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A059100
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a(n) = n^2 + 2.
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74
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2, 3, 6, 11, 18, 27, 38, 51, 66, 83, 102, 123, 146, 171, 198, 227, 258, 291, 326, 363, 402, 443, 486, 531, 578, 627, 678, 731, 786, 843, 902, 963, 1026, 1091, 1158, 1227, 1298, 1371, 1446, 1523, 1602, 1683, 1766, 1851, 1938, 2027, 2118, 2211, 2306, 2403, 2502, 2603
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OFFSET
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0,1
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COMMENTS
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Let s(n) = Sum_{k>=1} 1/n^(2^k). Then I conjecture that the maximum element in the continued fraction for s(n) is n^2 + 2. -Benoit Cloitre,Aug 15 2002
1/a(n) = R(n)/r with R(n) the n-th radius of the Pappus chain of the symmetric arbelos with semicircle radii r, r1 = r/2 = r2. See the MathWorld link for Pappus chain (there are two of them, a left and a right one. In this case these two chains are congruent). -Wolfdieter Lang,Mar 01 2013
a(n) is the number of election results for an election with n+2 candidates, say C1, C2,..., and C(n+2), and with only two voters (each casting a single vote) that have C1 and C2 receiving the same number of votes. See link below. -Dennis P. Walsh,May 08 2013
This sequence gives the set of values such that for sequences b(k+1) = a(n)*b(k) - b(k-1), with initial values b(0) = 2, b(1) = a(n), all such sequences are invariant under this transformation: b(k) = (b(j+k) + b(j-k))/b(j), except where b(j) = 0, for all integer values of j and k, including negative values. Examples are: at n=0, b(k) = 2 for all k; at n=1, b(k) =A005248;at n=2, b(k) = 2*A001541;at n=3, b(k)=A057076;at n=4, b(k) = 2*A023039.This b(k) family are also the transformation results for all related b'(k) (i.e., those with different initial values) including non-integer values. Further, these b(k) are also the bisections of the transformations of sequences of the form G(k+1) = n * G(k) + G(k-1), and those bisections are invariant for all initial values of g(0) and g(1), including non-integer values. For n = 1 this g(k) family includes Fibonacci and Lucas, where the invariant bisection is b(k) =A005248.The applicable bisection for this transformation of g(k) is for the odd values of k, and applies for all n. Also seeA000032for a related family of sequences. -Richard R. Forberg,Nov 22 2014
Also the number of maximum matchings in the n-gear graph. -Eric W. Weisstein,Dec 31 2017
Numbers of the form n^2+2 have no factors that are congruent to 7 (mod 8). -Gordon E. Michaels,Sep 12 2019
For n >= 1, the continued fraction expansion of sqrt(a(n)) is [n; {n, 2n}]. -Magus K. Chu,Sep 10 2022
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LINKS
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Eric Weisstein's World of Mathematics,Matching.
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FORMULA
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G.f.: (2 - 3*x + 3*x^2)/(1 - x)^3. -R. J. Mathar,Oct 05 2008
a(3*n) mod 9 = 2.
Sum_{n >= 1} 1/a(n) = Pi * coth(sqrt(2)*Pi) / 2^(3/2) - 1/4. -Vaclav Kotesovec,May 01 2018
Sum_{n>=0} (-1)^n/a(n) = (1 + sqrt(2)*Pi*(csch(sqrt(2)*Pi)))/4.
Product_{n>=0} (1 + 1/a(n)) = sqrt(3/2)*csch(sqrt(2)*Pi)*sinh(sqrt(3)*Pi).
Product_{n>=0} (1 - 1/a(n)) = csch(sqrt(2)*Pi)*sinh(Pi)/sqrt(2). (End)
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EXAMPLE
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For n = 2, a(2) = 6 since there are 6 election results in a 4-candidate, 2-voter election that have candidates c1 and c2 tied. Letting <i,j> denote voter 1 voting for candidate i and voter 2 voting for candidate j, the six election results are <1,2>, <2,1>, <3,3>, <3,4>, <4,3>, and <4,4>. -Dennis P. Walsh,May 08 2013
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MAPLE
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with(combinat, fibonacci):seq(fibonacci(3, i)+1, i=0..49); #Zerinvary Lajos,Mar 20 2008
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MATHEMATICA
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CoefficientList[Series[(-2 + 3 x - 3 x^2)/(-1 + x)^3, {x, 0, 20}], x] (*Eric W. Weisstein,Dec 31 2017 *)
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PROG
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(Sage) [lucas_number1(3, n, -2) for n in range(0, 50)] #Zerinvary Lajos,May 16 2009
(PARI) { for (n = 0, 1000, write( "b059100.txt", n, "", n^2+2); ) } \\Harry J. Smith,Jun 24 2009
(Haskell)
a059100 = (+ 2). (^ 2)
a059100_list = scanl (+) (2) [1, 3..]
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CROSSREFS
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Cf.A000032,A000196,A000290,A001541,A002522,A005248,A008865,A056105,A056109,A056899,A057076,A069987,A081908,A114964,A156798,A166464.
Apart from initial terms, same asA010000.
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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