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A111088 a(0) = a(1) = 1, a(2) = x, a(3) = 2x^2, a(n) = x*(n-1)*a(n-1) + Sum_{j=2..n-2} (j-1)*a(j)*a(n-j), n>=4 and for x = 2. 10
1, 1, 2, 8, 52, 464, 5184, 68928, 1057584, 18345536, 354570112, 7551674624, 175700025728, 4433961734656, 120642462777344, 3520972469815296, 109731998026937088, 3637456413350962176, 127800512612435896320 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,3
COMMENTS
For x = 1, this is : 1, 1, 1, 2, 7, 34, 206, 1476, 12123, ..., see A075834.
For x = 0, this is : 1, 1, 0, 0, 0, 0, 0, 0, 0, ...
For x = -1, this is : 1, 1, -1, 2, -5, 14, -42, 132, -429, ...,((-1)^(n+1)* A000108(n)).
a(n)*2^(-n) is the coefficient at the x^(n-1) term in the series reversal of the asymptotic expansion of 2 * DawsonF(sqrt(x))/sqrt(x) = sqrt(Pi) * exp(-x) * erfi(sqrt(x)) / sqrt(x) for x -> inf. - Vladimir Reshetnikov, Apr 23 2016
LINKS
FORMULA
O.g.f. A(x) satisfies:
(1) A(x) = x / Series_Reversion(x*G(x)) where G(x) = A(x*G(x)) and A(x) = G(x/A(x)) such that G(x) is the g.f. of the double factorials (A001147). - Paul D. Hanna, Jul 09 2006
(2) A(x) = Sum_{n>=0} A001147(n) * x^n / A(x)^n, where A001147(n) = (2*n)!/(n!*2^n). - Paul D. Hanna, Aug 02 2014
(3) A(x) = 1 + x * (A(x) + x*A'(x)) / (A(x) - x*A'(x)). - Paul D. Hanna, Aug 02 2014
(4) [x^(n+1)] A(x)^n = 2*n*([x^n] A(x)^n) for n>=0. - Paul D. Hanna, Aug 02 2014
a(n) ~ 2^(n+1/2) * n^n / exp(n+1/2). - Vaclav Kotesovec, Aug 02 2014
EXAMPLE
From Paul D. Hanna, Aug 02 2014: (Start)
O.g.f.: A(x) = 1 + x + 2*x^2 + 8*x^3 + 52*x^4 + 464*x^5 + 5184*x^6 +...
where A(x) = x/Series_Reversion(x + x^2 + 3*x^3 + 15*x^4 + 105*x^5 + 945*x^6 +...)
and thus
A(x) = 1 + x/A(x) + 3*x^2/A(x)^2 + 15*x^3/A(x)^3 + 105*x^4/A(x)^4 + 945*x^5/A(x)^5 +...
Illustration of the initial terms:
a(2) = 2;
a(3) = 2*2^2 = 8;
a(4) = 2*3*8 + 1*2*2 = 52;
a(5) = 2*4*52 + 1*2*8 + 2*8*2 = 464;
a(6) = 2*5*464 + 1*2*52 + 2*8*8 + 3*52*2 = 5184; ...
To illustrate formula: [x^(n+1)] A(x)^n = 2*n*([x^n] A(x)^n), form a table of coefficients of x^k in A(x)^n:
n=1: [1, 1, 2, 8, 52, 464, 5184, 68928, 1057584, ...];
n=2: [1, 2, 5, 20, 124, 1064, 11568, 150912, 2283888, ...];
n=3: [1, 3, 9, 37, 222, 1836, 19412, 248256, 3703536, ...];
n=4: [1, 4, 14, 60, 353, 2824, 29032, 363696, 5345040, ...];
n=5: [1, 5, 20, 90, 525, 4081, 40810, 500480, 7241460, ...];
n=6: [1, 6, 27, 128, 747, 5670, 55205, 662460, 9431172, ...];
n=7: [1, 7, 35, 175, 1029, 7665, 72765, 854197, 11958758, ...];
n=8: [1, 8, 44, 232, 1382, 10152, 94140, 1081080, 14876033, ...]; ...
then we can see that the diagonals are related in the following way:
[2, 20, 222, 2824, 40810, 662460, 11958758, ...]
= [2*1, 4*5, 6*37, 8*353, 10*4081, 12*55205, 14*854197, ...].
Also, the diagonal
[1, 5, 37, 353, 4081, 55205, 854197, 14876033, ...]
is the logarithmic derivative of the g.f. of the double factorials.
Further, the main diagonal in the above table equals:
[1, 2*1, 3*3, 4*15, 5*105, 6*945, 7*10395, 8*135135, ...].
(End)
MATHEMATICA
x = 2; a[0] = a[1] = 1; a[2] = x; a[3] = 2x^2; a[n_] := a[n] = x*(n - 1)*a[n - 1] + Sum[(j - 1)*a[j]*a[n - j], {j, 2, n - 2}]; Table[ a[n], {n, 0, 18}] (* Robert G. Wilson v *)
Module[{max = 20, s}, s = InverseSeries[Series[2 DawsonF[Sqrt[x]]/Sqrt[x], {x, Infinity, max + 1}][[2, 2, 2]]]; Table[SeriesCoefficient[s, n-1] 2^n, {n, 0, max}]] (* Vladimir Reshetnikov, Apr 23 2016 *)
PROG
(PARI) a(n)=Vec(x/serreverse(x*Ser(vector(n+1, k, (2*(k-1))!/(k-1)!/2^(k-1)))))[n+1] /* Paul D. Hanna, Jul 09 2006 */
(PARI) /* From o.g.f. A = 1 + x*(A + x*A')/(A - x*A'): */
{a(n)=local(A=1+x); for(i=1, n, A=1 + x*(A+x*A')/(A-x*A' +x*O(x^n))); polcoeff(A, n)}
for(n=0, 20, print1(a(n), ", ")) /* Paul D. Hanna, Aug 02 2014 */
CROSSREFS
Sequence in context: A358031 A368453 A007832 * A367371 A006351 A300697
KEYWORD
nonn
AUTHOR
Philippe Deléham, Oct 10 2005
EXTENSIONS
More terms from Robert G. Wilson v, Oct 12 2005
STATUS
approved

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Last modified April 24 19:59 EDT 2024. Contains 371963 sequences. (Running on oeis4.)