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 A263883 Ivanov's number a(n) = i*(n+2-i) where i = min{j > 0 | j*(j+1) >= 2*(n-1)}. 3
 1, 2, 3, 6, 8, 12, 15, 18, 24, 28, 32, 36, 45, 50, 55, 60, 65, 78, 84, 90, 96, 102, 108, 126, 133, 140, 147, 154, 161, 168, 192, 200, 208, 216, 224, 232, 240, 248, 279, 288, 297, 306, 315, 324, 333, 342, 351, 390, 400, 410, 420, 430, 440, 450, 460, 470, 480, 528, 539, 550, 561, 572, 583, 594, 605, 616, 627, 638, 696, 708, 720 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS The maximum number of regions into which n lines can divide the plane is A000124(n) = n(n+1)/2 + 1. Let m(n) be the least number such that every integer in the interval [m(n),n(n+1)/2 + 1] occurs as the number of regions into which n lines can divide the plane. Ivanov (2010, Theorem, p. 888) proved the upper bound m(n) <= a(n). Ivanov's upper bound is sharp, i.e., m(n) = a(n), at least for n <= 6. For example, the numbers of regions into which some configuration of 6 lines divides the plane are 7, 12, 15, 16, 17, 18, 19, 20, 21, 22, 22 (see A177862), so m(6) = 15 = a(6). Subsequence of A177862. LINKS O. A. Ivanov, Making Mathematics Come to Life: A Guide for Teachers and Students, American Mathematical Society, Providence, RI, 2009; see p. 11. O. A. Ivanov, On the number of regions into which n straight lines divide the plane, Amer. Math. Monthly, 117 (2010), 881-888. EXAMPLE If n = 6, then i = min{j > 0 | j*(j+1) >= 2*(6-1) = 10} = 3, so a(6) = 3*(6+2-3) = 15. MATHEMATICA i[n_] := (j = 1; While[j (j + 1) < 2 (n - 1), j++]; j); Table[i[n] (n + 2 - i[n]), {n, 0, 70}] PROG (PARI) a(n)=if(n<3, n+1, my(i=(sqrtint(8*n-8)+1)\2); (n+2-i)*i) \\ Charles R Greathouse IV, Nov 12 2016 CROSSREFS Cf. A000124, A055503, A177862. Sequence in context: A308843 A086552 A098393 * A103567 A277913 A131723 Adjacent sequences:  A263880 A263881 A263882 * A263884 A263885 A263886 KEYWORD nonn,easy AUTHOR Jonathan Sondow, Nov 30 2015 STATUS approved

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Last modified July 5 00:40 EDT 2020. Contains 335457 sequences. (Running on oeis4.)