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A259361
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n occurs 2n+2 times.
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6
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0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8
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OFFSET
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0,7
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COMMENTS
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Define the oblong root obrt(x) to be the (larger) solution of y * (y+1) = x; i.e., obrt(x) = sqrt(x+1/4) - 1/2. So obrt(x) is an integer iff x is an oblong number (A002378). Then a(n) = floor(obrt(n)).
a(n) gives (from the preceding comment) also the maximal number of parts of partitions of n with no part 1 and difference of parts at least two. See A003106, with the combinatorial interpretation of the sum of the Rogers-Ramanujan identity. - Wolfdieter Lang, Oct 29 2016
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LINKS
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FORMULA
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a(n) = floor((-1 + sqrt(1+4*n))/2). See the first comment above. - Wolfdieter Lang, Oct 29 2016
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MATHEMATICA
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Flatten[Table[PadLeft[{}, 2n + 2, n], {n, 0, 8}]] (* Alonso del Arte, Jun 30 2015 *)
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PROG
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(Haskell)
a259361 = floor . subtract (1 / 2) . sqrt . (+ 1 / 4) . fromIntegral
a259361_list = concat xss where
xss = iterate (\(xs@(x:_)) -> map (+ 1) (x : x : xs)) [0, 0]
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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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