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A077591
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Maximum number of regions into which the plane can be divided using n (concave) quadrilaterals.
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14
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1, 2, 18, 50, 98, 162, 242, 338, 450, 578, 722, 882, 1058, 1250, 1458, 1682, 1922, 2178, 2450, 2738, 3042, 3362, 3698, 4050, 4418, 4802, 5202, 5618, 6050, 6498, 6962, 7442, 7938, 8450, 8978, 9522, 10082, 10658, 11250, 11858, 12482, 13122, 13778
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OFFSET
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0,2
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COMMENTS
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Sequence found by reading the segment (1, 2) together with the line from 2, in the direction 2, 18, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Sep 05 2011
For a(n) > 1, a(n) are the numbers such that phi(sum of the odd divisors of a(n)) = phi(sum of even divisors of a(n)). - Michel Lagneau, Sep 14 2011
Let f(n) = 4*n^2 - 5, then (x, y, z) = (a(n+1), -f(n), -f(n + 1)) are solutions of the Diophantine equation x^3 + 4*y^3 + 4*z^3 = 512. - XU Pingya, Apr 25 2022
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LINKS
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FORMULA
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a(n) = 8*n^2 - 8*n + 2 = 2*(2*n-1)^2, n > 0, a(0)=1. [It would be nice to have a proof, or even a reference to a proof. - N. J. A. Sloane, Nov 30 2017]
Represent the configuration of n concave quadrilaterals by a planar graph with a node for each vertex of the quadrilaterals and for each intersection point. Let there be v_n nodes and e_n edges. By Euler's formula for planar graphs, a(n) = e_n - v_n + 2. When we go from n to n+1 quadrilaterals, each side of the new quadrilateral can meet each side of the existing quadrilaterals at most 4 times, so Dv_n := v_{n+1} - v_n <= 4*4n = 16n.
Each of these intersection points increases the number of edges in the graph by 2, so De_n := e_{n+1} - e_n = 4 + 2*Dv_n, Da_n := a(n+1) - a(n) = 4 + Dv_n <= 4+16*n.
These upper bounds can be achieved by taking n interwoven concave quadrilaterals (for n=1,2,3 see the attached Keyang Li links), and we achieve a(n) = 8n^2 - 8n + 2 (and v_n = 8n^2 - 4n, e_n = 4n*(4n-3)) for n > 0. QED (End)
G.f.: (1 - x + 15*x^2 + x^3)/(1-x)^3. - Colin Barker, Feb 23 2012
Sum_{n>=1} 1/a(n) = Pi^2/16.
Sum_{n>=1} (-1)^(n+1)/a(n) = G/2, where G is Catalan constant (A006752).
Product_{n>=1} (1 + 1/a(n)) = cosh(Pi/sqrt(8)).
Product_{n>=1} (1 - 1/a(n)) = cos(Pi/sqrt(8)). (End)
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EXAMPLE
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a(2) = 18 if you draw two concave quadrilaterals such that all four sides of one cross all four sides of the other.
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MAPLE
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MATHEMATICA
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Table[2*(4*n^2 - 4*n + 1), {n, 0, 50}] (* G. C. Greubel, Jul 15 2017 *)
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PROG
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(PARI) isok(n) = (sod = sumdiv(n, d, (d%2)*d)) && (sed = sumdiv(n, d, (1 - d%2)*d)) && (eulerphi(sod) == eulerphi(sed)); \\ from Michel Lagneau comment; Michel Marcus, Mar 15 2014
(GAP) Concatenation([1], List([1..2000], n->8*n^2 - 8*n + 2)); # Muniru A Asiru, Jan 29 2018
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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Joshua Zucker and the Castilleja School MathCounts club, Nov 07 2002
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STATUS
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approved
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