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A112934
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a(0) = 1; a(n+1) = Sum_{k=0..n} a(k)*A001147(n-k), where A001147 = double factorial numbers.
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26
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1, 1, 2, 6, 26, 158, 1282, 13158, 163354, 2374078, 39456386, 737125446, 15279024026, 347786765150, 8621313613954, 231139787526822, 6663177374810266, 205503866668090750, 6751565903597571842
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OFFSET
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0,3
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LINKS
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FORMULA
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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).
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) 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)
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
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)
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EXAMPLE
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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
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MAPLE
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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;
# 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
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MATHEMATICA
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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 *)
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PROG
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(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 */
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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