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A233824
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A recurrent sequence in Panaitopol's formula for pi(x), where pi(x) is the number of primes <= x.
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5
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0, 1, 3, 13, 71, 461, 3447, 29093, 273343, 2829325, 31998903, 392743957, 5201061455, 73943424413, 1123596277863, 18176728317413, 311951144828863, 5661698774848621, 108355864447215063, 2181096921557783605
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OFFSET
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0,3
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COMMENTS
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Sum_{k=0..n} k!*a(n-k) = n*n!.
Panaitopol proved that x/pi(x) = log(x) - 1 - Sum_{k=1..m} a(k)/log(x)^k + O(1/log(x)^{m+1}) for m > 0.
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LINKS
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G. C. Greubel, Table of n, a(n) for n = 0..445
M. Hassani, Some remarks on a determinant related to the prime counting function, The 8th Seminar on Linear Algebra and its Applications, 13-14th May 2015, University of Kurdistan, Iran.
A. Ivić, Review of "A formula for pi(x) applied to a result of Koninck-Ivić" by L. Panaitopol, Zbl 0982.11003.
A. Kourbatov, Upper Bounds for Prime Gaps Related to Firoozbakht's Conjecture, J. Integer Sequences, 18 (2015), Article 15.11.2. arXiv:1506.03042
A. Kourbatov, On the geometric mean of the first n primes, arXiv:1603.00855 [math.NT], 2016.
L. Panaitopol, A formula for pi(x) applied to a result of Koninck-Ivić, Nieuw Arch. Wiskd., (5) 1, No. 1 (2000), 55-56.
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FORMULA
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a(n) = n*n! - Sum_{k=1..n-1} k!*a(n-k).
a(n) = A003319(n+1) if n > 0. (Proof. Set b(n) = A003319(n), so that b(n) = n! - Sum_{k=1..n-1} k!*b(n-k). To get b(n+1) = a(n) for n > 0, induct on n, use (n+1)! = n*n! + n!, and replace k with k+1 in the sum.)
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EXAMPLE
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0!*a(0) = a(0) = 0*0!, so a(0) = 0.
0!*a(1) + 1!*a(0) = a(1) + a(0) = 1*1!, so a(1) = 1.
0!*a(2) + 1!*a(1) + 2!*a(0) = a(2) + a(1) + 2*a(0) = 2*2!, so a(2) = 4 - 1 = 3.
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MATHEMATICA
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a[0] = 0; a[n_] := a[n] = n*n! - Sum[ k! a[n - k], {k, n - 1}]; Table[a@ n, {n, 0, 19}] (* Michael De Vlieger, Mar 26 2016 *)
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CROSSREFS
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Cf. A000720, A003319, A062049.
Sequence in context: A272428 A167894 A158882 * A003319 A192239 A192936
Adjacent sequences: A233821 A233822 A233823 * A233825 A233826 A233827
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KEYWORD
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nonn
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AUTHOR
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Jonathan Sondow, Dec 17 2013
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STATUS
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approved
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