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A112934 a(0) = 1; a(n+1) = Sum_{k, 0<=k<=n} a(k)*A001147(n-k), where A001147 = double factorial numbers. 17
1, 1, 2, 6, 26, 158, 1282, 13158, 163354, 2374078, 39456386, 737125446, 15279024026, 347786765150, 8621313613954, 231139787526822, 6663177374810266, 205503866668090750, 6751565903597571842 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..200

FORMULA

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<=k<=n}A111106(n,k). - Philippe Deléham, Jun 20 2006

a(n) = 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

...

- Gary W. Adamson, Jul 08 2011

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),... - Gary W. Adamson, Jul 21 2016

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

G.f. 1/(G(0)-x) where G(k) = 1 - x*(k+1)/G(k+1); (continued fraction, 1-step). - Sergei N. Gladkovskii, Aug 11 2012

G.f.  1 + x/(G(0)-x) where G(k) = 1 - x*(k+1)/G(k+1); (continued fraction, 1-step). - Sergei N. Gladkovskii, Aug 12 2012

G.f.: A(x) = 1 + x/(G(0) - x) where G(k) =  1 + (2*k+1)*x - x*(2*k+2)/G(k+1); (recursively defined continued fraction). - Sergei N. Gladkovskii, Dec 26 2012.

G.f.: Q(0) where Q(k) = 1 - x*(2*k-1)/(1 - x*(2*k+2)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 20 2013

G.f.: 2/G(0), where G(k)= 1 + 1/(1 - x/(x + 1/(2*k-1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 02 2013

G.f.: 3*x - G(0), where G(k)= 3*x - 2*x*k - 1 - x*(2*k-1)/G(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Aug 14 2013

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) ))); (continued fraction). - Sergei N. Gladkovskii, Oct 22 2013

a(n) ~ n^(n-1) * 2^(n-1/2) / exp(n). - Vaclav Kotesovec, Feb 22 2014

EXAMPLE

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 to 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

MATHEMATICA

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 *)

PROG

(PARI) {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))}

(PARI) {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 */

CROSSREFS

Cf. A001147, A112935 (log derivative); A112936, A112937, A112938, A112939, A112940, A112941, A112942, A112943.

Sequence in context: A218691 A099758 A099760 * A135922 A213430 A103367

Adjacent sequences:  A112931 A112932 A112933 * A112935 A112936 A112937

KEYWORD

nonn

AUTHOR

Philippe Deléham and Paul D. Hanna, Oct 09 2005

STATUS

approved

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Last modified December 7 05:39 EST 2016. Contains 278841 sequences.