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A059233
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Number of rows in which n appears in Pascal's triangle A007318.
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9
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1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
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OFFSET
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2,5
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COMMENTS
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Central binomial coefficients c = A000984(n) > 1 appear once in the middle column C(2n, n), and thereafter in one or more later rows to the left as C(r,k) and to the right as C(r, r-k), k < r/2; the last time in row r = c = C(c,1) = C(c,c-1). For these, a(n) = (A003016(n)+1)/2. For all other numbers n > 1, a(n) = A003016(n)/2. - M. F. Hasler, Mar 01 2023
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REFERENCES
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L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 93, #47.
C. S. Ogilvy, Tomorrow's Math. 2nd ed., Oxford Univ. Press, 1972, p. 96.
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LINKS
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FORMULA
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EXAMPLE
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6 appears in both row 4 and row 6 in Pascal's triangle, therefore a(6) = 2.
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MATHEMATICA
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nmax = 101; A007318 = Table[Binomial[n, k], {n, 0, nmax}, {k, 0, n}]; a[n_] := Position[A007318, n][[All, 1]] // Union // Length; Table[a[n], {n, 2, nmax}] (* Jean-François Alcover, Sep 09 2013 *)
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PROG
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(Haskell)
a059233 n = length $ filter (n `elem`) $
take (fromInteger n) $ tail a007318_tabl
a059233_list = map a059233 [2..]
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CROSSREFS
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KEYWORD
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easy,nice,nonn
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AUTHOR
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STATUS
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approved
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