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A233824 A recurrent sequence in Panaitopol's formula for pi(x), where pi(x) is the number of primes <= x. 6
0, 1, 3, 13, 71, 461, 3447, 29093, 273343, 2829325, 31998903, 392743957, 5201061455, 73943424413, 1123596277863, 18176728317413, 311951144828863, 5661698774848621, 108355864447215063, 2181096921557783605 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,3
COMMENTS
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.
LINKS
Safia Aoudjit and Djamel Berkane, Explicit Estimates Involving the Primorial Integers and Applications, J. Int. Seq., Vol. 24 (2021), Article 21.7.8.
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. 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.
FORMULA
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.)
EXAMPLE
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.
MATHEMATICA
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 *)
CROSSREFS
Sequence in context: A272428 A167894 A158882 * A003319 A192239 A192936
KEYWORD
nonn
AUTHOR
Jonathan Sondow, Dec 17 2013
STATUS
approved

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Last modified April 20 15:59 EDT 2024. Contains 371844 sequences. (Running on oeis4.)