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%I #21 Nov 02 2019 09:29:47
%S 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,
%T 229,229,51,7,1,1,8,73,496,1313,496,73,8,1,1,9,99,919,4581,4581,919,
%U 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
%N Array T(k,n) read by antidiagonals: expansion of exp(x+y)/(1-xy).
%C Rows are polynomials in n whose coefficients are in A099599.
%C From _Peter Bala_, Aug 19 2013: (Start)
%C 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:
%C 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) ))))).
%C 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)
%H E. W. Weisstein, <a href="http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html">Modified Bessel Function of the First Kind</a>
%F 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
%F Recurrence equation: T(n,k) = n*k*T(n-1,k-1) + 1 with boundary conditions T(n,0) = T(0,n ) = 1.
%F Main subdiagonal and main superdiagonal [1, 3, 19, 229, ...] is A228229. - _Peter Bala_, Aug 19 2013
%e 1, 1, 1, 1, 1, 1,
%e 1, 2, 3, 4, 5, 6,
%e 1, 3, 9, 19, 33, 51,
%e 1, 4, 19, 82, 229, 496,
%e 1, 5, 33, 229, 1313, 4581,
%e 1, 6, 51, 496, 4581, 32826,
%p #A099597
%p T := proc(n,k) option remember;
%p if n = 0 then 1 elif k = 0 then 1
%p else n*k*thisproc(n-1,k-1) + 1
%p fi
%p end:
%p # Diplay entries by antidiagonals
%p seq(seq(T(n-k,k), k = 0..n), n = 0..10);
%p # _Peter Bala_, Aug 19 2013
%t T[_, 0] = T[0, _] = 1;
%t T[n_, k_] := T[n, k] = n k T[n - 1, k - 1] + 1;
%t Table[T[n - k, k], {n, 0, 11}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Nov 02 2019 *)
%Y Rows include A000012, A000027, A058331. Main diagonal is A006040. Antidiagonal sums are in A099598. Cf. A099599.
%Y Cf. A088699. A228229 (main super and subdiagonal).
%K nonn,easy,tabl
%O 0,5
%A _Ralf Stephan_, Oct 28 2004
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