A111530 -id:A111530 - cp's OEIS frontend

A001715 a(n) = n!/6.

1, 4, 20, 120, 840, 6720, 60480, 604800, 6652800, 79833600, 1037836800, 14529715200, 217945728000, 3487131648000, 59281238016000, 1067062284288000, 20274183401472000, 405483668029440000, 8515157028618240000, 187333454629601280000, 4308669456480829440000

Offset: 3

N. J. A. Sloane

The numbers (4, 20, 120, 840, 6720, ...) arise from the divisor values in the general formula a(n) = n*(n+1)*(n+2)*(n+3)* ... *(n+k)*(n*(n+k) + (k-1)*k/6)/((k+3)!/6) (which covers the following sequences: A000578, A000537, A024166, A101094, A101097, A101102). - Alexander R. Povolotsky, May 17 2008
a(n) is also the number of decreasing 3-cycles in the decomposition of permutations as product of disjoint cycles, a(3)=1, a(4)=4, a(5)=20. - Wenjin Woan, Dec 21 2008
Equals eigensequence of triangle A130128 reflected. - Gary W. Adamson, Dec 23 2008
a(n) is the number of n-permutations having 1, 2, and 3 in three distinct cycles. - Geoffrey Critzer, Apr 26 2009
From Johannes W. Meijer, Oct 20 2009: (Start)
The asymptotic expansion of the higher order exponential integral E(x,m=1,n=4) ~ exp(-x)/x*(1 - 4/x + 20/x^2 - 120/x^3 + 840/x^4 - 6720/x^5 + 60480/x^6 - 604800/x^7 + ...) leads to the sequence given above. See A163931 and A130534 for more information.
(End)

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 3..200
Somaya Barati, Beáta Bényi, Abbas Jafarzadeh, and Daniel Yaqubi, Mixed restricted Stirling numbers, arXiv:1812.02955 [math.CO], 2018.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 263.
Wolfdieter Lang, On generalizations of Stirling number triangles, J. Integer Seqs., Vol. 3 (2000), Article 00.2.4.
D. S. Mitrinovic and R. S. Mitrinovic, Tableaux d'une classe de nombres reliés aux nombres de Stirling, Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. Fiz. No. 77 1962, 77 pp.
Alexsandar Petojevic, The Function vM_m(s; a; z) and Some Well-Known Sequences, Journal of Integer Sequences, Vol. 5 (2002), Article 02.1.7.
A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.
Index entries for sequences related to factorial numbers.
Index to divisibility sequences.

Cf. A049352, A049458, A049460.
Cf. A034472, A130128.
Cf. A245334, A000142, A111530.
Cf. A001710, A001720, A007759, A094638.
Cf. A094587, A138533, A173333, A213936.

a001715 = (flip div 6) . a000142 -- Reinhard Zumkeller, Aug 31 2014

[Factorial(n)/6: n in [3..30]]; // Vincenzo Librandi, Jun 20 2011

f := proc(n) n!/6; end;
BB:= [S, {S = Prod(Z,Z,C), C = Union(B,Z,Z), B = Prod(Z,C)}, labelled]: seq(combstruct[count](BB, size=n)/12, n=3..20); # Zerinvary Lajos, Jun 19 2008
G(x):=1/(1-x)^4: f[0]:=G(x): for n from 1 to 18 do f[n]:=diff(f[n-1],x) od: x:=0: seq(f[n],n=0..16); # Zerinvary Lajos, Apr 01 2009

a[n_]:=n!/6; (*Vladimir Joseph Stephan Orlovsky, Dec 13 2008 *)
Range[3,30]!/6 (* Harvey P. Dale, Aug 12 2012 *)

a(n)=n!/6 \\ Charles R Greathouse IV, Jan 12 2012

a(n) = A049352(n-2, 1) (first column of triangle).
E.g.f. if offset 0: 1/(1-x)^4.
a(n) = A173333(n,3). - Reinhard Zumkeller, Feb 19 2010
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x/(x + 1/(k+4)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
G.f.: W(0), where W(k) = 1 - x*(k+4)/( x*(k+4) - 1/(1 - x*(k+1)/( x*(k+1) - 1/W(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Aug 26 2013
a(n) = A245334(n,n-3) / 4. - Reinhard Zumkeller, Aug 31 2014
From Peter Bala, May 22 2017: (Start)
The o.g.f. A(x) satisfies the Riccati equation x^2*A'(x) + (4*x - 1)*A(x) + 1 = 0.
G.f. as an S-fraction: A(x) = 1/(1 - 4*x/(1 - x/(1 - 5*x/(1 - 2*x/(1 - 6*x/(1 - 3*x/(1 - ... - (n + 3)*x/(1 - n*x/(1 - ... ))))))))) (apply Stokes, 1982).
A(x) = 1/(1 - 3*x - x/(1 - 4*x/(1 - 2*x/(1 - 5*x/(1 - 3*x/(1 - 6*x/(1 - ... - n*x/(1 - (n+3)*x/(1 - ... ))))))))). (End)
H(x) = (1 - (1 + x)^(-3)) / 3 = x - 4 x^2/2! + 20 x^3/3! - ... is an e.g.f. of the signed sequence (n!/4!), which is the compositional inverse of G(x) = (1 - 3*x)^(-1/3) - 1, an e.g.f. for A007559. Cf. A094638, A001710 (for n!/2!), and A001720 (for n!/4!). Cf. columns of A094587, A173333, and A213936 and rows of A138533.- Tom Copeland, Dec 27 2019
E.g.f.: x^3 / (3! * (1 - x)). - Ilya Gutkovskiy, Jul 09 2021
From Amiram Eldar, Jan 15 2023: (Start)
Sum_{n>=3} 1/a(n) = 6*e - 15.
Sum_{n>=3} (-1)^(n+1)/a(n) = 3 - 6/e. (End)

More terms from Harvey P. Dale, Aug 12 2012

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)

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)

A089949 Triangle T(n,k), read by rows, given by [0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...] DELTA [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, ...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 0, 1, 6, 6, 0, 1, 12, 34, 24, 0, 1, 20, 110, 210, 120, 0, 1, 30, 270, 974, 1452, 720, 0, 1, 42, 560, 3248, 8946, 11256, 5040, 0, 1, 56, 1036, 8792, 38338, 87504, 97296, 40320, 0, 1, 72, 1764, 20580, 129834, 463050, 920184, 930960, 362880

Offset: 0

Philippe Deléham, Jan 11 2004

Row reverse appears to be A111184. - Peter Bala, Feb 17 2017

			Triangle begins:
  1;
  0, 1;
  0, 1,  2;
  0, 1,  6,   6;
  0, 1, 12,  34,  24;
  0, 1, 20, 110, 210,  120;
  0, 1, 30, 270, 974, 1452, 720; ...

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

Cf. A084938, A111184.
Diagonals: A000007, A000012, A002378, A000142.
Row sums: A003319.

m = 10;
gf = (1/x)*(1-1/(1+Sum[Product[(1+k*y), {k, 0, n-1}]*x^n, {n, 1, m}]));
CoefficientList[#, y]& /@ CoefficientList[gf + O[x]^m, x] // Flatten (* Jean-François Alcover, May 11 2019 *)

PARI
```
T(n,k)=if(nPaul D. Hanna, Aug 16 2005
```

Sum_{k=0..n} x^(n-k)*T(n,k) = A111528(x, n); see A000142, A003319, A111529, A111530, A111531, A111532, A111533 for x = 0, 1, 2, 3, 4, 5, 6. - Philippe Deléham, Aug 09 2005
Sum_{k=0..n} T(n,k)*3^k = A107716(n). - Philippe Deléham, Aug 15 2005
Sum_{k=0..n} T(n,k)*2^k = A000698(n+1). - Philippe Deléham, Aug 15 2005
G.f.: A(x, y) = (1/x)*(1 - 1/(1 + Sum_{n>=1} [Product_{k=0..n-1}(1+k*y)]*x^n )). - Paul D. Hanna, Aug 16 2005

A111544 Triangular matrix T, read by rows, that satisfies: SHIFT_LEFT(column 0 of T^p) = p(column p+3 of T), or [T^p](m,0) = pT(p+m,p+3) for all m>=1 and p>=-3.

Original entry on oeis.org

1, 1, 1, 5, 2, 1, 33, 9, 3, 1, 261, 57, 15, 4, 1, 2361, 441, 99, 23, 5, 1, 23805, 3933, 783, 165, 33, 6, 1, 263313, 39249, 7083, 1383, 261, 45, 7, 1, 3161781, 430677, 71415, 13083, 2361, 393, 59, 8, 1, 40907241, 5137641, 789939, 136863, 23805, 3861, 567, 75, 9, 1

Offset: 0

Paul D. Hanna, Aug 07 2005

Column 0 equals A111530 (related to log of factorial series). Column 3 (A111547) equals SHIFT_LEFT(column 0 of log(T)), where the matrix logarithm, log(T), equals the integer matrix A111549.

			SHIFT_LEFT(column 0 of T^-3) = -3*(column 0 of T);
SHIFT_LEFT(column 0 of T^-2) = -2*(column 1 of T);
SHIFT_LEFT(column 0 of T^-1) = -1*(column 2 of T);
SHIFT_LEFT(column 0 of log(T)) = column 3 of T;
SHIFT_LEFT(column 0 of T^1) = 1*(column 4 of T);
where SHIFT_LEFT of column sequence shifts 1 place left.
Triangle T begins:
1;
1,1;
5,2,1;
33,9,3,1;
261,57,15,4,1;
2361,441,99,23,5,1;
23805,3933,783,165,33,6,1;
263313,39249,7083,1383,261,45,7,1;
3161781,430677,71415,13083,2361,393,59,8,1; ...
After initial term, column 2 is 3 times column 0.
Matrix inverse T^-1 = A111548 starts:
1;
-1,1;
-3,-2,1;
-15,-3,-3,1;
-99,-15,-3,-4,1;
-783,-99,-15,-3,-5,1;
-7083,-783,-99,-15,-3,-6,1; ...
where columns are all equal after initial terms;
compare columns of T^-1 to column 2 of T.
Matrix logarithm log(T) = A111549 is:
0;
1,0;
4,2,0;
23,6,3,0;
165,32,9,4,0;
1383,222,47,13,5,0;
13083,1824,321,70,18,6,0; ...
compare column 0 of log(T) to column 3 of T.

Paul Barry, A note on number triangles that are almost their own production matrix, arXiv:1804.06801 [math.CO], 2018.

Cf. A111545 (column 1), A111546 (column 2), A111547 (column 3), A111552 (row sums), A111548 (matrix inverse), A111549 (matrix log); related tables: A111528, A104980, A111536, A111553.

T[n_, k_] := T[n, k] = If[nJean-François Alcover, Aug 09 2018, from PARI *)

PARI
```
T(n,k)=if(n
				
```

T(n, k) = k*T(n, k+1) + Sum_{j=0..n-k-1} T(j+2, 2)*T(n, j+k+1) for n>k>0, with T(n, n) = 1, T(n+1, n) = n+1, T(n+3, 2) = 3*T(n+1, 0), T(n+4, 4) = T(n+1, 0), for n>=0.

A111546 Column 2 of triangle A111544.

Original entry on oeis.org

1, 3, 15, 99, 783, 7083, 71415, 789939, 9485343, 122721723, 1701224775, 25156450179, 395362560303, 6583219735563, 115817825451735, 2147443419579219, 41868118883289663, 856527397513863003, 18350158259899381095, 410942059850878349859, 9603217302778609785423

Offset: 0

Paul D. Hanna, Aug 07 2005

nonn

Also forms the columns of triangle A111548, which is the matrix inverse of triangle A111544.

Reinhard Zumkeller, Table of n, a(n) for n = 0..250
Glenn Bruda, Asymptotic expansions for the reciprocal Hardy-Littlewood logarithmic integrals, arXiv:2412.19866 [math.CO], 2024. See page 4.

Cf. A111544, A111530, A111548, A230253.

Cf. A005412.

a111546 n = a111546_list !! n
a111546_list = 1 : f 2 [1] where
   f v ws@(w:_) = y : f (v + 1) (y : ws) where
                  y = v * w + (sum $ zipWith (*) ws $ reverse ws)
-- Reinhard Zumkeller, Jan 24 2014

{a(n)=if(n<0,0,(matrix(n+3,n+3,m,j,if(m==j,1,if(m==j+1,-m+1, -(m-j-1)*polcoeff(log(sum(i=0,m,(i+2)!/2!*x^i)),m-j-1))))^-1)[n+3,3])}

a(n)=(1/2)*((n+3)!-3*(n+2)!-sum(k=0,n-2,(n+1-k)!*a(k+1))) \\ Formula by Groux Roland, implemented & checked to conform to given terms by M. F. Hasler, Dec 12 2010

G.f.: log(Sum_{n>=0} (n+2)!/2!*x^n) = Sum_{n>=1} a(n)*x^n/n. a(n) = 3*A111530(n) = -A111548(n+1, 0) for n>0.
a(n+1) = (1/2)*((n+4)!-3*(n+3)!-Sum_{k=0..n-1} (n+2-k)!*a(k+1)).
a(n+1) is the moment of order n for the measure of density: 2*x^2*exp(-x)/((x^2*exp(-x)*Ei(x)-x-1)^2+Pi^2*x^4*exp(-2*x)), on the interval 0..infinity. - Groux Roland, Dec 10 2010
a(n) = Sum_{k=0..n} A200659(n,k)*2^k. - Philippe Deléham, Nov 21 2011
G.f.: 1/(1-3x/(1-2x/(1-4x/(1-3x/(1-5x/(1-4x/(1-...(continued fraction). - Philippe Deléham, Nov 21 2011
G.f. 2 - U(0) where U(k)= 1 - x*(k+1)/(1 - x*(k+3)/U(k+1)); (continued fraction, 2-step). - Sergei N. Gladkovskii, Jun 29 2012
G.f. -1/G(0) where G(k) = x - 1 - k*x - x*(k+2)/G(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Aug 13 2012
G.f.: A(x) = 1/(G(0) - x) where G(k) = 1 + (k+1)*x - x*(k+3)/G(k+1) ; (recursively defined continued fraction). - Sergei N. Gladkovskii, Dec 26 2012
G.f.: 1/Q(0), where Q(k)= 1 - x + k*x - x*(k+2)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 03 2013
G.f.: 1/x -2 -2/(x*G(0)), where G(k)= 1 + 1/(1 - x*(k+3)/(x*(k+3) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 06 2013
G.f.: 1/x - 2 - 1/(x*W(0)), where W(k) = 1 - x*(k+3)/( x*(k+3) - 1/(1 - x*(k+1)/( x*(k+1) - 1/W(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Aug 25 2013
G.f.: W(0), where W(k) = 1 - x*(k+3)/( x*(k+3) - 1/(1 - x*(k+2)/( x*(k+2) - 1/W(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Aug 26 2013
a(n) ~ n! * n^3 / 2. - Vaclav Kotesovec, May 24 2025

A111534 Main diagonal of table A111528.

Original entry on oeis.org

1, 1, 4, 33, 416, 7045, 149472, 3804353, 112784896, 3812791581, 144643185600, 6081135558817, 280510445260800, 14080668974435141, 763890295406672896, 44529851124925034625, 2775373003913373810688, 184147301185264051623181

Offset: 0

Paul D. Hanna, Aug 06 2005

nonn

For n>0, a(n) is divisible by n: a(n)/n = A111535(n).

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

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[n, n];
Table[a[n], {n, 0, 17}] (* Jean-François Alcover, Aug 09 2018 *)

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

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

cp's OEIS Frontend

A001715 a(n) = n!/6.

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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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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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A089949 Triangle T(n,k), read by rows, given by [0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...] DELTA [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, ...] where DELTA is the operator defined in A084938.

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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.

A111544 Triangular matrix T, read by rows, that satisfies: SHIFT_LEFT(column 0 of T^p) = p(column p+3 of T), or [T^p](m,0) = pT(p+m,p+3) for all m>=1 and p>=-3.