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A259667
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Catalan numbers mod 6.
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5
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1, 1, 2, 5, 2, 0, 0, 3, 2, 2, 2, 4, 4, 4, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 2, 4, 4, 1, 0, 0, 0, 4, 4, 4, 2, 2, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 2, 4, 4, 4, 0, 0, 0, 4, 4, 4, 2, 2, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 4, 4, 2, 2, 2, 0, 0, 0, 2, 2, 2, 4
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OFFSET
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0,3
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COMMENTS
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The only odd terms are those with indices n = 2^k - 1 (k = 0, 1, 2, 3, ...); see also A038003.
It is conjectured that the only k which yield a(2^k-1) = 1 are k = 0, 1 and 5. Are there other k than 2 and 8 that yield a(2^k-1) = 5? Otherwise said, is a(2^k-1) = 3 for all k > 8?
The question is equivalent to: does 2^k - 1 always contain a digit 2 when converted into base 3 for all k > 8? Similar conjecture has been proposed for 2^k, see A004642. - Jianing Song, Sep 04 2018
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LINKS
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FORMULA
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MATHEMATICA
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Mod[CatalanNumber[Range[0, 120]], 6] (* Harvey P. Dale, Oct 24 2020 *)
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PROG
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(PARI) a(n)=binomial(2*n, n)/(n+1)%6
(PARI) A259667(n)=lift(if(n%3!=1, binomod(2*n+1, n, 6)/(2*n+1), if(bittest(n, 0), binomod(2*n, n-1, 6)/n, binomod(2*n, n, 6)/(n+1)))) \\ using binomod.gp by M. Alekseyev, cf. Links.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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