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A007949
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Greatest k such that 3^k divides n. Or, 3-adic valuation of n.
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37
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0, 0, 1, 0, 0, 1, 0, 0, 2, 0, 0, 1, 0, 0, 1, 0, 0, 2, 0, 0, 1, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 1, 0, 0, 2, 0, 0, 1, 0, 0, 1, 0, 0, 2, 0, 0, 1, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 1, 0, 0, 2, 0, 0, 1, 0, 0, 1, 0, 0, 2, 0, 0, 1, 0, 0, 1, 0, 0, 4, 0, 0, 1, 0, 0, 1, 0, 0, 2, 0, 0, 1, 0, 0, 1, 0, 0, 2, 0, 0, 1, 0, 0, 1
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,9
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COMMENTS
| a(n) mod 2 = 1 - A014578(n). [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Oct 04 2008]
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REFERENCES
| K. Atanassov, On the 61-st, 62-nd and the 63-rd Smarandache Problems, Notes on Number Theory and Discrete Mathematics, Sophia, Bulgaria, Vol. 4 (1998), No. 4, 175-182.
K. Atanassov, On Some of Smarandache's Problems, American Research Press, 1999, 16-21.
F. Q. Gouvea, p-Adic Numbers, Springer-Verlag, 1993; see p. 23.
F. Smarandache, Only Problems, not Solutions!, Xiquan Publ., Phoenix-Chicago, 1993.
M. Vassilev-Missana and K. Atanassov, Some Representations related to n!, Notes on Number Theory and Discrete Mathematics, Vol. 4 (1998), No. 4, 148-153.
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LINKS
| T. D. Noe, Table of n, a(n) for n=1..1000
K. Atanassov, On Some of Smarandache's Problems
M. L. Perez et al., eds., Smarandache Notions Journal
F. Smarandache, Only Problems, Not Solutions!.
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FORMULA
| a(n) = if n > 0 modulo 3 then 0 else 1 + a(n/3). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Aug 12 2001
a(n)=A051064(n)-1. G.f.: Sum(k>=1, x^3^k/(1-x^3^k)))). - Ralf Stephan (ralf(AT)ark.in-berlin.de), Apr 12 2002
Fixed point of the morphism : 0 -> 001; 1 -> 002; 2 -> 003; 3 -> 004; 4 -> 005; etc...; starting from a(1) = 0. - DELEHAM Philippe (kolotoko(AT)wanadoo.fr), Mar 29 2004
Totally additive with a(p) = 1 if p = 3, 0 otherwise.
$v_{m}(n)=\displaystyle\sum\limits_{r=1}^{\infty}\frac{r}{m^{r+1}} \sum_{j=1}^{(m-1)} \sum\limits_{k=0}^{(m^{r+1}-1)}e^{\frac{2k\pi i(n+(m-j)m^{r})}{m^{r+1}}}$ - This formula is for the general case, for this specific one set $m=3$ [From A.Neves (ammdneves(AT)gmail.com), Oct 04 2010]
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MAPLE
| A007949 := proc(n) option remember; if n mod 3 > 0 then 0 else A007949(n/3)+1; fi; end;
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MATHEMATICA
| p=3; Array[ If[ Mod[ #, p ]==0, Select[ FactorInteger[ # ], Function[ q, q[ [ 1 ] ]==p ], 1 ][ [ 1, 2 ] ], 0 ]&, 81 ]
Nest[ Function[ l, {Flatten[(l /. {0 -> {0, 0, 1}, 1 -> {0, 0, 2}, 2 -> {0, 0, 3}, 3 -> {0, 0, 4}}) ]}], {0}, 5] (from Robert G. Wilson v Mar 03 2005)
Contribution from Zak Seidov (zakseidov(AT)yahoo.com), Apr 15 2010: (Start)
(*1*)IntegerExponent[Range[200], 3]
(*2*)Table[If[Mod[n, 3]>0, 0, 1+b[n/3]], {n, 200}] (End)
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PROG
| (PARI) a(n)=valuation(n, 3)
(Haskell)
a007949 n = length $ takeWhile ((== 0) . (mod n)) $ iterate (* 3) 3
-- Reinhard Zumkeller, May 14 2011
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CROSSREFS
| Partial sums give A054861. Cf. A080278, A001511.
Cf. A122841, A007814, A112765.
Sequence in context: A203945 A015692 A016232 * A191265 A078595 A078128
Adjacent sequences: A007946 A007947 A007948 * A007950 A007951 A007952
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KEYWORD
| nonn,easy
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AUTHOR
| R. Muller
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