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A099597
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Array T(k,n) read by antidiagonals: expansion of exp(x+y)/(1-xy).
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10
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1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 9, 4, 1, 1, 5, 19, 19, 5, 1, 1, 6, 33, 82, 33, 6, 1, 1, 7, 51, 229, 229, 51, 7, 1, 1, 8, 73, 496, 1313, 496, 73, 8, 1, 1, 9, 99, 919, 4581, 4581, 919, 99, 9, 1, 1, 10, 129, 1534, 11905, 32826, 11905, 1534, 129, 10, 1, 1, 11, 163, 2377, 25733, 137431, 137431, 25733, 2377, 163, 11, 1
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OFFSET
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0,5
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COMMENTS
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Rows are polynomials in n whose coefficients are in A099599.
The k-th superdiagonal sequence of this square array occurs as the sequence of numerators in the convergents to a certain continued fraction representation of the constant BesselI(k,2), where BesselI(k,x) is a modified Bessel function of the first kind:
Let d_k(n) = T(n,n+k) = n!*(n+k)!*sum {i = 0..n} 1/(i!*(i+k)!) denote the sequence of entries on the k-th superdiagonal. It satisfies the first-order recurrence equation d_k(n) = n*(n+k)*d_k(n-1) + 1 with d_k(0) = 1 and also the second-order recurrence d_k(n) = (n*(n+k)+1)*d_k(n-1) - (n-1)(n-1+k)*d_k(n-2) with initial conditions d_k(0) = 1 and d_k(1) = k+2. This latter recurrence is also satisfied by the sequence n!*(n+k)!. From this observation we obtain the finite continued fraction expansion d_k(n) = n!*(n+k)!*(1/(k! - k!/((k+2) -(k+1)/((2*k+5) - 2*(k+2)/((3*k+10) - ... - n*(n+k)/(((n+1)*(n+k+1)+1) ))))).
Taking the limit as n - > infinity produces a continued fraction representation for the modified Bessel function value BesselI(k,2) = sum {i = 0..inf} 1/(i!*(i+k)!) = 1/(k! - k!/((k+2) -(k+1)/((2*k+5) - 2*(k+2)/((3*k+10) - ... - n*(n+k)/(((n+1)*(n+k+1)+1) - ...))))). See A070910 for the case k = 0 and A096789 for the case k = 1. (End)
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LINKS
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FORMULA
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T(n,k) = sum {i=0..min(n,k)} C(n,i)*C(k,i)*i!^2. The LDU factorization of this square array is P * D * transpose(P), where P is Pascal's triangle A007318 and D = diag(0!^2, 1!^2, 2!^2, ... ). Compare with A088699. - Peter Bala, Nov 06 2007
Recurrence equation: T(n,k) = n*k*T(n-1,k-1) + 1 with boundary conditions T(n,0) = T(0,n ) = 1.
Main subdiagonal and main superdiagonal [1, 3, 19, 229, ...] is A228229. - Peter Bala, Aug 19 2013
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EXAMPLE
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1, 1, 1, 1, 1, 1,
1, 2, 3, 4, 5, 6,
1, 3, 9, 19, 33, 51,
1, 4, 19, 82, 229, 496,
1, 5, 33, 229, 1313, 4581,
1, 6, 51, 496, 4581, 32826,
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MAPLE
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T := proc(n, k) option remember;
if n = 0 then 1 elif k = 0 then 1
else n*k*thisproc(n-1, k-1) + 1
fi
end:
# Diplay entries by antidiagonals
seq(seq(T(n-k, k), k = 0..n), n = 0..10);
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MATHEMATICA
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T[_, 0] = T[0, _] = 1;
T[n_, k_] := T[n, k] = n k T[n - 1, k - 1] + 1;
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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