

A263883


Ivanov's number a(n) = i*(n+2i) where i = min{j > 0  j*(j+1) >= 2*(n1)}.


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
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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

Table of n, a(n) for n=0..70.
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), 881888.


EXAMPLE

If n = 6, then i = min{j > 0  j*(j+1) >= 2*(61) = 10} = 3, so a(6) = 3*(6+23) = 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*n8)+1)\2); (n+2i)*i) \\ Charles R Greathouse IV, Nov 12 2016


CROSSREFS

Cf. A000124, A055503, A177862.
Sequence in context: A140496 A086552 A098393 * A103567 A277913 A131723
Adjacent sequences: A263880 A263881 A263882 * A263884 A263885 A263886


KEYWORD

nonn,easy


AUTHOR

Jonathan Sondow, Nov 30 2015


STATUS

approved



