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A141827
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a(n) = (n^3*a(n-1) - 1)/(n - 1) for n >= 2, with a(0) = 1, a(1) = 4.
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2
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1, 4, 31, 418, 8917, 278656, 12037939, 688168846, 50334635593, 4586743668412, 509638185379111, 67832842473959674, 10655922890454756061, 1950921882527424922168, 411794588127327229725307
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OFFSET
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0,2
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COMMENTS
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For related recurrences of the form a(n) = (n^k*a(n-1)-1)/(n-1) see A001339, A007808 (both k = 2) and A141828 (k = 4). 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.
For k = 1,2,3,... and x in Z, the recurrence equation a(n) = (n^k*a(n-1) - 1)/(n - 1) with starting value a(1) = x produces an integer sequence. This is because the sequence also satisfies the second-order recurrence a(n) = (1 + (n^k - 1)/(n - 1))*a(n-1) - (n - 1)^(k-1)*a(n-2) with integer starting values a(1) = x, a(2) = x*2^k - 1. Here we take k = 3 and x = 4.
The solution to the recurrence is a(n) = n*n!^(k-1)*( x - Sum_{i = 2..n} 1/(i*(i-1)*i!^(k-1)) ). Hence limit (n -> inf) a(n)/(n*n!^(k-1)) equals the constant x - Sum_{i = 2..inf} 1/(i*(i-1)*i!^(k-1)). Note that the sequence b(n) := n*n!^(k-1) satisfies the same second-order recurrence but with starting values b(1) = 1, b(2) = 2^k. From this observation one can get a generalized continued fraction expansion for a(n)/b(n) and hence, by going to the limit, for the constant x - Sum_{i = 2..n} 1/(i*(i-1)*i!^(k-1)). See, for example, A130820. (End)
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LINKS
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FORMULA
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Sum_{n = 0..inf} a(n)*x^n/n!^2 = 1/(1 - x)^2*Sum_{n = 0..inf} (n + 1)*x^n/n!^2.
a(n) = n!^2*Sum_{k = 0..n} (n - k + 1)(k + 1)/k!^2.
a(n) := n*n!^2*(4 - Sum_{k = 2..n} 1/(k!^2*k*(k - 1)).
Congruence property: a(n) == (1 + n + n^2) (mod n^3).
The recurrence a(n) = (n^2 + n + 2)*a(n-1) - (n - 1)^2*a(n-2), n >= 2, shows that a(n) is always a positive integer. The sequence b(n) := n*n!^2 also satisfies the same recurrence with b(0) = 0, b(1) = 1. Hence we obtain the finite continued fraction expansion a(n)/(n*n!^2) = 4 - 1^2/(8 - 2^2/(14 - 3^2/(22 -...-(n - 1)^2/(n^2 + n + 2)))), for n > 1. a(n)*b(n+1) - b(n)*a(n+1) = n!^2.
Limit_{n -> infinity} a(n)/(n*n!^2) = Sum_{n = 0..inf} (n + 1)/n!^2 = BesselI(0,2) + BesselI(1,2) = 3.87022 21569 ..., using the values of the modified Bessel function, BesselI(0, 2) = 2.27958 53023 ... and BesselI(1, 2) = 1.59063 68546 ... (see A070910 and A096789; Cf. A130820). This yields the continued fraction expansion BesselI(0,2) + BesselI(1,2) = 4 - 1^2/(8 - 2^2/(14 - 3^2/(22 -...-(n - 1)^2/(n^2 + n + 2 - ... )))).
Limit_{n -> infinity} a(n)/(n*n!^2) = Sum_{n = 1..inf} (n + n^2)/n!^2 = Sum_{n = 1..inf} n^3/n!^2 = 1/2 * Sum_{n = 1..inf} n^4/n!^2.
Limit_{n -> infinity} a(n)/(n*n!^2) = Sum_{n = 0..inf} A001405 (n)/n!.
Limit_{n -> infinity} a(n)/(n*n!^2) = 1 + Sum_{n = 0..inf} 1/(Product_{k = 0..n} (A008619(k)).
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MAPLE
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a(n) := n -> n!^2*sum((n-k+1)*(k+1)/k!^2, k = 0..n): seq(a(n), n = 0..16);
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MATHEMATICA
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a[0] = 1; a[1] = 4; a[n_] := a[n] = (n^3 a[n - 1] - 1)/(n - 1); Table[a@ n, {n, 0, 14}] (* or *)
Table[n!^2 Sum[(n - k + 1) (k + 1)/k!^2, {k, 0, n}], {n, 0, 14}] (* or *)
Table[n n!^2 (4 - Sum[ 1/(k!^2*k*(k - 1)), {k, 2, n}]), {n, 0, 14}] /. 0 -> 1 (* Michael De Vlieger, Jul 03 2016 *)
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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