A111106 -id:A111106 - cp's OEIS frontend

A112934 a(0) = 1; a(n+1) = Sum_{k=0..n} a(k)*A001147(n-k), where A001147 = double factorial numbers.

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

A111146 Triangle T(n,k), read by rows, given by [0, 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, ...] DELTA [1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 0, 2, 0, 0, 1, 4, 0, 0, 2, 5, 8, 0, 0, 6, 15, 17, 16, 0, 0, 24, 62, 68, 49, 32, 0, 0, 120, 322, 359, 243, 129, 64, 0, 0, 720, 2004, 2308, 1553, 756, 321, 128, 0, 0, 5040, 14508, 17332, 11903, 5622, 2151, 769, 256, 0, 0, 40320, 119664

Offset: 0

Philippe Deléham, Oct 19 2005

Let R(m,n,k), 0<=k<=n, the Riordan array (1, x*g(x)) where g(x) is g.f. of the m-fold factorials . Then Sum_{k, 0<=k<=n} = R(m,n,k) = Sum_{k, 0<=k<=n} T(n,k)*m^(n-k).
For m = -1, R(-1,n,k) is A026729(n,k).
For m = 0, R(0,n,k) is A097805(n,k).
For m = 1, R(1,n,k) is A084938(n,k).
For m = 2, R(2,n,k) is A111106(n,k).

			Triangle begins:
.1;
.0, 1;
.0, 0, 2;
.0, 0, 1, 4;
.0, 0, 2, 5, 8;
.0, 0, 6, 15, 17, 16;
.0, 0, 24, 62, 68, 49, 32;
.0, 0, 120, 322, 359, 243, 129, 64;
.0, 0, 720, 2004, 2308, 1553, 756, 321, 128;
.0, 0, 5040, 14508, 17332, 11903, 5622, 2151, 769, 256;
.0, 0, 40320, 119664, 148232, 105048, 49840, 18066, 5756, 1793, 512;
....................................................................
At y=2: Sum_{k=0..n} 2^k*T(n,k) = A113327(n) where (1 + 2*x + 8*x^2 + 36*x^3 +...+ A113327(n)*x^n +..) = 1/(1 - 2/1!*x*(1! + 2!*x + 3!*x^2 + 4!*x^3 +..) ).
At y=3: Sum_{k=0..n} 3^k*T(n,k) = A113328(n) where (1 + 3*x + 18*x^2 + 117*x^3 +...+ A113328(n)*x^n +..) = 1/(1 - 3/2!*x*(2! + 3!*x + 4!*x^2 + 5!*x^3 +..) ).
At y=4: Sum_{k=0..n} 4^k*T(n,k) = A113329(n) where (1 + 4*x + 32*x^2 + 272*x^3 +...+ A113329(n)*x^n +..) = 1/(1 - 4/3!*x*(3! + 4!*x + 5!*x^2 + 6!*x^3 +..) ).

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

Cf. A000045, A026729, A051295, A084938, A097805, A111106, A112934.
Cf. m-fold factorials : A000142, A001147, A007559, A007696, A008548, A008542.
Cf. A113326, A113327 (y=2), A113328 (y=3), A113329 (y=4), A113330 (y=5), A113331 (y=6).

T[n_, k_] := Module[{x = X + X*O[X]^n, y = Y + Y*O[Y]^k}, A = 1/(1 - x*y*Sum[x^j*Product[y + i, {i, 0, j - 1}], {j, 0, n}]); Coefficient[ Coefficient[A, X, n], Y, k]];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 26 2019, from PARI *)

{T(n,k)=local(x=X+X*O(X^n),y=Y+Y*O(Y^k)); A=1/(1-x*y*sum(j=0,n,x^j*prod(i=0,j-1,y+i))); return(polcoeff(polcoeff(A,n,X),k,Y))} (Hanna)

Sum_{k, 0<=k<=n} (-1)^(n-k)*T(n, k) = A000045(n+1), Fibonacci numbers.
Sum_{k, 0<=k<=n} T(n, k) = A051295(n).
Sum_{k, 0<=k<=n} 2^(n-k)*T(n, k) = A112934(n).
T(0, 0) = 1, T(n, n) = 2^(n-1).
G.f.: A(x, y) = 1/(1 - x*y*Sum_{j>=0} (y-1+j)!/(y-1)!*x^j ). - Paul D. Hanna, Oct 26 2005

A113129 Triangle T(n,k), 0<=k<=n, of coefficients of polynomials P_n(x) related to convolution of the k-fold factorials.

Original entry on oeis.org

1, 0, 1, 0, 0, 2, 0, 0, 1, 6, 0, 0, 0, 10, 24, 0, 0, 0, 4, 82, 120, 0, 0, 0, 0, 84, 672, 720, 0, 0, 0, 0, 27, 1236, 5820, 5040, 0, 0, 0, 0, 0, 930, 16328, 54288, 40320, 0, 0, 0, 0, 0, 248, 20850, 211080, 548496, 362880, 0, 0, 0, 0, 0, 0, 12452, 396528, 2775432

Offset: 0

Philippe Deléham and Paul D. Hanna, Oct 28 2005

Let R(m,n,k), 0<=k<=n, the Riordan array (1,x*g(x)) where g(x) is g.f. of the m-fold factorials . Then R(m,n,k) = R(m,n-1,k-1) + Sum_{j, 0<=j<=n-1-k} R(m,n-1,k+j)*P_m(j), R(m,n,0) = 0^n and R(m,0,k) = 0 if k>n.

			Triangle begins:
.1;
.0, 1;
.0, 0, 2;
.0, 0, 1, 6;
.0, 0, 0, 10, 24;
.0, 0, 0, 4, 82, 120;
.0, 0, 0, 0, 84, 672, 720;
.0, 0, 0, 0, 27, 1236, 5820, 5040;
.0, 0, 0, 0, 0, 930, 16328, 54288, 40320;
.0, 0, 0, 0, 0, 248, 20850, 211080, 548496, 362880;
.0, 0, 0, 0, 0, 0, 12452, 396528, 2775432, 6003360, 362880;
.0, 0, 0, 0, 0, 0, 2830, 38732, 7057308, 37831752, 71019360, 39916800;

Cf. A000007, A000108, A000142, A000699.

R(m, n, k) : A097805 (m=0), A084938 (m=1), A111106 (m=2), A113333 (column sums).

P_0(x) = 1, P_1(x) = x, P_2(x) = 2*x^2, P_ n(x) = n*x*P_(n-1)(x) + Sum_{j, 1<=j<=n-1} j*P_j(x)*P_(n-1-j)(x).
P_n(x) = Sum_{k, 0<=k<=n} T(n, k)*x^k.
P_n(0) = A000007(n).
P_n(x) = A075834(n+1), A111088(n+1), A113130(n+1), A113131(n+1), A113132(n+1), A113133(n+1), A113134(n+1), A113135(n+1) for x = 1, 2, 3, 4, 5, 6, 7, 8 respectively.
P_n(-1) = (-1)^n*A000108(n), signed Catalan numbers.
T(n, n) = n! = A000142(n).
T(2*n+1, n+1) = A000699(n+1) (number of irreducible diagrams with 2n+2 nodes).
T(2*n+2, n+2) = A113332(n) = A000699(n+2)*(2*n+3)*(n+2)/(3*(n+1)).

Corrected by Philippe Deléham, Dec 18 2008

A168441 Expansion of 1/(1-x/(1-2x/(1-4x/(1-6x/(1-8x/(1-.... (continued fraction).

Original entry on oeis.org

1, 1, 3, 17, 155, 2025, 34819, 743329, 18937707, 560071193, 18844479635, 710440531665, 29654234779771, 1357326276747721, 67589738142784803, 3637403230889380097, 210358430818676801675, 13009719599952748481145

Offset: 0

Paul Barry, Nov 25 2009

Hankel transform is A168442.

Cf. A004277, A111106, A168442.

nmax = 20; CoefficientList[1 + x*Series[1/(1 - x + ContinuedFractionK[-2*k*x, 1, {k, 1, nmax}]), {x, 0, nmax}], x] (* Vaclav Kotesovec, Jan 23 2024 *)

G.f.: 1/(1-x-2x^2/(1-6x-24x^2/(1-14x-80x^2/(1-22x-168x^2/(1-30x-288x^2/(1-... (continued fraction).
a(n) = Sum_{k=0..n} A111106(n,k)*2^(n-k). - Philippe Deléham, Nov 28 2009
a(n) = upper left term of M^n, M = an infinite square production matrix as follows:
  1, 1, 0, 0, 0, 0, ...
  2, 2, 2, 0, 0, 0, ...
  4, 4, 4, 4, 0, 0, ...
  6, 6, 6, 6, 6, 0, ...
  8, 8, 8, 8, 8, 8, ...
  ...
(where the series (1,2,4,6,8,...) = A004277, positive even integers prefaced with a 1). - Gary W. Adamson, Jul 19 2011
G.f. 1 + x/(G(0)-x) where G(k) = 1 - x*(2*k+2)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 28 2012
a(n) ~ 2^(2*n - 3/2) * n^(n-1) / exp(n). - Vaclav Kotesovec, Jan 23 2024

cp's OEIS Frontend

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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A111146 Triangle T(n,k), read by rows, given by [0, 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, ...] DELTA [1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is the operator defined in A084938.

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A113129 Triangle T(n,k), 0<=k<=n, of coefficients of polynomials P_n(x) related to convolution of the k-fold factorials.

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A168441 Expansion of 1/(1-x/(1-2x/(1-4x/(1-6x/(1-8x/(1-.... (continued fraction).

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