A054861 Greatest k such that 3^k divides n!.
0, 0, 0, 1, 1, 1, 2, 2, 2, 4, 4, 4, 5, 5, 5, 6, 6, 6, 8, 8, 8, 9, 9, 9, 10, 10, 10, 13, 13, 13, 14, 14, 14, 15, 15, 15, 17, 17, 17, 18, 18, 18, 19, 19, 19, 21, 21, 21, 22, 22, 22, 23, 23, 23, 26, 26, 26, 27, 27, 27, 28, 28, 28, 30, 30, 30, 31, 31, 31, 32, 32, 32, 34, 34, 34, 35, 35
Offset: 0
Examples
a(100) = 48. a(10^3) = 498. a(10^4) = 4996. a(10^5) = 49995. a(10^6) = 499993. a(10^7) = 4999994. a(10^8) = 49999990. a(10^9) = 499999993.
Links
- Hieronymus Fischer, Table of n, a(n) for n = 0..10000 (first 1000 terms from T. D. Noe)
- S-C Liu and J. C.-C. Yeh, Catalan numbers modulo 2^k, J. Int. Seq. 13 (2010), 10.5.4, Eq. (5).
- A. M. Oller-Marcen and J. Maria Grau, On the Base-b Expansion of the Number of Trailing Zeros of b^k!, J. Int. Seq. 14 (2011) 11.6.8
Crossrefs
Programs
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Magma
[Valuation(Factorial(n), 3): n in [0..80]]; // Bruno Berselli, Aug 05 2013
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Maple
A054861 := proc(n) (n - convert(convert(n, base, 3), `+`))/2 ; end proc: seq(A054861(n),n=0..1000) ; # Robert Israel, Jul 17 2014
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Mathematica
(Plus@@Floor[#/3^Range[Length[IntegerDigits[#,3]]-1]]&)/@Range[0,100] (* Peter J. C. Moses, Apr 07 2012 *) FoldList[Plus, 0, IntegerExponent[Range[100], 3]] (* T. D. Noe, Apr 10 2012 *) Table[IntegerExponent[n!,3],{n,0,80}] (* Harvey P. Dale, Feb 05 2015 *)
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PARI
a(n)=my(s);while(n\=3,s+=n);s \\ Charles R Greathouse IV, Jul 25 2011
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PARI
a(n)=(n - vecsum(digits(n,3)))\2; \\ Gheorghe Coserea, Jan 01 2018
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Sage
def A054861(n): A004128 = lambda n: A004128(n//3) + n if n > 0 else 0 return A004128(n//3) [A054861(i) for i in (0..76)] # Peter Luschny, Nov 16 2012
Formula
a(n) = floor(n/3) + floor(n/9) + floor(n/27) + floor(n/81) + ... .
a(n) = (n - A053735(n))/2.
a(n+1) = Sum_{k=1..n} A007949(k). - Benoit Cloitre, Mar 24 2002
From Hieronymus Fischer, Jun 18, Jun 25 and Aug 14 2007: (Start)
G.f.: (1/(1-x))*Sum_{k>0} x^(3^k)/(1-x^(3^k)).
a(n) = Sum_{k=3..n} Sum_{j>=3, j|k} (floor(log_3(j)) - floor(log_3(j-1))).
G.f.: L[b(k)](x)/(1-x), where L[b(k)](x) = Sum_{k>=0} b(k)*x^k/(1-x^k) is a Lambert series with b(k) = 1, if k>1 is a power of 3, otherwise b(k)=0.
G.f.: (1/(1-x))*Sum_{k>0} c(k)*x^k, where c(k) = Sum_{j>1, j|k} (floor(log_3(j)) - floor(log_3(j-1))).
Recurrence:
a(n) = floor(n/3) + a(floor(n/3));
a(3*n) = n + a(n);
a(n*3^m) = n*(3^m-1)/2 + a(n).
a(k*3^m) = k*(3^m-1)/2, for 0 <= k < 3, m >= 0.
Asymptotic behavior:
a(n) = n/2 + O(log(n)),
a(n+1) - a(n) = O(log(n)); this follows from the inequalities below.
a(n) <= (n-1)/2; equality holds for powers of 3.
a(n) >= (n-2)/2 - floor(log_3(n)); equality holds for n = 3^m - 1, m > 0.
lim inf (n/2 - a(n)) = 1/2 for n->oo.
lim sup (n/2 - log_3(n) - a(n)) = 0 for n->oo.
lim sup (a(n+1) - a(n) - log_3(n)) = 0 for n->oo. (End)
a(n) = A007949(n!). - R. J. Mathar, Sep 03 2016
From R. J. Mathar, Jul 08 2021: (Start)
a(n) = A122841(n!).
Partial sums of A007949. (End)
Extensions
Examples added by Hieronymus Fischer, Jun 06 2012
New name by David A. Corneth, Nov 02 2023
Comments