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%I #6 Mar 12 2017 15:39:22
%S 1,5,79,3199,272981,42653281,11055730435,4424134795739,
%T 2588750874763849,2123099311165701661,2358999234628557401111,
%U 3453810779419670890966615,6510747302004208690462157149,15496121141045183700690805861049
%N a(n) = (n^4*a(n-1)-1)/(n-1) for n >= 2, with a(0) = 1, a(1) = 5.
%C For related recurrences of the form a(n) = (n^k*a(n-1)-1)/(n-1) see A001339, A007808 (both k = 2) and A141827 (k = 3). 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.
%H Harvey P. Dale, <a href="/A141828/b141828.txt">Table of n, a(n) for n = 0..181</a>
%F Sum {n = 0..inf} a(n)*x^n/n!^3 = 1/(1-x)^2*sum {n = 0..inf} (n^2+n+1)*x^n/n!^3.
%F a(n) = n!^3*sum {k = 0..n} (n-k+1)(k^2+k+1)/k!^3.
%F a(n) := n* n!^3*(5 - sum{k = 2..n} 1/(k!^2*k*(k-1)).
%F Congruence property: a(n) == (1+n+n^2+n^3) (mod n^4).
%F The recurrence a(n) = (n^3+n^2+n+2)*a(n-1) - (n-1)^3*a(n-2), n >= 2, shows that a(n) is always a positive integer. The sequence b(n) := n*n!^3 also satisfies the same recurrence with b(0) = 0, b(1) = 1. Hence we obtain the finite continued fraction expansion a(n)/(n*n!^3) = 5 - 1^3/(16 - 2^3/(41 - 3^3/(86 -...-(n-1)^3/(n^3+n^2+n+2)))), for n >= 1. a(n)*b(n+1) - b(n)*a(n+1) = n!^3.
%F Lim n -> infinity a(n)/(n*n!^3) = sum {n = 0..inf} (n^2+n+1)/n!^3 = 4.93672 23378... .
%F Lim n -> infinity a(n)/(n*n!^3) = 1 + sum {n = 0..inf} 1/(prod{k = 0..n}(A008620 (k)).
%p a := n -> n!^3*add((n-k+1)*(k^2+k+1)/k!^3, k = 0..n): seq(a(n), n = 0..16);
%t nxt[{n_,a_}]:={n+1,((n+1)^4*a-1)/n}; Join[{1},NestList[nxt,{1,5},15][[All,2]]] (* _Harvey P. Dale_, Mar 12 2017 *)
%Y Cf. A001339, A007808, A141827.
%K easy,nonn
%O 0,2
%A _Peter Bala_, Jul 09 2008
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