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A352284
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a(n) is 2^(2*n) times the derivative of order 2*n of the logarithm of I_0(x) (the modified Bessel function of the first kind of order zero) evaluated at zero.
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2
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0, 2, -6, 80, -2310, 114912, -8741040, 943113600, -136984998150, 25771171025600, -6096112602836256, 1770904261963549440, -619781244413394663600, 257206839532730597664000, -124885627021632853758096000, 70139282938382254349182510080, -45116676897361263845107791884550
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OFFSET
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0,2
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COMMENTS
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The sequence results from some interesting combinatorics that come from the interaction of
1) the quotient rule,
2) the derivative recurrence relation for modified Bessel functions, and
3) the fact that I_n(0) = 0 unless n = 0 and I_0(0) = 1.
At each stage the numerator is the sum of products of Bessel functions. Each derivative introduces new terms which also (by the Bessel derivative recurrence relation) increment or decrement the Bessel function orders in the terms of the numerator. a(n) is the coefficient in front of the term (I_0(x))^(2^n) (if it exists). There are complicated combinatorics involved in what contributes to that coefficient.
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LINKS
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FORMULA
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a(n) = 2^(2n)*(d^(2n)/dx^(2n))(log(I_0(x)))|_x=0.
An expression which does not require derivatives is:
a(n) = 2^(2n)*Sum_{k=1..2n} (-1)^(k-1)(k-1)!*Y_{2n,k}(d_i), where Y_{n,k} is the partial Bell polynomial with inputs given by d_{2i} = 2^(-2i)*binomial(2i,i) and d_{2i+1} = 0.
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EXAMPLE
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For n = 0:
1*(d^0/dx^0)log(I_0(x)) = log(I_0(0)) = 0.
For n = 1:
4*(d^2/dx^2)log(I_0(x)) = (I_{-2}I_0 + 2I_0^2 + I_2I_0 - I_{-1}I_{-1} -
2I_{-1}I_1 + I_1 I_1)/(I_0)^4 which evaluates to 2 (because of the 2I_0^2).
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MAPLE
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A := proc(n) option remember; ifelse(n=2, (-BesselI(1, x)^2*x + BesselI(0, x)^2*x - BesselI(1, x)* BesselI(0, x))/(BesselI(0, x)^2*x), simplify(diff(A(n-1), x))) end:
a := n -> ifelse(n=0, 0, 2^n*limit(A(n), x=0)):
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MATHEMATICA
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Table[2^n*Sum[(-1)^(k-1)*Factorial[k-1]*BellY[n, k, Flatten[Table[{0, 2^{-(j + 1)}* Binomial[j+1, (j+1)/2]}, {j, 1, n-k+1, 2}]]], {k, 1, n}], {n, 0, 32, 2}]
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CROSSREFS
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KEYWORD
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sign,easy
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AUTHOR
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STATUS
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approved
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