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 A226314 Triangle read by rows: T(i,j) = j+(i-j)/gcd(i,j) (1<=i<=j). 9
 1, 1, 2, 1, 2, 3, 1, 3, 3, 4, 1, 2, 3, 4, 5, 1, 4, 5, 5, 5, 6, 1, 2, 3, 4, 5, 6, 7, 1, 5, 3, 7, 5, 7, 7, 8, 1, 2, 7, 4, 5, 8, 7, 8, 9, 1, 6, 3, 7, 9, 8, 7, 9, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 1, 7, 9, 10, 5, 11, 7, 11, 11, 11, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 1, 8, 3, 9, 5, 10, 13, 11, 9, 12, 11, 13, 13, 14 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS The triangle of fractions A226314(i,j)/A054531(i,j) is an efficient way to enumerate the rationals [Fortnow]. Sum(A226314(n,k)/A054531(n,k): 1<=k<=n) = A226555(n)/A040001(n). - Reinhard Zumkeller, Jun 10 2013 LINKS Reinhard Zumkeller, Rows n = 1..120 of triangle, flattened Lance Fortnow, Counting the Rationals Quickly, Computational Complexity Weblog, Monday, March 01, 2004. Yoram Sagher, Counting the rationals, Amer. Math. Monthly, 96 (1989), p. 823. Math. Rev. 90i:04001. EXAMPLE Triangle begins: [1] [1, 2] [1, 2, 3] [1, 3, 3, 4] [1, 2, 3, 4, 5] [1, 4, 5, 5, 5, 6] [1, 2, 3, 4, 5, 6, 7] [1, 5, 3, 7, 5, 7, 7, 8] [1, 2, 7, 4, 5, 8, 7, 8, 9] [1, 6, 3, 7, 9, 8, 7, 9, 9, 10] ... The resulting triangle of fractions begins: 1, 1/2, 2, 1/3, 2/3, 3, 1/4, 3/2, 3/4, 4, 1/5, 2/5, 3/5, 4/5, 5, ... MAPLE f:=(i, j) -> j+(i-j)/gcd(i, j); g:=n->[seq(f(i, n), i=1..n)]; for n from 1 to 20 do lprint(g(n)); od: PROG (Haskell) a226314 n k = n - (n - k) `div` gcd n k a226314_row n = a226314_tabl !! (n-1) a226314_tabl = map f \$ tail a002262_tabl where    f us'@(_:us) = map (v -) \$ zipWith div vs (map (gcd v) us)      where (v:vs) = reverse us' -- Reinhard Zumkeller, Jun 10 2013 CROSSREFS Cf. A002262. Cf. A037161, A037162, A066657, A066658. Sequence in context: A286343 A075106 A196935 * A036995 A225597 A116908 Adjacent sequences:  A226311 A226312 A226313 * A226315 A226316 A226317 KEYWORD nonn,frac,tabl AUTHOR N. J. A. Sloane, Jun 09 2013 STATUS approved

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Last modified May 25 08:37 EDT 2020. Contains 334587 sequences. (Running on oeis4.)