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A054861 Greatest k such that 3^k divides n!. 51
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,7
COMMENTS
Also the number of trailing zeros in the base-3 representation of n!. - Hieronymus Fischer, Jun 18 2007
Also the highest power of 6 dividing n!. - Hieronymus Fischer, Aug 14 2007
A column of A090622. - Alois P. Heinz, Oct 05 2012
The 'missing' values are listed in A096346. - Stanislav Sykora, Jul 16 2014
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
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)
a(n) = A007949(A000142(n)). - David A. Corneth, Nov 02 2023
EXAMPLE
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.
MAPLE
A054861 := proc(n)
(n - convert(convert(n, base, 3), `+`))/2 ;
end proc:
seq(A054861(n), n=0..1000) ; # Robert Israel, Jul 17 2014
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 *)
PROG
(PARI) a(n)=my(s); while(n\=3, s+=n); s \\ Charles R Greathouse IV, Jul 25 2011
(PARI) a(n)=(n - vecsum(digits(n, 3)))\2; \\ Gheorghe Coserea, Jan 01 2018
(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
(Magma) [Valuation(Factorial(n), 3): n in [0..80]]; // Bruno Berselli, Aug 05 2013
CROSSREFS
Cf. A011371 (for analog involving powers of 2). See also A027868.
Cf. A004128 (for a(3n)).
Sequence in context: A113402 A238726 A240321 * A187324 A334207 A323094
KEYWORD
nonn,easy
AUTHOR
Henry Bottomley, May 22 2000
EXTENSIONS
Examples added by Hieronymus Fischer, Jun 06 2012
New name by David A. Corneth, Nov 02 2023
STATUS
approved

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Last modified March 19 06:47 EDT 2024. Contains 370953 sequences. (Running on oeis4.)