvedi, Apr 01 2021 *)

from math import prod
def A168467(n): return prod(((m:=k+1<<1)*(m+1))**(n-k) for k in range(1,n+1))*3**n<Chai Wah Wu, Nov 26 2023

G.f.: Q(0)/(2*x) -1/x, where Q(k) = 1  + 1/(1 -(2*k+1)!*x/((2*k+1)!*x + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 17 2013
a(n) = Product_{k=1..n} (2*k+1)!. - Vladimir Reshetnikov, Sep 06 2016
a(n) ~ A^(-1/2) * 2^(n^2 + 3*n + 53/24) * exp((-3/2)*n^2 + (-5/2)*n + 1/24) * n^(n^2 + (5/2)*n + 35/24) * Pi^((n+1)/2), where A = A074962 is the Glaisher-Kinkelin constant. - Vladimir Reshetnikov, Sep 06 2016
a(n) = A000178(2*n + 1) / A098694(n). - Vaclav Kotesovec, Oct 28 2017
a(n) = A202768(n)*A000142(n). - Federico Provvedi, Feb 01 2021
For n > 0, a(n) = n * (2*n+1) * sqrt(BarnesG(2*n)) * Gamma(2*n)^2 / (sqrt(Gamma(n)) * 2^((n-3)/2)). - Vaclav Kotesovec, Nov 27 2024

A112938 INVERT transform (with offset) of quadruple factorials (A008545), where g.f. satisfies: A(x) = 1 + x[d/dx xA(x)^4]/A(x)^4.

Original entry on oeis.org

1, 1, 4, 28, 292, 4156, 75844, 1694812, 44835172, 1369657468, 47422855300, 1834403141788, 78377228106148, 3664969183404220, 186134931067171012, 10201887125268108508, 600142156513333537252, 37713563573426417361148

Offset: 0

Paul D. Hanna, Oct 09 2005

nonn

			A(x) = 1 + x + 4*x^2 + 28*x^3 + 292*x^4 + 4156*x^5 + ...
1/A(x) = 1 - x - 3*x^2 - 21*x^3 - 231*x^4 -... -A008545(n)*x^(n+1)-...

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

Cf. A008545, A112939 (log derivative); A112934, A112935, A112936, A112937, A112940, A112941, A112942, A112943.

CoefficientList[Series[1/(1 + 1/4*ExpIntegralE[3/4,-1/(4*x)]/E^(1/(4*x))), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 22 2014 *)

{a(n)=local(F=1+x+x*O(x^n));for(i=1,n,F=1+x+4*x^2*deriv(F)/F); return(polcoeff(F,n,x))}

G.f. satisfies: A(x) = 1+x + 4*x^2*[d/dx A(x)]/A(x) (log derivative).
G.f.: A(x) = 1+x +4*x^2/(1-7*x -4*2*3*x^2/(1-15*x -4*3*7*x^2/(1-23*x -4*4*11*x^2/(1-31*x -... -4*n*(4*n-5)*x^2/(1-(8*n-1)*x -...)))) (continued fraction).
G.f.: A(x) = 1/(1-1*x/(1 -3*x/(1-4*x/(1 -7*x/(1-8*x/(1 -11*x/(1-12*x/(1 -...)))))))) (continued fraction).
G.f.: Q(0) where Q(k) = 1 - x*(4*k-1)/(1 - x*(4*k+4)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 20 2013
G.f.: 1 + 2*x/G(0), where G(k)= 1 + 1/(1 - 2*x*(4*k+4)/(2*x*(4*k+4) - 1 + 2*x*(4*k+3)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 31 2013
a(n) ~ (n-1)! * 4^(n-1) / (GAMMA(3/4) * n^(1/4)). - Vaclav Kotesovec, Feb 22 2014

A112940 INVERT transform (with offset) of quintuple factorials (A008546), where g.f. satisfies: A(x) = 1 + x[d/dx xA(x)^5]/A(x)^5.

Original entry on oeis.org

1, 1, 5, 45, 605, 11045, 257005, 7288245, 243870205, 9401560645, 410141056205, 19966451812245, 1072718714991005, 63033317759267045, 4020725747388170605, 276661592017425909045, 20424931173615717011005

Offset: 0

Paul D. Hanna, Oct 09 2005

nonn

			A(x) = 1 + x + 5*x^2 + 45*x^3 + 605*x^4 + 11045*x^5 +...
1/A(x) = 1 - x - 4*x^2 - 36*x^3 - 504*x^4 -... -A008546(n)*x^(n+1) -...

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

Cf. A008546, A112941 (log derivative); A112934, A112935, A112936, A112937, A112938, A112939, A112942, A112943.

CoefficientList[Series[1/(1 + 1/5*ExpIntegralE[4/5, -1/(5*x)]/E^(1/(5*x))), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 22 2014 *)

{a(n)=local(F=1+x+x*O(x^n));for(i=1,n,F=1+x+5*x^2*deriv(F)/F); return(polcoeff(F,n,x))}

G.f. satisfies: A(x) = 1+x + 5*x^2*[d/dx A(x)]/A(x) (log derivative). G.f.: A(x) = 1+x +5*x^2/(1-9*x -5*2*4*x^2/(1-19*x -5*3*9*x^2/(1-29*x -5*4*13*x^2/(1-39*x -... -5*n*(5*n-6)*x^2/(1-(10*n-1)*x -...)))) (continued fraction). G.f.: A(x) = 1/(1-1*x/(1 -4*x/(1-5*x/(1 -9*x/(1-10*x/(1 -14*x/(1-15*x/(1 -...)))))))) (continued fraction).
G.f.: 1 + x/( Q(0) - x ) where Q(k) =  1 - x*(5*k+4)/(1 - x*(5*k+5)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 20 2013
a(n) ~ (n-1)! * 5^(n-1) / (GAMMA(4/5) * n^(1/5)). - Vaclav Kotesovec, Feb 22 2014

A112942 INVERT transform (with offset) of sextuple factorials (A008543), where g.f. satisfies: A(x) = 1 + x[d/dx xA(x)^6]/A(x)^6.

Original entry on oeis.org

1, 1, 6, 66, 1086, 24186, 684006, 23506626, 951191646, 44281107066, 2330310876486, 136747268000706, 8851092668419326, 626304664252772346, 48092138192079689766, 3982448437177141451586, 353746119265020213643806

Offset: 0

Paul D. Hanna, Oct 09 2005

nonn

Generally, if g.f. satisfies: A(x) = 1 + x*[d/dx x*A(x)^p]/A(x)^p, then a(n) ~ (n-1)! * p^(n-1) / (Gamma((p-1)/p) * n^(1/p)). - Vaclav Kotesovec, Feb 22 2014

			A(x) = 1 + x + 6*x^2 + 66*x^3 + 1086*x^4 + 24186*x^5 +...
1/A(x) = 1 - x - 5*x^2 - 55*x^3 - 935*x^4 -... -A008543(n)*x^(n+1)-...

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

Cf. A008543, A112943 (log derivative); A112934, A112935, A112936, A112937, A112938, A112939, A112940, A112941.

CoefficientList[Series[1/(1 + 1/6*ExpIntegralE[5/6,-1/(6*x)]/E^(1/(6*x))), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 22 2014 *)

{a(n)=local(F=1+x+x*O(x^n));for(i=1,n,F=1+x+6*x^2*deriv(F)/F); return(polcoeff(F,n,x))}

G.f. satisfies: A(x) = 1+x + 6*x^2*[d/dx A(x)]/A(x) (log derivative).
G.f.: A(x) = 1+x+6*x^2/(1-11*x-6*2*5*x^2/(1-23*x-6*3*11*x^2/(1-35*x -6*4*17*x^2/(1-47*x- ... -6*n*(6*n-7)*x^2/(1-(12*n-1)*x - ...)))) (continued fraction).
G.f.: A(x) = 1/(1-1*x/(1-5*x/(1-6*x/(1-11*x/(1-12*x/(1-17*x/(1-18*x/(1 -...)))))))) (continued fraction).
G.f.: G(0) where G(k) = 1 - x*(6*k-1)/( 1 - 6*x*(k+1)/G(k+1) ); (continued fraction ). - Sergei N. Gladkovskii, Mar 24 2013
a(n) ~ (n-1)! * 6^(n-1) / (Gamma(5/6) * n^(1/6)). - Vaclav Kotesovec, Feb 22 2014

cp's OEIS Frontend

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.

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A355721 Square table, read by antidiagonals: the g.f. for row n is given recursively by (2*n-1)*x*R(n,x) = 1 + (2*n-3)*x - 1/R(n-1,x) for n >= 1 with the initial value R(0,x) = Sum_{k >= 0} A112934(k+1)*x^k.

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A028310 Expansion of (1 - x + x^2) / (1 - x)^2 in powers of x.

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A098694 Double-superfactorials: a(n) = Product_{k=1..n} (2k)!.

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A112936 INVERT transform (with offset) of triple factorials (A008544), where g.f. satisfies: A(x) = 1 + x*[d/dx x*A(x)^3]/A(x)^3.

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A111146 Triangle T(n,k), read by rows, given by [0, 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, ...] DELTA [1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is the operator defined in A084938.

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A168467 a(n) = Product_{k=0..n} ((2*k+2)*(2*k+3))^(n-k).

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A355721 Square table, read by antidiagonals: the g.f. for row n is given recursively by (2n-1)xR(n,x) = 1 + (2n-3)x - 1/R(n-1,x) for n >= 1 with the initial value R(0,x) = Sum_{k >= 0} A112934(k+1)x^k.

A112936 INVERT transform (with offset) of triple factorials (A008544), where g.f. satisfies: A(x) = 1 + x[d/dx xA(x)^3]/A(x)^3.

A168467 a(n) = Product_{k=0..n} ((2k+2)(2*k+3))^(n-k).

A112938 INVERT transform (with offset) of quadruple factorials (A008545), where g.f. satisfies: A(x) = 1 + x[d/dx xA(x)^4]/A(x)^4.

A112940 INVERT transform (with offset) of quintuple factorials (A008546), where g.f. satisfies: A(x) = 1 + x[d/dx xA(x)^5]/A(x)^5.

A112942 INVERT transform (with offset) of sextuple factorials (A008543), where g.f. satisfies: A(x) = 1 + x[d/dx xA(x)^6]/A(x)^6.