

A099597


Array T(k,n) read by antidiagonals: expansion of exp(x+y)/(1xy).


8



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

0,5


COMMENTS

Rows are polynomials in n whose coefficients are in A099599.
From Peter Bala, Aug 19 2013: (Start)
The kth 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 kth superdiagonal. It satisfies the firstorder recurrence equation d_k(n) = n*(n+k)*d_k(n1) + 1 with d_k(0) = 1 and also the secondorder recurrence d_k(n) = (n*(n+k)+1)*d_k(n1)  (n1)(n1+k)*d_k(n2) 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)


LINKS

Table of n, a(n) for n=0..77.
E. W. Weisstein, Modified Bessel Function of the First Kind


FORMULA

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(n1,k1) + 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


EXAMPLE

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,


MAPLE

#A099597
T := proc(n, k) option remember;
if n = 0 then 1 elif k = 0 then 1
else n*k*thisproc(n1, k1) + 1
fi
end:
# Diplay entries by antidiagonals
seq(seq(T(nk, k), k = 0..n), n = 0..10);
# Peter Bala, Aug 19 2013


MATHEMATICA

T[_, 0] = T[0, _] = 1;
T[n_, k_] := T[n, k] = n k T[n  1, k  1] + 1;
Table[T[n  k, k], {n, 0, 11}, {k, 0, n}] // Flatten (* JeanFrançois Alcover, Nov 02 2019 *)


CROSSREFS

Rows include A000012, A000027, A058331. Main diagonal is A006040. Antidiagonal sums are in A099598. Cf. A099599.
Cf. A088699. A228229 (main super and subdiagonal).
Sequence in context: A156354 A295205 A297020 * A283113 A123610 A209631
Adjacent sequences: A099594 A099595 A099596 * A099598 A099599 A099600


KEYWORD

nonn,easy,tabl


AUTHOR

Ralf Stephan, Oct 28 2004


STATUS

approved



