A111534 -id:A111534 - cp's OEIS frontend

A111535 a(n) = A111534(n)/n = A111528(n,n)/n for n>=1.

1, 2, 11, 104, 1409, 24912, 543479, 14098112, 423643509, 14464318560, 552830505347, 23375870438400, 1083128382648857, 54563592529048064, 2968656741661668975, 173460812744585863168, 10832194187368473624893

Offset: 1

Paul D. Hanna, Aug 06 2005

nonn

Vaclav Kotesovec, Table of n, a(n) for n = 1..364

Cf: A111528 (table), A003319 (row 1), A111529 (row 2), A111530 (row 3), A111531 (row 4), A111532 (row 5), A111533 (row 6).

{a(n)=if(n<1, 0, (1/n)*polcoeff(log(sum(m=0, n, (n-1+m)!/(n-1)!*x^m) + x*O(x^n)), n))}

a(n) = [x^n] (1/n)*Log( Sum_{m=0..n} (n-1+m)!/(n-1)!*x^m ) for n>=1.

a(n) ~ n! * 4^(n-1) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Jul 27 2015

PARI program fixed by Vaclav Kotesovec, Jul 27 2015

A111528 Square table, read by antidiagonals, where the g.f. for row n+1 is generated by: xR_{n+1}(x) = (1+nx - 1/R_n(x))/(n+1) with R_0(x) = Sum_{n>=0} n!*x^n.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 3, 6, 1, 1, 4, 13, 24, 1, 1, 5, 22, 71, 120, 1, 1, 6, 33, 148, 461, 720, 1, 1, 7, 46, 261, 1156, 3447, 5040, 1, 1, 8, 61, 416, 2361, 10192, 29093, 40320, 1, 1, 9, 78, 619, 4256, 23805, 99688, 273343, 362880, 1, 1, 10, 97, 876, 7045, 48096, 263313

Offset: 0

Paul D. Hanna, Aug 06 2005

			Table begins:
  1, 1,  2,   6,   24,   120,    720,    5040,     40320, ...
  1, 1,  3,  13,   71,   461,   3447,   29093,    273343, ...
  1, 1,  4,  22,  148,  1156,  10192,   99688,   1069168, ...
  1, 1,  5,  33,  261,  2361,  23805,  263313,   3161781, ...
  1, 1,  6,  46,  416,  4256,  48096,  591536,   7840576, ...
  1, 1,  7,  61,  619,  7045,  87955, 1187845,  17192275, ...
  1, 1,  8,  78,  876, 10956, 149472, 2195208,  34398288, ...
  1, 1,  9,  97, 1193, 16241, 240057, 3804353,  64092553, ...
  1, 1, 10, 118, 1576, 23176, 368560, 6262768, 112784896, ...
Rows are generated by logarithms of factorial series:
log(1 + x + 2*x^2 + 6*x^3 + 24*x^4 + ... n!*x^n + ...) = x + (3/2)*x^2 + (13/3)*x^3 + (71/4)*x^4 + (461/5)*x^5 + ...
(1/2)*log(1 + 2*x + 6*x^2 + ... + ((n+1)!/1!)*x^n + ...) = x + (4/2)*x^2 + (22/3)*x^3 + (148/4)*x^4 + (1156/5)*x^5 + ...
(1/3)*log(1 + 3*x + 12*x^2 + 60*x^3 + ... + ((n+2)!/2!)*x^n + ...) = x + (5/2)*x^2 + (33/3)*x^3 + (261/4)*x^4 + (2361/5)*x^5 +...
G.f. of row n may be expressed by the continued fraction:
R_n(x) = 1/(1+n*x - (n+1)*x/(1+(n+1)*x - (n+2)*x/(1+(n+2)*x -...
or recursively by: R_n(x) = 1/(1+n*x - (n+1)*x*R_{n+1}(x)).

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

Cf: A003319 (row 1), A111529 (row 2), A111530 (row 3), A111531 (row 4), A111532 (row 5), A111533 (row 6), A111534 (diagonal).

Similar recurrences: A124758, A243499, A284005, A329369, A341392.

T := (n, k) -> coeff(series(hypergeom([n+1, 1], [], x)/hypergeom([n, 1], [], 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); # Peter Bala, Jul 16 2022

T[n_, k_] := T[n, k] = Which[n < 0 || k < 0, 0, k == 0 || k == 1, 1, n == 0, k!, True, (T[n - 1, k + 1] - T[n - 1, k])/n - Sum[T[n, j]*T[n - 1, k - j], {j, 1, k - 1}]]; Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 18 2018 *)

{T(n,k)=if(n<0||k<0,0,if(k==0||k==1,1,if(n==0,k!, (T(n-1,k+1)-T(n-1,k))/n-sum(j=1,k-1,T(n,j)*T(n-1,k-j)))))}
for(n=0,10,for(k=0,10,print1(T(n,k),", ")); print(""))

{T(n,k)=if(n<0||k<0,0,if(k==0,1,if(n==0,k!, k/n*polcoeff(log(sum(m=0,k,(n-1+m)!/(n-1)!*x^m)),k))))}
for(n=0,10,for(k=0,10,print1(T(n,k),", ")); print(""))

T(n, 0) = 1, T(0, k) = k!, otherwise for n>=1 and k>=1:
T(n, k) = (T(n-1, k+1) - T(n-1, k))/n - Sum_{j=1..k-1} T(n, j)*T(n-1, k-j).
T(n, k) = (k/n)*[x^k] log(Sum_{m=0..k} (n-1+m)!/(n-1)!*x^m).
T(n, k) = Sum_{j = 0..k} A089949(k, j)*n^(k-j). - Philippe Deléham, Aug 08 2005
R_n(x) = -((n-1)!/n)/Sum_{i>=1} (i+n-2)!*x^i, n > 0. - Vladeta Jovovic, May 06 2006
G.f. of row R may be expressed by the continued fraction: W(0), where W(k) = 1 - x*(k+1)/( x*(k+1) - 1/(1 - x*(k+1+R)/( x*(k+1+R) - 1/W(k+1) ))). - Sergei N. Gladkovskii, Aug 26 2013
Conjecture: T(n, k) = b(2^(k-1) - 1, n) for k > 0 with T(n, 0) = 1 where b(n, m) = b(floor(n/2), m) + b(floor((2n - 2^A007814(n))/2), m) + m*b(A025480(n-1), m) for n > 0 with b(0, m) = 1. - Mikhail Kurkov, Dec 16 2021
From Peter Bala, Jul 11 2022: (Start)
O.g.f. for row n, n >= 1: R(n,x) = ( Sum_{k >= 0} (n+k)!/n!*x^k )/( Sum_{k >= 0} (n-1+k)!/(n-1)!*x^k ).
R(n,x)/(1 - n*x*R(n,x)) = Sum_{k >= 0} (n+k)!/n!*x^k.
For n >= 0, R(n,x) satisfies the Riccati equation x^2*d/dx(R(n,x)) + n*x*R(n,x)^2 - (1 + (n-1)*x)*R(n,x) + 1 = 0 with R(n,0) = 1.
Apply Stokes 1982 to find that for n >= 0, R(n,x) = 1/(1 - x/(1 - (n+1)*x/(1 - 2*x/(1 - (n+2)*x/(1 - 3*x/(1 - (n+3)*x/(1 - 4*x/(1 - (n+4)*x/(1 - ...))))))))), a continued fraction of Stieltjes type. (End)

A111529 Row 2 of table A111528.

Original entry on oeis.org

1, 1, 4, 22, 148, 1156, 10192, 99688, 1069168, 12468208, 157071424, 2126386912, 30797423680, 475378906432, 7793485765888, 135284756985472, 2479535560687360, 47860569736036096, 970606394944476160, 20635652201785613824, 459015456156148876288, 10662527360021306782720

Offset: 0

Paul D. Hanna, Aug 06 2005

			(1/2)*log(1 + 2*x + 6*x^2 + ... + ((n+1)!/1!)*x^n + ...)
= x + (4/2)*x^2 + (22/3)*x^3 + (148/4)*x^4 + (1156/5)*x^5 + ...

Robert Israel, Table of n, a(n) for n = 0..410
Paul Barry, A note on number triangles that are almost their own production matrix, arXiv:1804.06801 [math.CO], 2018.
Richard J. Martin and Michael J. Kearney, Integral representation of certain combinatorial recurrences, Combinatorica: 35:3 (2015), 309-315.
A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.

Cf. A111528 (table), A003319 (row 1), A111530 (row 3), A111531 (row 4), A111532 (row 5), A111533 (row 6), A111534 (diagonal).

Cf. A077607, A089949, A260491.

N:= 30: # to get a(0) to a(N)
g:= 1/2*log(add((n+1)!*x^n,n=0..N+1)):
S:= series(g,x,N+1);
1, seq(j*coeff(S,x,j),j=0..N); # Robert Israel, Jul 10 2015

T[n_, k_] := T[n, k] = Which[n<0 || k<0, 0, k==0 || k==1, 1, n==0, k!, True, (T[n-1, k+1]-T[n-1, k])/n - Sum[T[n, j] T[n-1, k-j], {j, 1, k-1}]];
a[n_] := T[2, n];
Table[a[n], {n, 0, 21}] (* Jean-François Alcover, Aug 09 2018 *)

{a(n)=if(n<0,0,if(n==0,1, (n/2)*polcoeff(log(sum(m=0,n,(m+1)!/1!*x^m)),n)))}

G.f.: (1/2)*log(Sum_{n >= 0} (n+1)!*x^n) = Sum_{n >= 1} a(n)*x^n/n.
G.f.: 1/(1+2*x - 3*x/(1+3*x - 4*x/(1+4*x - ... (continued fraction).
a(n) = Sum_{k = 0..n} 2^(n-k)*A089949(n,k). - Philippe Deléham, Oct 16 2006
G.f. 1/(2*x-G(0)) where G(k) = 2*x - 1 - k*x - x*(k+1)/G(k+1); G(0)=x (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Aug 14 2012
G.f.: 1/(2*x) - 1/(G(0) - 1) where G(k) = 1 + x*(k+1)/(1 - 1/(1 + 1/G(k+1)));(continued fraction, 3-step). - Sergei N. Gladkovskii, Nov 20 2012
G.f.: 1 + x/(G(0)-2*x) where G(k) = 1 + (k+1)*x - x*(k+3)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 26 2012
G.f.: (1 + 1/Q(0))/2, where Q(k) = 1 + k*x - x*(k+2)/Q(k+1); (continued fraction). In general, the g.f. for row (r+2) is (r + 1 + 1/Q(0))/(r + 2). - Sergei N. Gladkovskii, May 04 2013
G.f.: W(0), where W(k) = 1 - x*(k+1)/( x*(k+1) - 1/(1 - x*(k+3)/( x*(k+3) - 1/W(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Aug 26 2013
a(n) ~ n! * n^2/2 * (1 - 1/n - 2/n^2 - 8/n^3 - 52/n^4 - 436/n^5 - 4404/n^6 - 51572/n^7 - 683428/n^8 - 10080068/n^9 - 163471284/n^10), where the coefficients are given by (n+2)*(n+1)/n^2 * Sum_{k>=0} A260491(k)/(n+2)^k. - Vaclav Kotesovec, Jul 27 2015
a(n) = -A077607(n+2)/2. - Vaclav Kotesovec, Jul 29 2015
From Peter Bala, Jul 12 2022: (Start)
O.g.f: A(x) = ( Sum_{k >= 0} ((k+2)!/2!)*x^k )/( Sum_{k >= 0} (k+1)!*x^k ).
A(x)/(1 - 2*x*A(x)) = Sum_{k >= 0} ((k+2)!/2!)*x^k.
Riccati differential equation: x^2*A'(x) + 2*x*A^2(x) - (1 + x)*A(x) + 1 = 0.
Apply Stokes 1982 to find that A(x) = 1/(1 - x/(1 - 3*x/(1 - 2*x/(1 - 4*x/(1 - 3*x/(1 - 5*x/(1 - ... - n*x/(1 - (n+2)*x/(1 - ...))))))))), a continued fraction of Stieltjes type. (End)

A111530 Row 3 of table A111528.

Original entry on oeis.org

1, 1, 5, 33, 261, 2361, 23805, 263313, 3161781, 40907241, 567074925, 8385483393, 131787520101, 2194406578521, 38605941817245, 715814473193073, 13956039627763221, 285509132504621001, 6116719419966460365

Offset: 0

Paul D. Hanna, Aug 06 2005

			(1/3)*(log(1 + 3*x + 12*x^2 + 60*x^3 + ... + (n+2)!/2!)*x^n + ...)
= x + 5/2*x^2 + 33/3*x^3 + 261/4*x^4 + 2361/5*x^5 + ...

Vaclav Kotesovec, Table of n, a(n) for n = 0..440
Paul Barry, A note on number triangles that are almost their own production matrix, arXiv:1804.06801 [math.CO], 2018.
A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.

Cf: A111528 (table), A003319 (row 1), A111529 (row 2), A111531 (row 4), A111532 (row 5), A111533 (row 6), A111534 (diagonal).

Cf. A001710, A001715.

T[n_, k_] := T[n, k] = Which[n<0 || k<0, 0, k==0 || k==1, 1, n==0, k!, True, (T[n-1, k+1]-T[n-1, k])/n - Sum[T[n, j]*T[n-1, k-j], {j, 1, k-1}]];
a[n_] := T[3, n];
Table[a[n], {n, 0, 18}] (* Jean-François Alcover, Aug 09 2018 *)

{a(n)=if(n<0,0,if(n==0,1, (n/3)*polcoeff(log(sum(m=0,n,(m+2)!/2!*x^m) + x*O(x^n)),n)))} \\ fixed by Vaclav Kotesovec, Jul 27 2015

G.f.: (1/3)*log(Sum_{n>=0} (n+2)!/2!*x^n) = Sum_{n>=1} a(n)*x^n/n.
G.f.: A(x) = 1/(1 + 3*x - 4*x/(1 + 4*x - 5*x/(1 + 5*x - ... (continued fraction).
a(n) = Sum_{k=0..n} 3^(n-k)*A089949(n,k). - Philippe Deléham, Oct 16 2006
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(k+1)/(x*(k-1/2) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 06 2013
G.f.: W(0), where W(k) = 1 - x*(k+1)/( x*(k+1) - 1/(1 - x*(k+1+R)/( x*(k+1+R) - 1/W(k+1) ))); R=3 is Row R of table A111528 (continued fraction). - Sergei N. Gladkovskii, Aug 26 2013
a(n) ~ n! * n^3/6 * (1 - 4/n^2 - 15/n^3 - 99/n^4 - 882/n^5 - 9531/n^6 - 119493/n^7 - 1693008/n^8 - 26638245/n^9 - 459682047/n^10). - Vaclav Kotesovec, Jul 27 2015
From Peter Bala, May 24 2017: (Start)
O.g.f. A(x) = ( Sum_{n >= 0} (n+3)!/3!*x^n ) / ( Sum_{n >= 0} (n+2)!/2!*x^n ).
1/(1 - 3*x*A(x)) = Sum_{n >= 0} (n+2)!/2!*x^n. Cf. A001710.
A(x)/(1 - 3*x*A(x)) = Sum_{n >= 0} (n+3)!/3!*x^n. Cf. A001715.
A(x) satisfies the Riccati equation x^2*A'(x) + 3*x*A^2(x) - (1 + 2*x)*A(x) + 1 = 0.
G.f. as an S-fraction: A(x) = 1/(1 - x/(1 - 4*x/(1 - 2*x/(1 - 5*x/(1 - 3*x/(1 - 6*x/(1 - ... - n*x/(1 - (n+3)*x/(1 - ... ))))))))), by Stokes 1982.
A(x) = 1/(1 + 3*x - 4*x/(1 - x/(1 - 5*x/(1 - 2*x/(1 - 6*x/(1 - 3*x/(1 - ... - (n + 3)*x/(1 - n*x/(1 - ... ))))))))). (End)

A111531 Row 4 of table A111528.

Original entry on oeis.org

1, 1, 6, 46, 416, 4256, 48096, 591536, 7840576, 111226816, 1680157056, 26918720896, 455971214336, 8143926373376, 153013563734016, 3017996904928256, 62369444355076096, 1348096649995841536, 30426167700424728576, 715935203128235401216

Offset: 0

Paul D. Hanna, Aug 06 2005

			(1/4)*(log(1 + 4*x + 20*x^2 + 120*x^3 + ... + (n+3)!/3!)*x^n + ...)
= x + 6/2*x^2 + 46/3*x^3 + 416/4*x^4 + 4256/5*x^5 + ...

Vaclav Kotesovec, Table of n, a(n) for n = 0..440
Paul Barry, A note on number triangles that are almost their own production matrix, arXiv:1804.06801 [math.CO], 2018.
A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.

Cf: A111528 (table), A003319 (row 1), A111529 (row 2), A111530 (row 3), A111532 (row 5), A111533 (row 6), A111534 (diagonal).

Cf. A001715, A001720.

T[n_, k_] := T[n, k] = Which[n<0 || k<0, 0, k==0 || k==1, 1, n==0, k!, True, (T[n-1, k+1]-T[n-1, k])/n-Sum[T[n, j]*T[n-1, k-j], {j, 1, k-1}]];
a[n_] := T[4, n];
a /@ Range[0, 19] (* Jean-François Alcover, Oct 01 2019 *)

{a(n)=if(n<0,0,if(n==0,1, (n/4)*polcoeff(log(sum(m=0,n,(m+3)!/3!*x^m) +x*O(x^n)),n)))}
for(n=0,20,print1(a(n),", "))

G.f.: (1/4)*log(Sum_{n>=0} (n+3)!/3!*x^n) = Sum_{n>=1} a(n)*x^n/n.
G.f.: A(x) = 1/(1 + 4*x - 5*x/(1 + 5*x - 6*x/(1 + 6*x - ... (continued fraction).
a(n) = Sum_{k=0..n} 4^(n-k)*A089949(n,k). - Philippe Deléham, Oct 16 2006
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(k+1)/(x*(k-1) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 05 2013
G.f.: W(0)/4 + 3/4, where W(k) = 1 - x*(k+4)/( x*(k+4) - 1/(1 - x*(k+2)/( x*(k+2) - 1/W(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Aug 26 2013
a(n) ~ n! * n^4/24 * (1 + 2/n - 5/n^2 - 30/n^3 - 184/n^4 - 1664/n^5 - 18688/n^6 - 245120/n^7 - 3641280/n^8 - 60090368/n^9 - 1086985152/n^10). - Vaclav Kotesovec, Jul 27 2015
From Peter Bala, May 25 2017: (Start)
O.g.f. A(x) = ( Sum_{n >= 0} (n+4)!/4!*x^n ) / ( Sum_{n >= 0} (n+3)!/3!*x^n ).
1/(1 - 4*x*A(x)) = Sum_{n >= 0} (n+3)!/3!*x^n. Cf. A001715.
A(x)/(1 - 4*x*A(x)) = Sum_{n >= 0} (n+4)!/4!*x^n. Cf. A001720.
A(x) satisfies the Riccati equation x^2*A'(x) + 4*x*A^2(x) - (1 + 3*x)*A(x) + 1 = 0.
G.f. as an S-fraction: A(x) = 1/(1 - x/(1 - 5*x/(1 - 2*x/(1 - 6*x/(1 - 3*x/(1 - 7*x/(1 - ... - n*x/(1 - (n+4)*x/(1 - ... ))))))))), by Stokes 1982.
A(x) = 1/(1 + 4*x - 5*x/(1 - x/(1 - 6*x/(1 - 2*x/(1 - 7*x/(1 - 3*x/(1 - ... - (n + 4)*x/(1 - n*x/(1 - ... ))))))))). (End)

A111532 Row 5 of table A111528.

Original entry on oeis.org

1, 1, 7, 61, 619, 7045, 87955, 1187845, 17192275, 264940405, 4326439075, 74593075525, 1353928981075, 25809901069525, 515683999204675, 10779677853137125, 235366439343773875, 5359766538695291125

Offset: 0

Paul D. Hanna, Aug 06 2005

			(1/5)*(log(1 + 5*x + 30*x^2 + 210*x^3 + ... + (n+4)!/4!)*x^n + ...)
= x + 7/2*x^2 + 61/3*x^3 + 619/4*x^4 + 7045/5*x^5 + ...

Vaclav Kotesovec, Table of n, a(n) for n = 0..440
A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.

Cf: A111528 (table), A003319 (row 1), A111529 (row 2), A111530 (row 3), A111531 (row 4), A111533 (row 6), A111534 (diagonal).

Cf. A001720, A001725.

m = 18; (-1/(5x)) ContinuedFractionK[-i x, 1 + i x, {i, 5, m+4}] + O[x]^m // CoefficientList[#, x]& (* Jean-François Alcover, Nov 02 2019 *)

{a(n)=if(n<0,0,if(n==0,1, (n/5)*polcoeff(log(sum(m=0,n,(m+4)!/4!*x^m) + x*O(x^n)),n)))} \\ fixed by Vaclav Kotesovec, Jul 27 2015

G.f.: (1/5)*log(Sum_{n>=0} (n+4)!/4!*x^n) = Sum_{n>=1} a(n)*x^n/n.
G.f.: 1/(1 + 5*x - 6*x/(1 + 6*x - 7*x/(1 + 7*x - ... (continued fraction).
a(n) = Sum_{k=0..n} 5^(n-k)*A089949(n,k). - Philippe Deléham, Oct 16 2006
G.f.: (4 + 1/Q(0))/5, where Q(k) = 1 - 3*x + k*x - x*(k+2)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 04 2013
a(n) ~ n! * n^5/5! * (1 + 5/n - 55/n^3 - 356/n^4 - 3095/n^5 - 35225/n^6 - 475000/n^7 - 7293775/n^8 - 124710375/n^9 - 2339428250/n^10). - Vaclav Kotesovec, Jul 27 2015
From Peter Bala, May 25 2017: (Start)
O.g.f.: A(x) = ( Sum_{n >= 0} (n+5)!/5!*x^n ) / ( Sum_{n >= 0} (n+4)!/4!*x^n ).
1/(1 - 5*x*A(x)) = Sum_{n >= 0} (n+4)!/4!*x^n. Cf. A001720.
A(x)/(1 - 5*x*A(x)) = Sum_{n >= 0} (n+5)!/5!*x^n. Cf. A001725.
A(x) satisfies the Riccati equation x^2*A'(x) + 5*x*A^2(x) - (1 + 4*x)*A(x) + 1 = 0.
G.f. as an S-fraction: A(x) = 1/(1 - x/(1 - 6*x/(1 - 2*x/(1 - 7*x/(1 - 3*x/(1 - 8*x/(1 - ... - n*x/(1 - (n+5)*x/(1 - ... ))))))))), by Stokes 1982.
A(x) = 1/(1 + 5*x - 6*x/(1 - x/(1 - 7*x/(1 - 2*x/(1 - 8*x/(1 - 3*x/(1 - ... - (n + 5)*x/(1 - n*x/(1 - ... ))))))))). (End)

A111533 Row 6 of table A111528.

Original entry on oeis.org

1, 1, 8, 78, 876, 10956, 149472, 2195208, 34398288, 571525200, 10022997888, 184897670112, 3578224662720, 72486450479808, 1534267158087168, 33877135427154048, 779208751651730688, 18645519786163266816

Offset: 0

Paul D. Hanna, Aug 06 2005

			(1/6)*(log(1 + 6*x + 42*x^2 + 336*x^3 + ... + (n+5)!/5!)*x^n + ...)
= x + 8/2*x^2 + 78/3*x^3 + 876/4*x^4 + 10956/5*x^5 + ...

Vaclav Kotesovec, Table of n, a(n) for n = 0..440
A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.

Cf: A111528 (table), A003319 (row 1), A111529 (row 2), A111530 (row 3), A111531 (row 4), A111532 (row 5), A111534 (diagonal).

Cf. A001725, A001730.

m = 18; (-1/(6x)) ContinuedFractionK[-i x, 1 + i x, {i, 6, m+5}] + O[x]^m // CoefficientList[#, x]& (* Jean-François Alcover, Nov 02 2019 *)

{a(n)=if(n<0,0,if(n==0,1, (n/6)*polcoeff(log(sum(m=0,n,(m+5)!/5!*x^m) + x*O(x^n)),n)))} \\ fixed by Vaclav Kotesovec, Jul 27 2015

G.f.: (1/6)*log(Sum_{n>=0} (n+5)!/5!*x^n) = Sum_{n>=1} a(n)*x^n/n.
G.f.: 1/(1 + 6*x - 7*x/(1 + 7*x - 8*x/(1 + 8*x -... (continued fraction).
a(n) = Sum_{k=0..n} 6^(n-k)*A089949(n,k). - Philippe Deléham, Oct 16 2006
G.f.: (5 + 1/Q(0))/6, where Q(k) = 1 - 4*x + k*x - x*(k+2)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 04 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(k+1)/(x*(k-2) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 05 2013
a(n) ~ n! * n^6/6! * (1 + 9/n + 19/n^2 - 69/n^3 - 704/n^4 - 5880/n^5 - 65736/n^6 - 896832/n^7 - 14068080/n^8 - 246800304/n^9 - 4760585136/n^10). - Vaclav Kotesovec, Jul 27 2015
From Peter Bala, May 25 2017: (Start)
O.g.f.: A(x) = ( Sum_{n >= 0} (n+6)!/6!*x^n ) / ( Sum_{n >= 0} (n+5)!/5!*x^n ).
1/(1 - 6*x*A(x)) = Sum_{n >= 0} (n+5)!/5!*x^n. Cf. A001725.
A(x)/(1 - 6*x*A(x)) = Sum_{n >= 0} (n+6)!/6!*x^n. Cf. A001730.
A(x) satisfies the Riccati equation x^2*A'(x) + 6*x*A^2(x) - (1 + 5*x)*A(x) + 1 = 0.
G.f. as an S-fraction: A(x) = 1/(1 - x/(1 - 7*x/(1 - 2*x/(1 - 8*x/(1 - 3*x/(1 - 9*x/(1 - ... - n*x/(1 - (n+6)*x/(1 - ... ))))))))), by Stokes 1982.
A(x) = 1/(1 + 6*x - 7*x/(1 - x/(1 - 8*x/(1 - 2*x/(1 - 9*x/(1 - 3*x/(1 - ... - (n + 6)*x/(1 - n*x/(1 - ... ))))))))). (End)

cp's OEIS Frontend

A111535 a(n) = A111534(n)/n = A111528(n,n)/n for n>=1.

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A111528 Square table, read by antidiagonals, where the g.f. for row n+1 is generated by: x*R_{n+1}(x) = (1+n*x - 1/R_n(x))/(n+1) with R_0(x) = Sum_{n>=0} n!*x^n.

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A111529 Row 2 of table A111528.

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A111530 Row 3 of table A111528.

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A111531 Row 4 of table A111528.

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A111532 Row 5 of table A111528.

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A111533 Row 6 of table A111528.

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A111528 Square table, read by antidiagonals, where the g.f. for row n+1 is generated by: xR_{n+1}(x) = (1+nx - 1/R_n(x))/(n+1) with R_0(x) = Sum_{n>=0} n!*x^n.