cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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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.

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

Views

Author

Paul D. Hanna, Aug 06 2005

Keywords

Examples

			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)).
		

Crossrefs

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.

Programs

  • Maple
    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
  • Mathematica
    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 *)
  • PARI
    {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(""))
    
  • PARI
    {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(""))

Formula

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

Views

Author

Paul D. Hanna, Aug 06 2005

Keywords

Examples

			(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 + ...
		

Crossrefs

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

Programs

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

Formula

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

Views

Author

Paul D. Hanna, Aug 06 2005

Keywords

Examples

			(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 + ...
		

Crossrefs

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

Programs

  • Mathematica
    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 *)
  • PARI
    {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

Formula

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)

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

Views

Author

Paul D. Hanna, Aug 06 2005

Keywords

Examples

			(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 + ...
		

Crossrefs

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

Programs

  • Mathematica
    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 *)
  • PARI
    {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

Formula

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

Views

Author

Paul D. Hanna, Aug 06 2005

Keywords

Examples

			(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 + ...
		

Crossrefs

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

Programs

  • Mathematica
    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 *)
  • PARI
    {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

Formula

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)

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.

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

Views

Author

Paul D. Hanna, Aug 07 2005

Keywords

Comments

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.

Examples

			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.
		

Crossrefs

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.

Programs

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

Formula

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

Views

Author

Philippe Deléham, Jan 11 2004

Keywords

Comments

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

Examples

			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; ...
		

Crossrefs

Row sums: A003319.

Programs

  • Mathematica
    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

Formula

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

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

Views

Author

Paul D. Hanna, Aug 06 2005

Keywords

Comments

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

Crossrefs

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

Programs

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

Formula

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

A111556 Column 3 of triangle A111553.

Original entry on oeis.org

1, 4, 24, 184, 1664, 17024, 192384, 2366144, 31362304, 444907264, 6720628224, 107674883584, 1823884857344, 32575705493504, 612054254936064, 12071987619713024, 249477777420304384, 5392386599983366144
Offset: 0

Views

Author

Paul D. Hanna, Aug 07 2005

Keywords

Comments

Also found in column 0 of triangle A111559, which equals the matrix inverse of A111553.

Crossrefs

Programs

  • PARI
    {a(n)=if(n<0,0,(matrix(n+4,n+4,m,j,if(m==j,1,if(m==j+1,-m+1, -(m-j-1)*polcoeff(log(sum(i=0,m,(i+3)!/3!*x^i)),m-j-1))))^-1)[n+4,4])}

Formula

G.f.: log(Sum_{n>=0} ((n+3)!/3!)*x^n) = Sum_{n>=1} a(n)*x^n/n.
a(n) = 4*A111531(n) for n>0.
From Groux Roland, Dec 10 2010: (Start)
6*a(n+1) = (n+5)! - 4*(n+4)! - Sum_{k=0..n-1} (n-k+3)!*a(k+1).
a(n+1) is the moment of order n for the density 6*x^3*exp(-x)/( (x^3*exp(-x)*Ei(x)-x^2-x-2)^2 + Pi^2*x^6*exp(-2*x) ) over the interval 0..infinity. (End)
a(n) = Sum_{k=0..n} A200659(n,k)*3^k. - Philippe Deléham, Nov 21 2011
G.f.: 1/(1-4x/(1-2x/(1-5x/(1-3x/(1-6x/(1-4x/(1-...(continued fraction). - Philippe Deléham, Nov 21 2011
G.f.: 1/Q(0), where Q(k) = 1 - 2*x + k*x - x*(k+2)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 03 2013
G.f.: 1/x - 3 - 2/(x*G(0)), where G(k) = 1 + 1/(1 - x*(k+4)/(x*(k+4) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 06 2013
G.f.: W(0), 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

A172455 The case S(6,-4,-1) of the family of self-convolutive recurrences studied by Martin and Kearney.

Original entry on oeis.org

1, 7, 84, 1463, 33936, 990542, 34938624, 1445713003, 68639375616, 3676366634402, 219208706540544, 14397191399702118, 1032543050697424896, 80280469685284582812, 6725557192852592984064, 603931579625379293509683
Offset: 1

Views

Author

N. J. A. Sloane, Nov 20 2010

Keywords

Examples

			G.f. = x + 7*x^2 + 84*x^3 + 1463*x^4 + 33936*x^5 + 990542*x^6 + 34938624*x^7 + ...
a(2) = 7 since (6*2 - 4) * a(2-1) - (a(1) * a(2-1)) = 7.
		

Crossrefs

Cf. A000079 S(1,1,-1), A000108 S(0,0,1), A000142 S(1,-1,0), A000244 S(2,1,-2), A000351 S(4,1,-4), A000400 S(5,1,-5), A000420 S(6,1,-6), A000698 S(2,-3,1), A001710 S(1,1,0), A001715 S(1,2,0), A001720 S(1,3,0), A001725 S(1,4,0), A001730 S(1,5,0), A003319 S(1,-2,1), A005411 S(2,-4,1), A005412 S(2,-2,1), A006012 S(-1,2,2), A006318 S(0,1,1), A047891 S(0,2,1), A049388 S(1,6,0), A051604 S(3,1,0), A051605 S(3,2,0), A051606 S(3,3,0), A051607 S(3,4,0), A051608 S(3,5,0), A051609 S(3,6,0), A051617 S(4,1,0), A051618 S(4,2,0), A051619 S(4,3,0), A051620 S(4,4,0), A051621 S(4,5,0), A051622 S(4,6,0), A051687 S(5,1,0), A051688 S(5,2,0), A051689 S(5,3,0), A051690 S(5,4,0), A051691 S(5,5,0), A053100 S(6,1,0), A053101 S(6,2,0), A053102 S(6,3,0), A053103 S(6,4,0), A053104 S(7,1,0), A053105 S(7,2,0), A053106 S(7,3,0), A062980 S(6,-8,1), A082298 S(0,3,1), A082301 S(0,4,1), A082302 S(0,5,1), A082305 S(0,6,1), A082366 S(0,7,1), A082367 S(0,8,1), A105523 S(0,-2,1), A107716 S(3,-4,1), A111529 S(1,-3,2), A111530 S(1,-4,3), A111531 S(1,-5,4), A111532 S(1,-6,5), A111533 S(1,-7,6), A111546 S(1,0,1), A111556 S(1,1,1), A143749 S(0,10,1), A146559 S(1,1,-2), A167872 S(2,-3,2), A172450 S(2,0,-1), A172485 S(-1,-2,3), A177354 S(1,2,1), A292186 S(4,-6,1), A292187 S(3, -5, 1).

Programs

  • Mathematica
    a[1] = 1; a[n_]:= a[n] = (6*n-4)*a[n-1] - Sum[a[k]*a[n-k], {k, 1, n-1}]; Table[a[n], {n, 1, 20}] (* Vaclav Kotesovec, Jan 19 2015 *)
  • PARI
    {a(n) = local(A); if( n<1, 0, A = vector(n); A[1] = 1; for( k=2, n, A[k] = (6 * k - 4) * A[k-1] - sum( j=1, k-1, A[j] * A[k-j])); A[n])} /* Michael Somos, Jul 24 2011 */
    
  • PARI
    S(v1, v2, v3, N=16) = {
      my(a = vector(N)); a[1] = 1;
      for (n = 2, N, a[n] = (v1*n+v2)*a[n-1] + v3*sum(j=1,n-1,a[j]*a[n-j])); a;
    };
    S(6,-4,-1)
    \\ test: y = x*Ser(S(6,-4,-1,201)); 6*x^2*y' == y^2 - (2*x-1)*y - x
    \\ Gheorghe Coserea, May 12 2017

Formula

a(n) = (6*n - 4) * a(n-1) - Sum_{k=1..n-1} a(k) * a(n-k) if n>1. - Michael Somos, Jul 24 2011
G.f.: x / (1 - 7*x / (1 - 5*x / (1 - 13*x / (1 - 11*x / (1 - 19*x / (1 - 17*x / ... )))))). - Michael Somos, Jan 03 2013
a(n) = 3/(2*Pi^2)*int((4*x)^((3*n-1)/2)/(Ai'(x)^2+Bi'(x)^2), x=0..inf), where Ai'(x), Bi'(x) are the derivatives of the Airy functions. [Vladimir Reshetnikov, Sep 24 2013]
a(n) ~ 6^n * (n-1)! / (2*Pi) [Martin + Kearney, 2011, p.16]. - Vaclav Kotesovec, Jan 19 2015
6*x^2*y' = y^2 - (2*x-1)*y - x, where y(x) = Sum_{n>=1} a(n)*x^n. - Gheorghe Coserea, May 12 2017
G.f.: x/(1 - 2*x - 5*x/(1 - 7*x/(1 - 11*x/(1 - 13*x/(1 - ... - (6*n - 1)*x/(1 - (6*n + 1)*x/(1 - .... Cf. A062980. - Peter Bala, May 21 2017
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