

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
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OFFSET

0,3


COMMENTS

Sum_{k=0..n} k!*a(nk) = 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

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, 1314th May 2015, University of Kurdistan, Iran.
A. Ivić, Review of "A formula for pi(x) applied to a result of KoninckIvić" 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 KoninckIvić, Nieuw Arch. Wiskd., (5) 1, No. 1 (2000), 5556.


FORMULA

a(n) = n*n!  Sum_{k=1..n1} k!*a(nk).
a(n) = A003319(n+1) if n > 0. (Proof. Set b(n) = A003319(n), so that b(n) = n!  Sum_{k=1..n1} k!*b(nk). 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

Cf. A000720, A003319, A062049.
Sequence in context: A272428 A167894 A158882 * A003319 A192239 A192936
Adjacent sequences: A233821 A233822 A233823 * A233825 A233826 A233827


KEYWORD

nonn


AUTHOR

Jonathan Sondow, Dec 17 2013


STATUS

approved



