A112936 -id:A112936 - cp's OEIS frontend

A112937 Logarithmic derivative of A112936 such that a(n)=(1/3)*A112936(n+1) for n>0, where A112936 equals the INVERT transform (with offset) of triple factorials A008544.

1, 5, 37, 377, 4981, 81305, 1580797, 35637377, 913115701, 26189790425, 830916198157, 28883617580177, 1091455878504421, 44541746007215945, 1952125704702209917, 91440056107001450177, 4558596081095404198741

Offset: 1

Paul D. Hanna, Oct 09 2005

nonn

			log(1+x + 3*x*[x + 5*x^2 + 37*x^3 + 377*x^4 + 4981*x^5 +...])
= x + 5/2*x^2 + 37/3*x^3 + 377/4*x^4 + 4981/5*x^5 + ...

Cf. A008544, A112936; A112934, A112935, A112938, A112939, A112940, A112941, A112942, A112943.

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

G.f.: log(1+x + 3*x*[Sum_{k>=1} a(n)]) = Sum_{k>=1} a(n)/n*x^n.

A355793 Square table, read by antidiagonals: the g.f. for row n is given recursively by (3n-1)xR(n,x) = 1 + (3n-4)x - 1/R(n-1,x) for n >= 1 with the initial value R(0,x) = Sum_{k >= 0} A112936(k+1)x^k.

Original entry on oeis.org

1, 1, 3, 1, 3, 15, 1, 3, 24, 111, 1, 3, 33, 282, 1131, 1, 3, 42, 507, 4236, 14943, 1, 3, 51, 786, 9609, 76548, 243915, 1, 3, 60, 1119, 17736, 212835, 1608864, 4742391, 1, 3, 69, 1506, 29103, 459768, 5350785, 38488152, 106912131, 1, 3, 78, 1947, 44196, 859143, 13333488

Offset: 0

Peter Bala, Jul 17 2022

Compare with A111528 and A355721, which have similar definitions and properties.

			Square array begins
1, 3, 15,  111,  1131,   14943,    243915,    4742391,    106912131, ...
1, 3, 24,  282,  4236,   76548,   1608864,   38488152,   1032125136, ...
1, 3, 33,  507,  9609,  212835,   5350785,  149961675,   4628365305, ...
1, 3, 42,  786, 17736,  459768,  13333488,  425600976,  14791250688, ...
1, 3, 51, 1119, 29103,  859143,  28091463, 1002057591,  38606468343, ...
1, 3, 60, 1506, 44196, 1458588,  52917360, 2080630776,  87823112496, ...
1, 3, 69, 1947, 63501, 2311563,  91949469, 3943276347, 180679742061, ...
1, 3, 78, 2442, 87504, 3477360, 150259200, 6970190160, 344116224960, ...

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

Cf. A112936 (row 0), A355794 (row 1), A355795 (row 2), A355796 (row 3), A355797 (row 4). Cf. A008544, A111528, A355721.

T := (n,k) -> coeff(series(hypergeom([n+2/3, 1], [], 3*x)/ hypergeom([n-1/3, 1], [], 3*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 t(n) = Product_{k = 1..n} 3*k-1 = A008544(n) (triple factorial numbers).
O.g.f. for row n >= 0: R(n,x) = ( Sum_{k >= 0} t(n+k)/t(n)*x^k )/( Sum_{k >= 0} t(n-1+k)/t(n-1)*x^k ).
R(n,x)/(1 - (3*n-1)*x*R(n,x)) = Sum_{k >= 0} t(n+k)/t(n)*x^k.
R(n,x) = 1/(1 + (3*n-1)*x - (3*n+2)*x/(1 + (3*n+2)*x - (3*n+5)*x/(1 + (3*n+5)*x - (3*n+8)*x/(1 + (3*n+8)*x - ... )))) (continued fraction).
R(n,x) satisfies the Riccati differential equation 3*x^2*d/dx(R(n,x)) + (3*n-1)*x*R(n,x)^2 - (1 + (3*n-4)*x)*R(n,x) + 1 = 0 with R(n,0) = 1.
Applying Stokes 1982 gives R(n,x) = 1/(1 - 3*x/(1 - (3*n+2)*x/(1 - 6*x/(1 - (3*n+5)*x/(1 - 9*x/(1 - (3*n+8)*x/(1 - 12*x/(1 - ...)))))))), a continued fraction of Stieltjes type.

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

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

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

A112939 Logarithmic derivative of A112938 such that a(n)=(1/4)*A112938(n+1) for n>0, where A112938 equals the INVERT transform (with offset) of quadruple factorials A008545.

Original entry on oeis.org

1, 7, 73, 1039, 18961, 423703, 11208793, 342414367, 11855713825, 458600785447, 19594307026537, 916242295851055, 46533732766792753, 2550471781317027127, 150035539128333384313, 9428390893356604340287, 630318228814408172573761

Offset: 1

Paul D. Hanna, Oct 09 2005

nonn

			log(1+x + 4*x*[x + 7*x^2 + 73*x^3 + 1039*x^4 + 18961*x^5 +...])
= x + 7/2*x^2 + 73/3*x^3 + 1039/4*x^4 + 18961/5*x^5 + ...

Cf. A008545, A112938; A112934, A112935, A112936, A112937, A112940, A112941, A112942, A112943.

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

G.f.: log(1+x + 4*x*[Sum_{k>=1} a(n)]) = Sum_{k>=1} a(n)/n*x^n.

G.f.: 1/x - G(0)/(2*x), where G(k)= 1 + 1/(1 - x*(4*k-1)/(x*(4*k-3) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 04 2013

A112941 Logarithmic derivative of A112940 such that a(n)=(1/5)*A112940(n+1) for n>0, where A112940 equals the INVERT transform (with offset) of quintuple factorials A008546.

Original entry on oeis.org

1, 9, 121, 2209, 51401, 1457649, 48774041, 1880312129, 82028211241, 3993290362449, 214543742998201, 12606663551853409, 804145149477634121, 55332318403485181809, 4084986234723143402201, 322064057582671115832449

Offset: 1

Paul D. Hanna, Oct 09 2005

nonn

			log(1+x + 5*x*[x + 9*x^2 + 121*x^3 + 2209*x^4 + 51401*x^5 +...])
= x + 9/2*x^2 + 121/3*x^3 + 2209/4*x^4 + 51401/5*x^5 + ...

Cf. A008546, A112940; A112934, A112935, A112936, A112937, A112938, A112939, A112942, A112943.

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

G.f.: log(1+x + 5*x*[Sum_{n>=1} a(n)]) = Sum_{n>=1} a(n)/n*x^n.

A112943 Logarithmic derivative of A112942 such that a(n)=(1/6)*A112942(n+1) for n>0, where A112942 equals the INVERT transform (with offset) of sextuple factorials A008543.

Original entry on oeis.org

1, 11, 181, 4031, 114001, 3917771, 158531941, 7380184511, 388385146081, 22791211333451, 1475182111403221, 104384110708795391, 8015356365346614961, 663741406196190241931, 58957686544170035607301

Offset: 1

Paul D. Hanna, Oct 09 2005

nonn

			log(1+x + 6*x*[x + 11*x^2 + 181*x^3 + 4031*x^4 + 114001*x^5 +...])
= x + 11/2*x^2 + 181/3*x^3 + 4031/4*x^4 + 114001/5*x^5 + ...

Cf. A008543, A112942; A112934, A112935, A112936, A112937, A112938, A112939, A112940, A112941.

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

G.f.: log(1+x + 6*x*[Sum_{n>=1} a(n)]) = Sum_{n>=1} a(n)/n*x^n.

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A112937 Logarithmic derivative of A112936 such that a(n)=(1/3)*A112936(n+1) for n>0, where A112936 equals the INVERT transform (with offset) of triple factorials A008544.

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A355793 Square table, read by antidiagonals: the g.f. for row n is given recursively by (3*n-1)*x*R(n,x) = 1 + (3*n-4)*x - 1/R(n-1,x) for n >= 1 with the initial value R(0,x) = Sum_{k >= 0} A112936(k+1)*x^k.

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

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

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

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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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A112939 Logarithmic derivative of A112938 such that a(n)=(1/4)*A112938(n+1) for n>0, where A112938 equals the INVERT transform (with offset) of quadruple factorials A008545.

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A355793 Square table, read by antidiagonals: the g.f. for row n is given recursively by (3n-1)xR(n,x) = 1 + (3n-4)x - 1/R(n-1,x) for n >= 1 with the initial value R(0,x) = Sum_{k >= 0} A112936(k+1)x^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.