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A067397
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Maximal power of 3 that divides n-th Catalan number.
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3
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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
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OFFSET
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0,15
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COMMENTS
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Let v(n) = A007949(n) be the 3-adic valuation of n. For n == 0 or 1 (mod 3), we have a(n) = v(binomial(2*n,n)/(n+1)) = v(binomial(2*n,n)) = A000989(n), so a(n) = 0 if and only if n is in A005836. For n == 0 or 2 (mod 3), we have a(n) = v(binomial(2*n+2,n+1)/(4*n+2)) = v(binomial(2*n+2,n+1)) = A000989(n+1), so a(n) = 0 if and only if n+1 is in A005836. In other words, the indices of 0 are precisely numbers of the form 3*k-1 (k>0), 3*k or 3*k+1 for k in A005836. - Jianing Song, Feb 29 2024
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LINKS
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FORMULA
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Let k=floor(log3(n)), i.e., 3^k<=n<3^(k+1): if (3/2)*3^k<n<(5/2)*3^k then a(n)=a(n-3^k)+1, if n=3*3^k-1 then a(n)=a(n-3^k)-1=0, otherwise a(n)=a(n-3^k) [starting with a(0)=0, so a(3^k)=0].
G.f.: Sum_{k>=1} (x^((3^k+1)/2) - x^(3^k-1))/((1-x^(3^k))*(1-x)). - Robert Israel, Sep 20 2015
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EXAMPLE
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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.
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MAPLE
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ListTools:-PartialSums([seq(padic:-ordp((2*n-1)/(n+1), 3), n=0..100)]); # Robert Israel, Sep 20 2015
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MATHEMATICA
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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 *)
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PROG
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(PARI) a(n) = (sumdigits(n, 3) + sumdigits(n+1, 3) - sumdigits(2*n, 3) - 1)/2 \\ Jianing Song, Feb 24 2024
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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