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A177885
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a(n) = (1-n)^(n-1).
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17
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1, 1, -1, 4, -27, 256, -3125, 46656, -823543, 16777216, -387420489, 10000000000, -285311670611, 8916100448256, -302875106592253, 11112006825558016, -437893890380859375, 18446744073709551616, -827240261886336764177
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OFFSET
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0,4
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COMMENTS
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LeClair gives an approximation z(n) for the location of the n-th nontrivial zero of the Riemann zeta function on the critical line, which can be expressed in terms of the exponential generating function of this sequence A(x) = x/LambertW(x) as follows: z(n) = 1/2 + 2*Pi*exp(1)*A((n - 11/8)/exp(1))*i. For example, working to 1 decimal place, z(1) = 1/2 + 14.5*i (the first nontrivial zero is at 1/2 + 14.1*i), z(10) = 1/2 + 50.2*i (the tenth nontrivial zero is at 1/2 + 49.8*i) and z(100) = 1/2 + 236*i (the hundredth nontrivial zero is at 1/2 + 236.5*i). [Peter Bala, Jun 12 2013]
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LINKS
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FORMULA
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E.g.f. satisfies A(x) = exp(x/A(x)).
E.g.f. A(x) = x/LambertW(x) = exp(LambertW(x)) = 1 + x - x^2/2! + 4*x^3/3! - 27*x^4/4! + .... - Peter Bala, Jun 12 2013
E.g.f.: 1 + Series_Reversion( (1+x)*log(1+x) ). - Paul D. Hanna, Aug 24 2016
E.g.f.: 1 + Series_Reversion( x + Sum_{n>=2} (-x)^n/(n*(n-1)) ). - Paul D. Hanna, Aug 24 2016
a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n-1, k-1)*n^(n-k), for n >= 1 and a(0) = 1, that is, Sum_{k=0..n}*A137452(n, k), for n >= 0. - Wolfdieter Lang, Apr 11 2023
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EXAMPLE
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E.g.f.: A(x) = 1 + x - x^2/2! + 4*x^3/3! - 27*x^4/4! + 256*x^5/5! - 3125*x^6/6! + 46656*x^7/7! - 823543*x^8/8! +...+ (1-n)^(n-1)*x^n/n! +...
Related series.
Series_Reversion(A(x) - 1) = x + x^2/2 - x^3/6 + x^4/12 - x^5/20 + x^6/30 - x^7/42 + x^8/56 - x^9/72 + x^10/90 +...+ (-x)^n/(n*(n-1)) +... (End)
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MATHEMATICA
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Join[{1, 1}, Table[(1-n)^(n-1), {n, 2, 20}]] (* Harvey P. Dale, Aug 10 2012 *)
nn = 18; Range[0, nn]! CoefficientList[ Series[ Exp[ ProductLog[ x]], {x, 0, nn}], x] (* Robert G. Wilson v, Aug 23 2012 *)
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PROG
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(PARI) {a(n) = my(A = 1 + serreverse( x + sum(m=2, n+2, (-x)^m/(m*(m-1)) +x^2*O(x^n)))); n!*polcoeff(A, n)}
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CROSSREFS
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KEYWORD
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sign,easy
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AUTHOR
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STATUS
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approved
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