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%I #43 Dec 30 2021 00:24:16
%S 0,1,3,13,71,461,3447,29093,273343,2829325,31998903,392743957,
%T 5201061455,73943424413,1123596277863,18176728317413,311951144828863,
%U 5661698774848621,108355864447215063,2181096921557783605
%N A recurrent sequence in Panaitopol's formula for pi(x), where pi(x) is the number of primes <= x.
%C Sum_{k=0..n} k!*a(n-k) = n*n!.
%C 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.
%H G. C. Greubel, <a href="/A233824/b233824.txt">Table of n, a(n) for n = 0..445</a>
%H Safia Aoudjit and Djamel Berkane, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL24/Berkane/berkane8.html">Explicit Estimates Involving the Primorial Integers and Applications</a>, J. Int. Seq., Vol. 24 (2021), Article 21.7.8.
%H M. Hassani, <a href="http://www.nlaa8.ir/EN-Final.pdf#page=147">Some remarks on a determinant related to the prime counting function</a>, The 8th Seminar on Linear Algebra and its Applications, 13-14th May 2015, University of Kurdistan, Iran.
%H A. Ivić, <a href="http://zbmath.org/?q=an:0982.11003">Review of "A formula for pi(x) applied to a result of Koninck-Ivić" by L. Panaitopol</a>, Zbl 0982.11003.
%H A. Kourbatov, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Kourbatov/kourb7.html">Upper Bounds for Prime Gaps Related to Firoozbakht's Conjecture</a>, J. Integer Sequences, 18 (2015), Article 15.11.2. arXiv:1506.03042
%H A. Kourbatov, <a href="http://arxiv.org/abs/1603.00855">On the geometric mean of the first n primes</a>, arXiv:1603.00855 [math.NT], 2016.
%H L. Panaitopol, <a href="http://www.nieuwarchief.nl/serie5/toonnummer.php?deel=01&nummer=1&taal=0">A formula for pi(x) applied to a result of Koninck-Ivić</a>, Nieuw Arch. Wiskd., (5) 1, No. 1 (2000), 55-56.
%F a(n) = n*n! - Sum_{k=1..n-1} k!*a(n-k).
%F 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.)
%e 0!*a(0) = a(0) = 0*0!, so a(0) = 0.
%e 0!*a(1) + 1!*a(0) = a(1) + a(0) = 1*1!, so a(1) = 1.
%e 0!*a(2) + 1!*a(1) + 2!*a(0) = a(2) + a(1) + 2*a(0) = 2*2!, so a(2) = 4 - 1 = 3.
%t 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 *)
%Y Cf. A000720, A003319, A062049.
%K nonn
%O 0,3
%A _Jonathan Sondow_, Dec 17 2013
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