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Sum_{k = 0..n} A094816(n,k)*x^k give A000522(n), A001339(n), A082030(n), A095000(n) for x = 1, 2, 3, 4 respectively.
Comments from Peter Bala (pbala(AT)toucansurf.com), Jul 10 2008 (Start): a(n) is a difference divisibility sequence, that is, the difference a(n) - a(m) is divisible by n - m for all n and m (provided n is not equal to m). See A000522 for further properties of difference divisibility sequences.
Recurrence relation: a(0) = 1, a(1) = 6, a(n) = (n+5)*a(n-1) - (n-1)*a(n-2) for n >= 2. Let p_4(n) = n^4+2*n^3+5*n^2+1 = n^(4)-4*n^(3)+6*n^(2)-4*n^(1)+1, where n^(k) denotes the rising factorial n*(n+1)*...*(n+k-1). The polynomial p_4(n) is an example of a Poisson-Charlier polynomial c_k(x;a) at k = 4, x = -n and a = -1.
The sequence b(n) := n!*p_4(n+1) = A001688(n) satisfies the same recurrence as a(n) but with the initial conditions b(0) = 9, b(1) = 53. This leads to the finite continued fraction expansion expansion a(n)/b(n) = 1/(9-1/(6-1/(7-2/(8-...-(n-1)/(n+5))))).
Lim n -> infinity a(n)/b(n) = e/24 = 1/(9-1/(6-1/(7-2/(8-...-n/((n+6)-...))))).
a(n) = b(n) * sum {k = 0..n} 1/(k!*p_4(k)*p_4(k+1)) - since the rhs satisfies the above recurrence with the same initial conditions. Hence e = 24 * sum {k = 0..inf} 1/(k!*p_4(k)p_4(k+1)).
For sequences satisfying the more general recurrence a(n) = (n+1+r)*a(n-1) - (n-1)*a(n-2), which yield series acceleration formulas for e/r! that involve the Poisson-Charlier polynomials c_r(-n;-1), refer to A000522 (r = 0), A001339 (r=1), A082030 (r=2), A095000 (r=3). (End)
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