A111528 -id:A111528 - cp's OEIS frontend

A111529 Row 2 of table A111528.

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)

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 ).

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

Original entry on oeis.org

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

A111553 Triangular matrix T, read by rows, that satisfies: SHIFT_LEFT(column 0 of T^p) = p(column p+4 of T), or [T^p](m,0) = pT(p+m,p+4) for all m>=1 and p>=-4.

Original entry on oeis.org

1, 1, 1, 6, 2, 1, 46, 10, 3, 1, 416, 72, 16, 4, 1, 4256, 632, 116, 24, 5, 1, 48096, 6352, 1016, 184, 34, 6, 1, 591536, 70912, 10176, 1664, 282, 46, 7, 1, 7840576, 864192, 113216, 17024, 2696, 416, 60, 8, 1, 111226816, 11371072, 1375456, 192384, 28792, 4256, 592, 76, 9, 1

Offset: 0

Paul D. Hanna, Aug 07 2005

Column 0 equals A111531 (related to log of factorial series). Column 4 (A111557) equals SHIFT_LEFT(column 0 of log(T)), where the matrix logarithm, log(T), equals the integer matrix A111560.

			SHIFT_LEFT(column 0 of T^-4) = -4*(column 0 of T);
SHIFT_LEFT(column 0 of T^-3) = -3*(column 1 of T);
SHIFT_LEFT(column 0 of T^-2) = -2*(column 2 of T);
SHIFT_LEFT(column 0 of T^-1) = -1*(column 3 of T);
SHIFT_LEFT(column 0 of log(T)) = column 4 of T;
SHIFT_LEFT(column 0 of T^1) = 1*(column 5 of T);
where SHIFT_LEFT of column sequence shifts 1 place left.
Triangle T begins:
1;
1,1;
6,2,1;
46,10,3,1;
416,72,16,4,1;
4256,632,116,24,5,1;
48096,6352,1016,184,34,6,1;
591536,70912,10176,1664,282,46,7,1;
7840576,864192,113216,17024,2696,416,60,8,1; ...
After initial term, column 3 is 4 times column 0.
Matrix inverse T^-1 = A111559 starts:
1;
-1,1;
-4,-2,1;
-24,-4,-3,1;
-184,-24,-4,-4,1;
-1664,-184,-24,-4,-5,1;
-17024,-1664,-184,-24,-4,-6,1; ...
where columns are all equal after initial terms;
compare columns of T^-1 to column 3 of T.
Matrix logarithm log(T) = A111560 is:
0;
1,0;
5,2,0;
34,7,3,0;
282,44,10,4,0;
2696,354,60,14,5,0;
28792,3328,470,84,19,6,0; ...
compare column 0 of log(T) to column 4 of T.

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

Cf. A111531 (column 0), A111554 (column 1), A111555 (column 2), A111556 (column 3), A111557 (column 4), A111558 (row sums), A111559 (matrix inverse), A111560 (matrix log); related tables: A111528, A104980, A111536, A111544.

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+3, 3)*T(n, j+k+1) for n>k>0, with T(n, n) = 1, T(n+1, n) = n+1, T(n+4, 3) = 4*T(n+1, 0), T(n+5, 5) = T(n+1, 0), for n>=0.

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.

cp's OEIS Frontend

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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A111534 Main diagonal of table A111528.

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

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A111553 Triangular matrix T, read by rows, that satisfies: SHIFT_LEFT(column 0 of T^p) = p*(column p+4 of T), or [T^p](m,0) = p*T(p+m,p+4) for all m>=1 and p>=-4.

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A111553 Triangular matrix T, read by rows, that satisfies: SHIFT_LEFT(column 0 of T^p) = p(column p+4 of T), or [T^p](m,0) = pT(p+m,p+4) for all m>=1 and p>=-4.

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.