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Search: id:A051890
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| 2, 2, 6, 14, 26, 42, 62, 86, 114, 146, 182, 222, 266, 314, 366, 422, 482, 546, 614, 686, 762, 842, 926, 1014, 1106, 1202, 1302, 1406, 1514, 1626, 1742, 1862, 1986, 2114, 2246, 2382, 2522, 2666, 2814, 2966, 3122, 3282, 3446, 3614, 3786, 3962
(list; graph; listen)
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OFFSET
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0,1
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COMMENT
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Draw n ellipses in the plane (n>0), any 2 meeting in 4 points; sequence gives number of regions into which the plane is divided.
Least k such that Z(k,2) <= Z(n,3) where Z(m,s) = sum(i>=m, 1/i^s) = zeta(s)-sum(i=1,m-1,1/i^s). - Benoit Cloitre (benoit7848c(AT)orange.fr), Nov 29 2002
For n>2, third diagonal of [A154685] [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Jan 25 2009]
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REFERENCES
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Parabola, vol. 20, no. 2, 1984, p. 27, Problem #Q607.
J. V. Post, "When Centered Polygonal Numbers are Perfect Squares" preprint.
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LINKS
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Parabola, Web site
Eric Weisstein's World of Mathematics, Plane Division by Ellipses
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FORMULA
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a(n)=4*binomial(n, 2)+2. - Francois Jooste (phukraut(AT)hotmail.com), Mar 05 2003
For n>2 nearest integer to sum(k>=n, 1/k^3)/sum(k>=n, 1/k^5) - Benoit Cloitre (benoit7848c(AT)orange.fr), Jun 12 2003
a(n) = 2*A002061(n). - Jonathan Vos Post (jvospost3(AT)gmail.com), Jun 19 2005
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MAPLE
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A051890 := n->2*(n^2-n+1);
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MATHEMATICA
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a=2; lst={}; Do[a+=n; AppendTo[lst, a], {n, 0, 6!, 4}]; lst...and/or... lst={}; Do[AppendTo[lst, 2*(n^2-n+1)], {n, 5!}]; lst [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Mar 01 2009]
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CROSSREFS
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Cf. A001844, A002061, A014206, A002061.
Cf. A154685 [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Jan 25 2009]
Adjacent sequences: A051887 A051888 A051889 this_sequence A051891 A051892 A051893
Sequence in context: A049952 A019100 A019101 this_sequence A071109 A005310 A002203
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KEYWORD
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nonn
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AUTHOR
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Antreas P. Hatzipolakis (xpolakis(AT)otenet.gr), Apr 30 2000
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