A059233 Number of rows in which n appears in Pascal's triangle A007318.
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
Offset: 2
Examples
6 appears in both row 4 and row 6 in Pascal's triangle, therefore a(6) = 2.
References
- L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 93, #47.
- C. S. Ogilvy, Tomorrow's Math. 2nd ed., Oxford Univ. Press, 1972, p. 96.
Links
- T. D. Noe, Table of n, a(n) for n=2..10000
- D. Singmaster, How often does an integer occur as a binomial coefficient?, Amer. Math. Monthly, 78 (1971), 385-386.
- Eric Weisstein's World of Mathematics, Pascal's Triangle
- Wikipedia, Singmaster's conjecture
- Index entries for triangles and arrays related to Pascal's triangle
Programs
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Haskell
a059233 n = length $ filter (n `elem`) $ take (fromInteger n) $ tail a007318_tabl a059233_list = map a059233 [2..] -- Reinhard Zumkeller, Dec 24 2012
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Mathematica
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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PARI
A059233(n)=A003016(n)\/2 \\ M. F. Hasler, Mar 01 2023
Formula
a(A180058(n)) = n and a(m) < n for m < A180058(n); a(A182237(n)) = 2; a(A098565(n)) = 3. - Reinhard Zumkeller, Dec 24 2012
a(n) = ceiling(A003016(n)/2). - M. F. Hasler, Mar 01 2023
Comments