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 A067397 Maximal power of 3 that divides n-th Catalan number. 2
 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 3, 3, 3, 2, 2, 2, 2, 2, 2, 3, 3, 3, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 1, 3, 3, 3, 2, 2, 2, 2, 2, 2, 3, 3, 3, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,15 LINKS Robert Israel, Table of n, a(n) for n = 0..10000 Henry Bottomley, Illustration for A067397. FORMULA Let k=floor(log3(n)), i.e., 3^k<=n<3^(k+1): if (3/2)*3^k=1} (x^((3^k+1)/2) - x^(3^k-1))/((1-x^(3^k))*(1-x)). - Robert Israel, Sep 20 2015 a(n) = A000989(n) - A007949(n+1). - Amiram Eldar, Feb 21 2021 EXAMPLE a(13)=0 since Catalan(13)=742900, which is not divisible by 3; a(14)=2 since Catalan(14)=2674440, which is divisible by 9 but not by 27. MAPLE ListTools:-PartialSums([seq(padic:-ordp((2*n-1)/(n+1), 3), n=0..100)]); # Robert Israel, Sep 20 2015 MATHEMATICA f[n_] := Block[{p = FactorInteger@ n}, Take[Last /@ p, Flatten@ Position[First /@ p, 3]]]; Table[f[(2 n)!/n!/(n + 1)!], {n, 104}] /. {} -> 0 // Flatten (* Michael De Vlieger, Sep 21 2015 *) IntegerExponent[#, 3]&/@CatalanNumber[Range[0, 110]] (* Harvey P. Dale, Oct 09 2015 *) CROSSREFS Cf. A000989, A007949, A000108, A048881. Sequence in context: A157343 A102679 A025146 * A080586 A152906 A128522 Adjacent sequences:  A067394 A067395 A067396 * A067398 A067399 A067400 KEYWORD nonn AUTHOR Henry Bottomley, Jan 22 2002 STATUS approved

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Last modified September 21 09:52 EDT 2021. Contains 347597 sequences. (Running on oeis4.)