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

0,3

Links

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.

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)

Example

(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 + ...

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 *)

Prog

(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)))}

Crossrefs

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

Cf. A077607, A089949, A260491.

Sequence in context: A112898 A368733 A253095 * A346764 A228883 A307439

Adjacent sequences: A111526 A111527 A111528 * A111530 A111531 A111532

Keyword

nonn,easy

Author

Paul D. Hanna, Aug 06 2005

Status

approved