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COMMENTS
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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
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EXAMPLE
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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)
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