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A206424
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The number of 1's in row n of Pascal's Triangle (mod 3)
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9
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1, 2, 2, 2, 4, 4, 2, 4, 5, 2, 4, 4, 4, 8, 8, 4, 8, 10, 2, 4, 5, 4, 8, 10, 5, 10, 14, 2, 4, 4, 4, 8, 8, 4, 8, 10, 4, 8, 8, 8, 16, 16, 8, 16, 20, 4, 8, 10, 8, 16, 20, 10, 20, 28, 2, 4, 5, 4, 8, 10, 5, 10, 14, 4, 8, 10, 8, 16, 20, 10, 20, 28, 5, 10, 14, 10, 20, 28
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OFFSET
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0,2
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COMMENTS
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LINKS
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FORMULA
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a(n) = (3^k + 1)*2^(y-1), where y = A062756(n) and k = A081603(n). [See e.g. Wells or Wilson references.]
(End)
Based on the first formula above, we have following identities:
a(3n) = a(n).
a(3n+1) = 2*a(n).
a(9n+4) = 4*a(n).
(End)
a(n) = (1/2)*Sum_{k = 0..n} mod(C(n,k) + C(n,k)^2, 3). - Peter Bala, Dec 17 2020
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EXAMPLE
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Example: Rows 0-8 of Pascal's Triangle (mod 3) are:
1 So a(0) = 1
1 1 So a(1) = 2
1 2 1 So a(2) = 2
1 0 0 1 .
1 1 0 1 1 .
1 2 1 1 2 1 .
1 0 0 2 0 0 1
1 1 0 2 2 0 1 1
1 2 1 2 1 2 1 2 1
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MATHEMATICA
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Table[Count[Mod[Binomial[n, Range[0, n]], 3], 1], {n, 0, 99}] (* Alonso del Arte, Feb 07 2012 *)
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PROG
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(Haskell)
a206424 = length . filter (== 1) . a083093_row
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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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